Download Maths – YR11 METHODS PROBABILITY

ChapTEr 11
Introductory probability
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10 Quick Questions
ChapTEr ConTEnTS
11a
11B
11C
11d
11E
11F
11G
11h
11i
Introduction to experimental probability
Calculating probabilities
Tree diagrams and lattice diagrams
The Addition Law of Probabilities
Karnaugh maps and probability tables
Conditional probability
Transition matrices and Markov chains
Independent events
Simulation
introduction to experimental
probability
11a
Tossing a fair coin or rolling a standard 6-sided die will result in a range of outcomes. The coin can land
Heads or Tails, and the number appearing uppermost on the die will be one of the numbers 1, 2, 3, 4,
5 or 6. Probability involves assigning a numerical value to the likelihood of such events occurring.
In this respect, certain events will clearly be more probable than others; for example, getting only 1 of
the required 6 numbers in Tattslotto is more likely than obtaining all 6 winning numbers.
A numerical value for the probability of an event can be established in a number of ways. It can
be based on results arising from experiments; alternatively, a reasoned estimate of the likelihood
of the event can be provided on the strength of personal experience and knowledge (the subjective
probability). A third way is to consider the ‘symmetry’ of the situation where the activity has
equiprobable or equally likely outcomes. For example, if we toss a coin 50 times and note how
many times it lands ‘Heads’ (a Head facing up), we may conclude (based on the experiment) that the
probability of a coin landing Heads up is half. We may also reason that a tossed coin has two equally
likely outcomes (a Head and a Tail), of which Heads is one possibility, so there is 1 chance in 2, or 1,
2
of a Head. However, deciding what the chances are of a runner winning her race will be subjective and
dependent on considerations such as the runner’s past performances, her current state of fitness and the
abilities of the other competitors.
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long run proportion
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one die
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Two dice
random outcome experiments
What is the probability of a fair coin landing Tails? For a single trial of this experiment (one toss of the
coin), we know the coin will land either Heads or Tails, but we cannot be sure the toss will produce a
favourable outcome (that is, a Tail). The result is a random outcome. (Closing one’s eyes and taking
out a marble from a box containing different coloured marbles, or shuffling a pack of playing cards and
choosing the topmost card, are also activities that produce random outcomes.)
For our example of the coin, if many trials are conducted we will observe that the ratio
number of Tails
, which is the experimental probability for the favourable outcomes, converges
total number of trials
(‘gets closer’) to a particular value. This particular value is known as the long-run proportion.
ChapTEr 11 • Introductory probability
475
number of Heads
Similarly, we observe that the ratio
, the experimental probability for a Head,
total
number
of
trials
converges to a particular value.
For a coin tossed many times the long-run proportion of a Head is 0.5 and the long-run proportion of a
Tail is 0.5.
Experimental probability and expected number
of outcomes
In general, the experimental probability is given by:
experimental probability =
number of favourable outcomes observed
total number of trials
The number of times an outcome of an activity is expected to occur is given by:
expected number of favourable outcomes
= experimental probability (long-run proportion) × number of trials
WorkEd ExamplE 1
A 6-sided die (not necessarily a fair one) was rolled 12 times and the number showing uppermost
was noted each time. The numbers uppermost on the die were:
2, 4, 1, 1, 5, 6, 4, 3, 4, 5, 6, 1.
Estimate the probability of rolling a 5 with this die.
Think
1
There are 2 favourable outcomes.
2
There are 12 outcomes altogether.
3
Use the formula: experimental probability
number of favourable outcomes observed
=
total number of trials
WriTE
Experimental probability =
=
2
12
1
6
WorkEd ExamplE 2
A fair 6-sided die is rolled 48 times. How many times is an even number expected to show
uppermost?
Think
1
2
WriTE
There are 6 equally likely outcomes for the
roll of the die and 3 favourable outcomes
corresponding to an even number.
Experimental probability of an even number
There are 48 trials. Use the formula
expected number of favourable outcomes =
experimental probability × number of trials.
Expected number of even numbers
= 36
= 12
= 12 × 48
= 24
WorkEd ExamplE 3
Inside a bag are 18 marbles, some white and the rest green. One marble is taken out without
looking, its colour is noted and the marble put back inside the bag. When this is done 30 times it
is found that a green marble was taken out 5 times. Estimate how many marbles of each colour
are in the bag.
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Maths Quest 11 Mathematical Methods CAS
Think
1
WriTE
A green marble was taken out 5 times and
a white marble 25 times. Work out the
experimental probabilities.
Experimental probability of green marble =
=
Experimental probability of white marble =
=
2
Calculate the expected number of each colour
marble. Use the formula
expected number of favourable outcomes =
experimental probability × number of trials.
5
30
1
6
25
30
5
6
Expected number of green marbles = 1 × 18
6
=3
Expected number of white marbles = 56 × 18
= 15
Estimated number of each type of marble:
3 green, 15 white
Exercise 11a
introduction to experimental probability
1 WE 1 A coin was tossed 10 times and the outcomes noted as H, T, H, H, T, T, T, T, H, T, where H is a
Head and T is a Tail. Find the experimental probability of a Tail.
2 Twenty letters were chosen at random from the alphabet and recorded as either consonant, c,
or vowel, v. The results were c, c, v, c, v, c, c, c, c, v, c, v, v, c, c, c, c, v, c, c. Calculate the
experimental probability of choosing a consonant.
3 A biased coin is tossed 50 times. The results were 33 Tails and 17 Heads.
a What is the experimental probability of tossing a Tail with this coin?
b What is the experimental probability of tossing a Head with this coin?
4 WE2 A die is tossed 96 times. How many times is an odd number expected to appear uppermost on
the die?
5 A coin is tossed 500 times. What is the expected number of Heads?
6 A die is rolled 300 times. How many odd numbers or the number 2 are expected to turn up?
7 mC A die is tossed 102 times. The number of times a number between 1 and 3 inclusive is expected to
appear uppermost on the die is:
a 51
B 34
C 20
d 64
E 68
8 mC A box contains 2 blue beads, 3 green beads and 1 yellow bead. One bead is taken out, its colour
is noted and it is put back in the box. This is repeated 246 times. The number of times a bead that is not
yellow is expected to be taken out of the box is:
a 41
B 82
C 205
d 123
E 164
9 WE3 Inside a box are 42 plastic shapes. Some of the shapes are squares and the remainder are circles.
One shape is taken out at random, its shape is noted and it is put back in the box. After this is repeated
84 times it is found that a square was taken out 36 times. Estimate how many squares and how many
circles are in the bag.
10 A closed box with a hole in one corner contains coloured marbles: 4 are red, 2 are blue, 3 are white and
1 is green. The box is shaken and 1 marble falls out. Its colour is recorded and it is placed back in the
box. This is done 200 times.
a How many times is a red or blue marble expected to fall out of the box after 200 trials?
b How many times is a marble that has a colour other than white expected to fall out of the box after
200 trials?
11 mC A moneybox contains 128 coins. There are 5-cent and 10-cent coins. The box is shaken, a coin
falls out, the value of the coin is noted and it is placed back inside the box. After this is repeated
96 times it is noted that a 5-cent coin fell out 60 times. The estimated number of 10-cent coins in the
moneybox is:
a 24
B 64
C 60
d 36
E 48
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477
12 During a period of one week 190 people telephoned Hot-Shot
Electrics with enquiries. During the same period Zap Inc
received 305 enquiries. Based on this information, how many
enquiries did Hot-Shot and Zap Inc each expect during a week
where the total number of phone calls made to the two
businesses was 650 (to the nearest call)?
13 The probability that the Last Legs football team can win a match
is 27. If the team is to play 35 matches during the season, how
many wins should it expect?
14 In a 9-game chess tournament, Adam won 6 games, lost 2 games
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WorkSHEET 11.1
and drew 1 game.
a Based on this information, if Adam is to play 108 games next
year, how many games should he expect to:
i win?
ii lose?
iii draw?
b Based on the fact that Adam won 81 of the 108 games, how
many games does he expect to lose or to draw in a tournament
comprising 16 games?
15 mC Inside a bag are red, blue and black marbles. Sally takes out one marble, notes its colour and puts
it back in the bag. When she has taken out a marble 360 times, she finds that a red marble was taken
out 140 times and a black marble 200 times. If Sally takes out a marble 270 times, the number of blue
marbles expected is:
a 15
B 90
C 105
d 150
E 125
16 A post-office has two letterboxes, Domestic and Overseas. Letters to be delivered within Australia are
placed in the Domestic box, and letters intended for overseas destinations are deposited in the Overseas
box. During the month of January there were 980 Domestic and 310 Overseas letters handled.
a Estimate the probability of a particular letter having an Australian destination.
b Estimate the probability of a particular letter having an overseas destination.
c During February there were 1580 letters posted in total. How many of these would you expect to
have been delivered: i within Australia?
ii overseas?
11B
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Extension
Sets and Venn
diagrams
Calculating probabilities
Many of the methods arising from a study of probability can be investigated by using set theory. A
review of the basic work on sets is provided in your eBookPLUS.
This section describes how to calculate exact theoretical probabilities rather than use experimental results
to estimate probabilities. We know that the theoretical probability of a fair coin coming up Heads is exactly 12.
We must remember that this does not mean that exactly half the tosses of the coin will be Heads, but rather
that the long-run proportion of Heads will approach 12 as the number of tosses becomes very large.
If n is the number of trials, then as n → ∞ (→ means ‘approaches’, or ‘gets closer to’): the proportion
of successes → the theoretical probability of a ‘success’.
Before describing how to calculate theoretical probability (and avoid the need to perform a large
number of trials), we need to discuss the ideas of event space and favourable outcome in more detail.
Event space
The event space (or sample space) consists of all possible outcomes of an experiment. The event space is
the universal set and is denoted by ξ.
WorkEd ExamplE 4
A 6-sided die is rolled. List the elements of the event space and state the cardinal number.
Think
478
WriTE
1
List the elements of the event space.
ξ = {1, 2, 3, 4, 5, 6}
2
Count the number of elements in the event space.
n(ξ) = 6
Maths Quest 11 Mathematical Methods CAS
WorkEd ExamplE 5
A coin and a die are tossed.
a List the elements of the event space.
b List the elements of the event E = ‘Head and a number greater than 4’.
Think
WriTE
a What are the different outcomes using a
a Let H be Head, T be Tail and H4 mean ‘Head
coin and a die together?
on the coin and a 4 on the die’.
Then ξ = {H1, H2, H3, H4, H5, H6, T1, T2, T3,
T4, T5, T6}.
b List all the possible ways of obtaining event
b Event E consists of 2 sample points:
E = ‘Head and a number greater than 4’.
E = {H5, H6}.
probabilities
The game of ‘Zilch’ involves tossing a fair 6-sided die and scoring points for rolling a 6 or a 1.
The events ‘rolling a 6’ and ‘rolling a 1’ are called ‘favourable outcomes’.
The total number of outcomes is 6 (a result of 1, 2, 3, 4, 5 or 6 could be rolled). Each outcome is
equally likely for a fair die.
Intuition may lead us to assert that the probability of scoring by rolling a die in a game of Zilch is 26 = 13.
More formally, for equally likely outcomes:
probability of a favourable outcome =
number of favourable outcomes
total number of possible outcome
or
Pr(favourable outcome) =
number of favourable outcomes
total number of possible outcome
So in our Zilch example,
Pr(scoring) =
number of favourable outcomes
total number of possible outcome
Pr(scoring) = 26
= 13 as before.
There are other, equivalent expressions for calculating probability, including
Pr(E) =
number of favourable outcomes in E
total number of possible outcomes
Or, using set notation,
n(E )
n(ξ )
where Pr(E ) is the probability of event E, n(E ) is the cardinal number of event E and n(ξ ) is the cardinal
number of the event space. The above Zilch example may be illustrated as follows, where E = getting a
6 or a 1.
Pr(E) =
ξ
2
3
E
5
6 1 4
Notice that Pr(E ′) + Pr(E ′) = 26 + 46
In general, if E and E ′ are complementary events,
= 1.
Pr(E) + Pr(E′) = 1
and
Pr(E′) = 1 − Pr(E)
ChapTEr 11 • Introductory probability
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WorkEd ExamplE 6
A number is randomly chosen from the first 12 positive integers. Find the probability of:
a choosing the number 8
b choosing any number except 8.
Think
WriTE
a 1 Pr(favourable outcome)
a
number of favourable outcomes
=
total number of possible outcomes
2
Pr(8) =
There is one favourable outcome (choosing an 8)
and 12 possible outcomes.
1
12
b Pr(not 8) = 1 −
b We require the complementary probability.
Pr(E ′ ) = 1 − Pr(E ).
= 11
12
1
12
range of probabilities
If there is no favourable outcome for event E, then n(E ) = 0, so:
n(E )
n(ξ )
0
=
n(ξ )
=0
Pr(E) =
We interpret this to mean that impossible events have a probability of zero.
If every outcome in the event space for E is a favourable outcome, then
n(E ) = n(ξ) and
n(E )
Pr(E) =
n(ξ )
=
n(ξ )
n(ξ )
= 1.
We interpret this to mean that events certain to happen have a probability of 1. Thus the range of
values for the probability of an event is given by 0 ≤ Pr(E ) ≤ 1. The probability line below illustrates the
range of probabilities and the likelihood of the event occurring.
0
1–
4
1–
2
3–
4
1
Impossible Unlikely Equally Likely Certain
likely
WorkEd ExamplE 7
A fair cubic die with faces numbered 1, 3, 4, 6, 8, 10 is rolled. Determine the probability that the
number appearing uppermost will be:
a even
b odd
c less than 1
d greater than or equal to 1.
Think
a 1 There are 6 possible outcomes when rolling the die.
2
Four of the outcomes correspond to an even
number.
WriTE
a n(ξ ) = 6
Pr(even number) = 4
6
=2
3
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Maths Quest 11 Mathematical Methods CAS
b Two of the outcomes correspond to an odd number.
b Pr(odd number) =
Also 1 − 2 = 1 since
3
3
Pr(odd number) + Pr(even number) = 1
c None of the outcomes correspond to a number
=
2
6
1
3
c Pr(number is less than 1) =
less than 1.
0
6
=0
d All 6 outcomes correspond to a number greater
d Pr(number is greater than or equal to 1) =
than or equal to 1.
6
6
=1
In worked example 7, notice that if A is the event ‘even number’ then the complement (A′) of A is the
event ‘odd number’ and
Pr(A) + Pr(A′) = 46 + 26
= 1.
Similarly, if B is the event ‘a number less than 1’ then B′ is the event ‘a number greater than or equal to 1’
so that
Pr(B) + Pr(B′) = 0 + 6
6
6
=1
WorkEd ExamplE 8
One letter is randomly selected from the letters in the sentence LITTLE MISS MUFFETT. Calculate
the probability that the letter is:
a a vowel
b a consonant other than a T
c a consonant.
Think
WriTE
a 1 Pr(favourable outcome)
a
number of favourable outcomes
=
total number of possible outcomes
2
A vowel is a favourable outcome. There are
17 possible outcomes (letters), of which 5
are vowels.
3
Substitute this information into the probability
formula.
b 1 There are 8 consonants other than T.
2
Write the probability.
Exercise 11B
Pr(consonant other than T) =
8
17
c
Pr(consonant) = 12
Calculating probabilities
1 WE4 A spinner is divided into 4 equal sections as shown at right.
For one spin:
a list the elements in the event space
b state the cardinal number of the event space.
5
17
b
Use the probability formula.
c 1 There are 12 consonants.
2
Pr(vowel) =
17
Red
Blue
Yellow Green
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2 A numberplate is made up of 3 letters followed by 3 numbers. What is the event space for the first
position on the numberplate?
3 A card is chosen from a pack of 52 playing cards. What is the cardinal number of the event space?
4 WE5 A card is chosen from a pack of 52 playing cards and its suit noted, then a coin is tossed.
a List the elements in the event space.
b List the elements in the event S = ‘a spade is chosen’.
5 A coin is tossed twice. List the elements in the event space.
6 A student is chosen at random from a class of 12 girls and 14 boys, then a chocolate bar is chosen from
a bag containing a Time Out, a Mars Bar and a Violet Crumble.
a List the elements in the event space.
b List the elements in the event M = ‘a Mars Bar is chosen’.
7 WE6 One player is chosen at random from the senior netball team to be the captain. If there are
7 players in the team, what is the probability the person who plays goal attack is:
a chosen?
b not chosen?
8 One Year 11 student must be chosen to represent the year level at a staff meeting. If all 81 girls’ and
73 boys’ names are put into a container and one name is chosen at random, find the probability that:
a a Year 11 student is chosen
b any particular Year 11 student is chosen
c a boy is chosen.
9 One card is chosen from a pack of 52 playing cards.
What is the probability that the card is:
a a queen?
c a picture card (J, Q, K)?
e red or black?
b a heart?
d not a picture card?
10 Four hundred thousand tickets are sold in a raffle. The winner of the
raffle will toss the coin at the AFL grand final. If you bought
10 tickets, what is the probability that you will win?
11 WE7 A standard die is rolled. What is the probability of rolling:
a an even number?
b a 5?
c a number from 2 to 4 inclusive?
d a number less than 7?
12 A bag has 20 marbles numbered 1, 2, 3, . . . , 20. One marble is randomly drawn. Find the probability
that the number on the marble is:
a even
b greater than 4
c a multiple of 4
d not even.
13 WE8 One letter is randomly selected from the letters of the palindrome ‘Madam, I’m Adam’.
Calculate the probability that the letter is:
a a vowel
b a consonant other than a D.
14 What is the probability of randomly choosing a consonant other than P from the letters of the
palindrome ‘A man, a plan, a canal, Panama’?
15 One letter is randomly selected from the words Mathematical Methods. What is the probability of
randomly selecting:
a the letter m?
b a vowel?
c a consonant?
d a letter from the first half of the alphabet?
16 A lolly shop has 85 different types of lollies
including Smarties in clear plastic containers.
Forty of the lollies contain chocolate. If one
container is chosen at random, what is the
probability it contains:
a a lolly containing chocolate?
b Smarties?
c a biscuit?
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Maths Quest 11 Mathematical Methods CAS
11C Tree diagrams
Tree diagrams
and lattice diagrams
A useful way of representing all possible outcomes for sequential activities is by means
of a tree diagram. A tree diagram consists of paths formed from branches. Each sample
point (possible outcome) corresponds to a unique path that is found by following the
branches. For example, a tree diagram could be drawn to show all possible outcomes
when a coin is tossed twice.
The first set of branches shows the possible outcome of the first activity, in this case
tossing the coin the first time.
The second set of branches is then joined onto the ends of the first
2nd coin
1st coin
set to show all outcomes of both tosses of the coin.
Note that the outcomes are written at the end of each path through
the tree diagram.
H
The cardinal number of the sample space is the total number of
end branches at the end of each path. If all outcomes are equally
likely, the probability can then be determined as before by using
T
n(E )
Pr(E ) =
.
n(ξ )
1st coin
H
T
Outcome
H HH
T HT
H TH
T TT
WorkEd ExamplE 9
A card is chosen from a pack of 52 playing cards and its suit noted; then
it is returned to the pack before another card is chosen.
a Draw a tree diagram showing all possible suit outcomes.
b Calculate the probability of choosing:
i two hearts
ii a diamond then a spade
iii a heart and a club.
Think
a Draw a tree diagram.
WriTE/draW
a
1st card
S — spade
2nd card
S
H
D
C
S
H
D
C
S
H
D
C
S
H
D
C
S
H — heart
H
D — diamond
C — club
D
C
b i Use the probability formula with one
favourable outcome (heart, heart) out of
16 possible outcomes.
ii Use the probability formula with one
favourable outcome (diamond, spade) out of
16 possible outcomes.
iii Use the probability formula with two
favourable outcomes (heart then club or club
then heart) out of 16 possible outcomes.
TUTorial
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Worked example 9
Outcome
SS
SH
SD
SC
HS
HH
HD
HC
DS
DH
DD
DC
CS
CH
CD
CC
1
b i Pr(HH) = 16
ii Pr(DS) =
1
16
iii Pr(HC or CH) =
=
2
16
1
8
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WorkEd ExamplE 10
Two letters are selected from the word BIRD.
a Draw a tree diagram to illustrate the event space.
b What is the probability that the second letter is a vowel or that the first letter is D?
Think
WriTE/draW
a 1 There are 4 letters to choose from as the first
a
letter of the pair of letters.
2
1st letter
2nd letter
Outcomes
B
I
R
D
BI
BR
BD
I
B
R
D
IB
IR
ID
R
B
I
D
RB
RI
RD
D
B
I
R
DB
DI
DR
For each letter chosen as the first letter, there
are 3 letters remaining to choose from.
b There are 5 favourable outcomes {BI, RI, DI, DB
b Pr(second letter is a vowel or first letter is D)
and DR} and 12 outcomes altogether.
5
= 12
lattice diagrams
6
7
8
9 10 11 12
5
5
6
7
8
4
4
5
6
7
9 10 11
8 9 10
3
4
5
6
7
8
9
2
2
3
4
5
6
7
8
1
1
2
3
4
5
6
7
1
2
3
4
5
6
3
1
2
3 4
Die 1
5
6
Die 1
WorkEd ExamplE 11
A die is rolled twice.
a Draw a lattice diagram to show all of the possible outcomes.
b Find the probability of rolling a 2, then a 1.
c Find the probability of getting a total of 7.
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Maths Quest 11 Mathematical Methods CAS
Coin
6
Die 2
Die 2
When showing all possible outcomes of two activities such as ‘a die is rolled twice’, a tree diagram can
become very large. An alternative method of showing all possible outcomes in this situation is a lattice
diagram.
A lattice diagram is a graphical representation in which the axes show the possible outcomes of each
activity. The ‘coordinates’ or points inside the graph show the possible outcomes from the combination
of both activities, for example a total. These can be written as dots, as totals or by using a symbol for
each outcome.
T
T1 T2 T3 T4 T5 T6
H
H1 H2 H3 H4 H5 H6
1
2
3 4
Die
5
6
Think
WriTE/draW
a The possible outcomes from each roll are
a
1, 2, 3, 4, 5, 6.
Put these numbers on each axis.
6
5
4
3
2
1
1
b The probability of obtaining a 2 on the first die and
b Pr(2, 1) =
a 1 on the second is shown by one outcome only.
There are 36 total possible outcomes.
2 numbers and the probability of a total of 7, so
write totals on the diagram to show the possible
outcomes of both events.
There are 6 ways of getting a total of 7 from
a total of 36 possible outcomes.
{1 and 6, 2 and 5, 3 and 4, 4 and 3, 5 and 2,
6 and 1}
2
Write the answer.
c
Die 2
c 1 The question asks for the probability of
2
3
4
5
6
1
36
6
5
4
7 8 9 10 11 12
6 7 8 9 10 11
5 6 7 8 9 10
3
2
1
4 5 6
3 4 5
2 3 4
7 8 9
6 7 8
5 6 7
1 2 3 4 5 6
Die 1
Pr(total = 7) =
6
36
= 16
Exercise 11C
Tree diagrams and lattice diagrams
1 WE9 A psychic powers test kit contains 10 blue, 10 red and 10 green cards, each without any
markings. In one particular test session, ‘Mental Mal’ selects a card, replaces it, and selects a card
again.
a Draw a tree diagram showing the possible colour outcomes at each stage.
b Calculate the probability of Mal choosing:
i two blue cards
ii a red card, then a green card
iii a green and a red card.
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Stirling’s formula
2 A coin is tossed together with a disc that is red on one side and white on its other side.
a Show all possible outcomes on a tree diagram.
b Calculate the probability that the coin lands Tails and the disc lands red.
3 Two letters from the word CAT are chosen.
a Show all possible outcomes on a tree diagram.
b Calculate the probability that the letter A is chosen first and the letter T is chosen second.
4 Two coins are tossed.
a Show all possible outcomes on a tree diagram.
b Find the probability that one head and one tail turned up.
5 The two spinners shown are spun and the colour on which each stops is noted.
Find the probability that the spinners land on:
a red and green
b yellow and blue
c yellow and green.
Spinner 1
Spinner 2
ChapTEr 11 • Introductory probability
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6 mC A coin is tossed and a wheel that is coloured blue, white and
yellow is spun.
The probability of getting Tails and the colour yellow is:
a
d
5
6
1
2
B
E
1
6
3
4
C
4
5
7 A pentagonal solid whose faces are numbered 2, 4, 6, 8, 10 is rolled and a disc that is red
on one side and blue on its other side is tossed. Draw a tree diagram and calculate the
probability that a number greater than 4 is rolled and the colour showing uppermost on the
disc is red.
8 WE 10 An integer from 2 to 3 inclusive is chosen from one hat and an integer from 4 to 6 inclusive is
chosen from another hat.
Draw a tree diagram showing the possible outcomes and determine the probability of selecting:
a two even numbers
b two odd numbers
c two even numbers or two odd numbers.
9 Peter chooses to wear a jacket and tie from the available
jackets and ties on his clothes rack, which is shown in
the photo at right. Use the photo to draw a tree diagram
showing the possible jacket and tie choices. Calculate the
probability of choosing the darker brown jacket with the red
and yellow tie.
10 Each of the smaller triangles formed by the intersection of
the diagonals of a square is painted using either red, green or
blue before covering each one with a low-sheen or full-gloss
varnish. If the colour of each triangle is chosen at random,
draw a suitable tree diagram and find the probability that the
triangle is not coloured red or green and is covered with fullgloss varnish.
11 A coin is tossed three times.
a Show all possible outcomes on a tree diagram.
b Find the probability of getting Head, Tail, Tail.
c What is the probability of getting at least two Tails?
12 Johnny wishes to try all combinations of a supercone ice-cream that has three
scoops of different flavours chosen from chocolate, vanilla, strawberry, lime
and banana. The middle scoop must be chocolate.
If Johnny randomly chooses his supercone ice-cream, show all possible
outcomes on a tree diagram.
13 Alan, Bjorn and Carl each toss a coin at the same time. Draw an appropriate tree
diagram and use it to find the probability that Alan’s and Carl’s result will both be Tails.
Chocolate
14 A consonant is selected from each of the words MATHS IS FUN.
a Show the possible outcomes on a tree diagram.
b Find the probability that the letters H and S will appear in the selection.
15 Two coins are tossed and a die is rolled. One of the coins is double-headed. Find the probability that
you get:
a two Heads and an even number
b a Head, a Tail and an odd number
c a Head, a Tail and a number less than 4.
16 Find the probability of obtaining an odd number and at least one Tail when a die and two coins are
tossed.
17 mC Three coins are tossed once. The probability that at least one coin shows Heads is:
a
d
486
3
4
7
8
Maths Quest 11 Mathematical Methods CAS
B
E
3
8
1
4
C
2
3
18 Four coins are tossed.
a Show all possible outcomes on a tree diagram.
b Find the probability of obtaining Head, Tail, Head, Tail in that order.
c Find the probability of obtaining two Heads and two Tails.
d Find the probability of obtaining at least two consecutive Tails.
19 WE 11 a Draw a lattice diagram to show all possible outcomes when two dice are rolled.
b Use the lattice diagram to find the probability that both the numbers appearing uppermost are odd
numbers.
c Find the probability of getting a total of 9.
20 A die is rolled and a coin is tossed.
a Draw a lattice diagram to show all of the possible outcomes.
b Find the probability of obtaining a 3 and a Tail.
c Find the probability of obtaining an even number and a Head.
21 Two dice are rolled. Find the probability:
a of obtaining two 6s
b of rolling a 3 and a 4
c that the sum of the numbers appearing uppermost is
less than 10
d that the first number is a 3 and the sum of the numbers
appearing uppermost is less than 8
e of rolling two multiples of 2.
22 A die labelled with the letters T, O, M, A, T, O and a die numbered 3, 4, 5, 6, 7, 8 are rolled together.
Determine the probability that the first die shows a vowel and the second die shows a number greater
than 6.
23 A diner orders an entree, main course and dessert from a lunch menu that offers 3 different entrees,
2 different main courses and 2 different desserts. Show these choices on a tree diagram and find the
probability that the diner orders a particular entree and main course.
11d
The addition law of probabilities
Recall from our review of set theory that:
n(A ∪ B) = n(A) + n(B) − n(A ∩ B)
We also know that:
Pr(A ∪ B) =
[1]
n(A ∪ B)
n(ξ )
[2]
Substituting [1] into [2], we get:
n(A) + n(B) − n(A ∩ B)
n(ξ )
n(A) n(B) n(A ∩ B)
=
+
−
n(ξ ) n(ξ )
n(ξ )
Pr(A ∪ B) =
So, Pr(A ∪ B) = Pr(A) + Pr(B) − Pr(A ∩ B).
Since we may equate ∪ with OR and ∩ with AND, we can say:
Pr(A or B) = Pr(A) + Pr(B) − Pr(A and B)
This is known as the Addition Law of Probabilities.
mutually exclusive events
If A ∩ B = ϕ, then A and B are mutually exclusive. That is, events A and B cannot happen at the same
time. If A and B are mutually exclusive, Pr(A ∩ B) = 0, the Addition Law becomes:
Pr(A ∪ B) = Pr(A) + Pr(B)
or
Pr(A or B) = Pr(A) + Pr(B)
ChapTEr 11 • Introductory probability
487
WorkEd ExamplE 12
If A and B are events such that Pr(A) = 0.8, Pr(B) = 0.2 and Pr(A ∩ B) = 0.1, calculate Pr(A ∪ B).
Think
Substitute the values for Pr(A), Pr(B) and
Pr(A ∩ B) in the Addition Law to find
Pr(A ∪ B).
WriTE
Pr(A ∪ B) = Pr(A) + Pr(B) − Pr(A ∩ B)
Pr(A ∪ B) = 0.8 + 0.2 − 0.1
Pr(A ∪ B) = 0.9
WorkEd ExamplE 13
If A and B are events such that Pr(A ∪ B) = 0.55, Pr(A) = 0.2 and Pr(B) = 0.45, calculate
Pr(A ∩ B).
Think
WriTE
1
Substitute the values for Pr(A ∪ B), Pr(A)
and Pr(B).
Pr(A ∪ B) = Pr(A) + Pr(B) − Pr(A ∩ B)
0.55 = 0.2 + 0.45 − Pr(A ∩ B)
2
Rearrange the expression to find Pr(A ∩ B).
0.55 = 0.65 − Pr(A ∩ B)
Pr(A ∩ B) = 0.1
WorkEd ExamplE 14
If Pr(A ∩ B) = 0.2 and Pr(A ∪ B) = 0.9, calculate Pr(A) and Pr(B) if events A and B are equally
likely to occur.
Think
WriTE
1
Use the Addition Law.
Pr(A ∪ B) = Pr(A) + Pr(B) − Pr(A ∩ B)
2
If events A and B are equally likely to occur,
then Pr(A) = Pr(B).
Let x represent Pr(A) and hence Pr(B).
3
Substitute the information into the
Addition Law and solve.
0.9 = x + x − 0.2
0.9 = 2x − 0.2
1.1 = 2x
so x = 0.55
Pr(A) = 0.55, Pr(B) = 0.55
WorkEd ExamplE 15
A box contains 16 marbles numbered 1, 2, 3, . . . , 16. One marble is
randomly selected.
Let A be the event ‘the marble selected is a prime number greater than 3’ and
let B be the event ‘the marble selected is an odd number’.
a Evaluate:
i Pr(A)
ii Pr(B)
iii Pr(A ∩ B)
iv Pr(A ∪ B).
b Are A and B mutually exclusive events?
Think
a 1 Write down the elements of A, B,
A ∩ B and A ∪ B.
2
488
What is n(ξ)?
Maths Quest 11 Mathematical Methods CAS
TUTorial
eles-1449
Worked example 15
WriTE
a A = {5, 7, 11, 13}
B = {1, 3, 5, 7, 9, 11, 13, 15}
A ∩ B = {5, 7, 11, 13}
A ∪ B = {1, 3, 5, 7, 9, 11, 13, 15}
n (ξ) = 16
3
Calculate the probability of A, B,
A ∩ B and A ∪ B.
Pr(A) =
i
4
16
=1
Pr(B) =
ii
=
iii
Pr(A ∩ B) =
=
iv Pr(A ∪ B) =
=
4
8
16
1
2
4
16
1
4
8
16
1
2
1
4
b From iii above, we see that Pr(A ∩ B) = .
b Since Pr(A ∩ B) ≠ 0, it follows that A and B
are not mutually exclusive.
As Pr(A ∩ B) ≠ 0, A and B are not mutually
exclusive.
Note: The Addition Law could also be used to determine any one of Pr(A ∪ B), Pr(A), Pr(B) or
Pr(A ∩ B) when the other three quantities are known.
For example, to find Pr(A ∪ B) we have:
Pr(A ∪ B) = Pr(A) + Pr(B) − Pr(A ∩ B)
=
=
4
16
1
2
+
8
16
−
4
16
The Venn diagram below left may be adapted to show probabilities rather than outcomes and used to
solve problems (below right).
ξ (A ∪ B)′
A
A ∩ B′
A′ ∩ B′
or
ξ Pr(A ∪ B)'
B
A∩B
A′ ∩ B
or
Pr(A' ∪ B')
A
B
Pr(A ∩ B')
Pr(A ∩ B)
Pr(A' ∩ B)
WorkEd ExamplE 16
An 8-sided die (numbered from 1 to 8) is rolled once. Find the probability that the number
appearing uppermost is: a an even number b an even number or a multiple of 3.
Think
a 1 Let E = even number = {2, 4, 6, 8}.
The probability of getting an even
n(E )
number =
.
n(ξ )
2
n(E) = 4, n(ξ ) = 8
WriTE
a Pr(E ) =
n(E )
n(ξ )
Pr(E) = 4
8
= 12
b 1 M = multiple of 3 = {3, 6}. The probability
b Pr(E ∪ M) = Pr(E) + Pr(M ) − Pr(E ∩ M )
of an even number or a multiple of
3 = Pr(E ∪ M ).
2
Pr(E ) = 12, Pr(M ) = 28 = 14,
E ∩ M = {6} so Pr(E ∩ M ) = 18.
Pr(E ∪ M ) = 12 + 14 − 18
= 58
ChapTEr 11 • Introductory probability
489
WorkEd ExamplE 17
If Pr(A) = 0.6, Pr(B) = 0.45 and Pr(A ∪ B) = 0.7, show this information on a Venn diagram and
calculate Pr(A ∪ B)′.
Think
WriTE/draW
1
Draw a 2-set Venn diagram with an overlapping
region.
2
Calculate the probability of the overlap, Pr(A ∩ B),
using the Addition Law.
Pr(A ∪ B) = Pr(A) + Pr(B) − Pr(A ∩ B)
0.7 = 0.6 + 0.45 − Pr(A ∩ B)
Pr(A ∩ B) = 1.05 − 0.7
= 0.35
3
Complete the Venn diagram using the available
information.
ξ (A ∪ B)'
A
B
0.25 0.35 0.1
4
Calculate Pr(A ∪ B)′.
Exercise 11d
1 WE12
Pr(A ∪ B) + Pr(A ∪ B)′ = 1
Pr(A ∪ B)′ = 1 − Pr(A ∪ B)
= 1 − 0.7
= 0.3
The addition law of probabilities
If Pr(A) = 0.4, Pr(B) = 0.5 and Pr(A ∩ B) = 0.2, what is Pr(A ∪ B)?
2 If Pr(A) = 0.65, Pr(B) = 0.25 and Pr(A ∩ B) = 0.22, what is Pr(A ∪ B)?
3 If A and B are mutually exclusive events and Pr(A) = 0.38, Pr(B) = 0.51, what is Pr(A ∪ B)?
4 WE13 If A and B are events such that Pr(A) = 0.4, Pr(B) = 0.5 and Pr(A ∪ B) = 0.6, calculate Pr(A ∩ B).
5 For events X, Y, if Pr(Y) = 0.44, Pr(X ∩ Y) = 0.16 and Pr(X ∪ Y) = 0.73, what is Pr(X)?
6
For events D and E, if Pr(D) = 0.76, Pr(D ∪ E) = 0.82 and Pr(D ∩ E) = 0.35, what is Pr(E)?
7 WE14 If Pr(A) = 2 × Pr(B), Pr(A ∩ B) = 0.23 and Pr(A ∪ B) = 0.94, determine the values of Pr(A)
and Pr(B).
8 If Pr(A ∪ B) = 0.75, Pr(A) = 0.28 and Pr(B) = 0.47, what can be concluded about the relationship
between A and B?
9 If Pr(A ∩ B) = Pr(A), what is the relationship between A and B?
10 WE15 A card is chosen at random from a pack of 52 playing cards. Let H be the event ‘choosing a
heart’ and P be the event ‘choosing a picture card (J, Q, K)’.
a Evaluate:
i Pr(H)
ii Pr(P)
iii Pr(H ∩ P)
iv Pr(H ∪ P).
b Are H and P mutually exclusive events?
11
490
A box of chocolates contains 12 with hard centres and 8 with soft centres. One chocolate is chosen
at random. Let H be the event ‘choosing a hard centre’ and S be the event ‘choosing a soft centre’.
a Evaluate:
i Pr(H)
ii Pr(S)
iii Pr(H ∩ S)
iv Pr(H ∪ S).
b Are H and S mutually exclusive events?
Maths Quest 11 Mathematical Methods CAS
12 From a group of 100 people, 25 said they drink tea, 40 said they drink coffee and 15 said they drink
both beverages. If one member of the group is randomly chosen, what is the probability that the person:
a drinks only tea?
b drinks neither tea nor coffee?
c drinks tea and coffee?
d drinks tea or coffee?
13 WE16 A box contains 20 marbles numbered 1 to 20. Find the probability of obtaining:
a an even number
b a multiple of 3
c a multiple of 2 or 3.
14 Find the probability of an odd number or a multiple of 4 appearing uppermost when a die is rolled.
15 Find the probability that a number divisible by 4 or 5 is drawn from a ‘lucky dip’ containing the first
50 natural numbers.
16 From a standard pack of 52 playing cards, one card is randomly drawn. State the probability that the card is:
a a ten
b a diamond
c a king or a jack
d a diamond, a spade or the ace of hearts.
17 A mixed bag of lollies contains 8 peppermint twists, 10 red jelly beans, 10 caramels, 18 chocolates,
4 peppermint twirls, 5 yellow jelly beans and 25 toffees. If Tara randomly selects one lolly, what is the
probability that it is:
a a peppermint or a jelly bean?
b not a toffee, a caramel or a jelly bean?
c a peppermint, given that the jelly beans are stuck together and cannot be selected?
18 Sarah is competing in a 400-metre race against 13 other runners. If each contestant has the same
probability of winning, find the probability that Sarah:
a wins the race
b comes first or second
c finishes in the top four
d does not qualify for the final 5.
19 A moneybox containing eight $1 coins, five $2 coins, nine 50c coins and two 20c coins is shaken and
one coin falls out. Assuming that each coin is equally likely to fall out, calculate the probability that the
coin’s value is:
a between 10c and $2 (not including 10c or $2)
b not 50c
c $1 or $2
d less than $1.
20 Inside a dresser drawer are 4 ties, 10 socks, 4 handkerchiefs and 2 towels. If Tony randomly takes out
one item, find the probability that it is:
a something to be worn
b not a towel and not a sock
c either a sock or not a sock
d either a towel or not a handkerchief.
21 A patron in a restaurant is presented with a fruit platter consisting of 6 whole apples, 8 slices of orange,
5 sliced pear pieces, 11 whole strawberries, 6 whole plums and 4 sliced apricot halves. The waiter
accidentally trips and a piece of fruit falls off the platter. Assuming that each piece of fruit was equally
likely to fall, state the probability that the fallen fruit is:
a not a plum and not an apricot
b not sliced
c sliced or is not a strawberry
d either a pear or an orange that has not been sliced.
22 The games Alotto, Blotto and Clotto involve guessing a number from 1 to 100 inclusive. To win Alotto
the number guessed must be a multiple of 3. To win Blotto the number must be a multiple of 5 or a
multiple of 8. To win Clotto the guessed number is to be between 10 and 20 or greater than 77.
Decide which game is easier to win.
ChapTEr 11 • Introductory probability
491
23 WE17 If Pr(A) = 0.3, Pr(B) = 0.4 and Pr(A ∪ B) = 0.65, show this information on a Venn diagram and
find Pr(A′ ∪ B).
24 Of 20 people interviewed, 7 stated that they use both a tram and a train to get to work, and
2 said they drive their own car. No other form of transport or combination of transport is
used. If 5 people travel only by train, find the probability that a person selected at random travels
by tram only.
25 The unusual dartboard shown below consists of 10 concentric circles, with 1024 points given for a dart
landing within the first (smallest) circle, 512 points for a hit within the area bounded by the first and
second circle, 256 points if the dart lands within the area bounded by the second and third circles, and so
on. The area bounded by any two consecutive circles is the same.
Area 1
1024 points
Area 2
512 points
Area 3
256 points
Area 4
128 points
diGiTal doC
doc-9807
WorkSHEET 11.2
a Find the probability that a dart randomly hitting the board will score:
i 64
ii a multiple of 128
iii a number from 16 to 256 inclusive
iv a number from 17 to 1023 inclusive or a number less than 256.
b Why is it necessary to state that the areas bound by any two consecutive circles are
the same?
karnaugh maps and probability
tables
11E
We have seen how Venn diagrams provide a visual representation of sets and probabilities. Another effective
approach is to display the information by means of a Karnaugh map. Consider a Venn diagram for two
sets A and B.
ξ
A
B
A ∩ B' A ∩ B A' ∩ B
A' ∩ B'
Notice that the Venn diagram consists of four mutually exclusive regions, A ∩ B′, A ∩ B, A′ ∩ B and
A′ ∩ B′. These four subsets of ξ can be presented as a Karnaugh map.
Row 1
Row 2
Row 3
492
Maths Quest 11 Mathematical Methods CAS
A
A′
Column 1
B
A∩B
A′ ∩ B
Column 2
B′
A ∩ B′
A′ ∩ B′
Column 3
Comparing the table entries with the Venn diagram provides equality relationships across rows and
down columns. That is, in terms of regions we can see that for column 1, (A ∩ B) ∪ (A′ ∩ B) = B, and
for column 2, (A ∩ B′) ∪ (A′ ∩ B′) = B′.
Similarly, for row 1, (A ∩ B) ∪ (A ∩ B′) = A, and for row 2, (A′ ∩ B) ∪ (A′ ∩ B′) = A′. The third row
and column can be used to check the sum totals of each row and column. This type of verification can be
useful in practical problems.
The probability table
We can present a Karnaugh map in terms of the probability of each of the four subsets A ∩ B′, A ∩ B,
A′ ∩ B and A′ ∩ B′ of ξ.
B
B′
A
Pr(A ∩ B)
Pr(A ∩ B′)
Pr(A)
A′
Pr(A′ ∩ B)
Pr(A′ ∩ B′)
Pr(A′)
Pr(B)
Pr(B′)
1
Note the value of 1 at the bottom right of the table. This is the sum of the probabilities across the last
row and the sum of the probabilities down the last column.
That is, Pr(B) + Pr(B′) = 1 and Pr(A) + Pr(A′) = 1.
Consider the following example. A survey of 1000 taxi drivers revealed that 450 of them drive
Falcons and 500 drive Commodores. It was also found that 350 taxi drivers have occasion to use
both types of car. This information can be represented as a Venn diagram, a Karnaugh map or a
probability table.
ξ
F
C
C
100 350 150
400
Venn
C
Row 1
F
350
C′
100
Row 2
F′
150
400
550
500
500
1000
Row 3
dia
gram
Karnaugh
450
F
0.35
C′
0.1
F′
0.15
0.4
0.55
0.5
0.5
1.0
m
ap
0.45
Probability able
t
The Karnaugh map provides the following information:
1. 350 drivers drive both a Falcon and a Commodore (row 1, column 1: F ∩ C ).
2. 100 drivers drive only a Falcon (row 1, column 2: F ∩ C ′).
3. The total number of Falcon drivers is 450 (350 + 100).
4. 150 drivers drive only a Commodore (row 2, column 1: F ′ ∩ C ).
5. 400 drivers do not drive either a Falcon or a Commodore (row 2, column 2: F ′ ∩ C ′).
6. There are 500 Commodore drivers altogether (350 + 150).
7. There are 1000 drivers in total (row 3, column 3).
WorkEd ExamplE 18
Complete the probability table shown below and represent the information
as a Venn diagram.
Row 1
A
Row 2
A′
Row 3
Column 1
Column 2
B
B′
Column 3
TUTorial
eles-1450
Worked example 18
0.3
0.25
0.65
1
ChapTEr 11 • Introductory probability
493
WriTE/draW
Think
1
Find the value for row 2, column 3 and
for row 3, column 2.
B
0.25
0.35
0.3
0.7
1
0.45
0.65
B′
0.1
0.25
0.35
0.3
0.7
1
B
0.2
0.45
0.65
B′
0.1
0.25
0.35
0.3
0.7
1
A′
0.65
2
Find the value for row 2, column 1 and for row 1,
column 2.
B
A
A′
3
Find the value for row 1, column 1.
A
A′
4
Represent the information as a Venn diagram.
B′
A
ξ
A
0.1
0.2
B
0.45
0.25
WorkEd ExamplE 19
Complete a probability table, given that Pr(A′ ∩ B) = 0.24, Pr(A) = 0.32 and Pr(B) = 0.35.
Think
1
WriTE
Place the known information in the
appropriate cells of the probability table.
B
B′
A
A′
0.32
0.24
0.35
2
Build up the table using the given information
and the fact that the probability totals 1.
B
A
0.11
A′
0.24
0.35
B′
0.32
B
1
A
0.11
B′
0.21
A′
0.24
0.44
0.68
0.35
0.65
1
0.32
0.68
0.65
1
WorkEd ExamplE 20
A group was surveyed in relation to their drinking of tea and coffee. From the results it was
established that if a member of the group is randomly chosen, the probability that that member
drinks tea is 0.5, the probability that they drink coffee is 0.6, and the probability that they drink
neither tea nor coffee is 0.1.
a Use the information to complete a probability table.
b Calculate the probability that a randomly selected person of the group:
i drinks tea but not coffee
ii drinks tea and coffee.
494
Maths Quest 11 Mathematical Methods CAS
Think
WriTE
a 1 Let T and C be the set of people who drink
a
C
tea and coffee respectively. Place the given
information in the table.
2
0.5
0.1
T′
Build up the table as shown.
C
0.6
C′
T
C′
T
1
0.5
0.1
0.5
0.6
0.4
1
C
C′
0.3
0.5
T
0.4
0.1
0.5
T′
0.6
0.4
1
T′
T
T′
C
0.2
C′
0.3
0.5
0.4
0.1
0.5
0.6
0.4
1
b i Pr(T ∩ C′ ) = 0.3
ii Pr(T ∩ C ) = 0.2
b 3 Use the appropriate probability from the
table.
karnaugh maps and probability tables
Exercise 11E
1 WE 18 Complete each Karnaugh map and represent the information as a Venn diagram.
B
a
B′
A
17
A′
B
b
25
A
33
13
A′
27
15
c
72
114
B
d
B′
A
0.3
0.57
0.4
A′
B′
B
B′
A
0.03
A′
0.22
0.36
2 mC Decide which of the following statements is true.
a U = 0.15
d V − X = W − 0.58
B
B V + W = 0.42
E U+Z=W−Z
A
0.31
B′
Y
0.75
A′
X
U
Z
V
0.58
W
C X + Y = 0.55
3 Complete a Karnaugh map given that n(A ∩ B) = 87, n(A′ ∩ B) = 13, n(A ∩ B′) = 63 and
n(ξ ) = 218.
4 Complete a Karnaugh map given that n(A ∩ B) = 35, n(A ∩ B′) = 29, n(A′ ∩ B′) = 44 and
n(A′ ∩ B) = 56.
5 Draw a Karnaugh map representing each Venn diagram.
a ξ
A
b ξ
B
A
B
0.27 0.61 0.12
0.14 0.45 0.3
0.11
ChapTEr 11 • Introductory probability
495
c ξ
A
10
d ξ
B
6
4
A
15
B
7
18
5
6 Determine the probability values and complete a probability table using the given information.
a ξ = {letters of the alphabet from a to k}, A = {a, b, c, d, e, f, g}, B = {e, f, g, h}
b ξ = {first 20 natural numbers}, A = {natural numbers less than 11}, B = {natural numbers from 8
to 15 inclusive}
7 mC If A = {2, 7, 8, 10}, B = {3, 5, 7, 9, 10} and ξ = {1, 2, . . . , 10}, then A ∩ B′ will contain the set:
a {3, 5, 7}
d {6, 7, 10}
B {1, 4}
E {2, 8}
C {2, 6, 9}
8 A survey of students revealed that 30 of them like football, 26 like soccer, 6 like both sports and 10
prefer a sport other than football or soccer. Represent this information as a:
a Venn diagram
b Karnaugh map.
9 Complete a probability table for the information in question 8.
10 mC Of a group of 200 people, 48% drink coffee (C ) each day and 39% drink tea (T ). If 38% of the
people do not drink tea or coffee, the probability table is:
a
T
T′
C
0.25
0.23
C′
0.14
0.38
T
T′
C
0.25
0.35
C′
0.48
0.38
T
T′
C
0.38
0.10
C′
0.01
0.51
C
E
11 WE 19
a
b
c
d
B
T
T′
C
0.23
0.14
C′
0.38
0.25
T
T′
C
0.39
0.38
C′
0.48
0.23
d
Complete a probability table, given:
Pr(A ∩ B) = 0.3, Pr(A′ ∩ B′) = 0.2 and Pr(A) = 0.6
Pr(B ∩ A′) = 0.7, Pr(B) = 0.8 and Pr(B′ ∩ A) = 0.1
Pr(A ∩ B) = 0.5, Pr(A′ ∩ B′) = 0.1 and Pr(B′) = 0.4
Pr(A′ ∩ B) = 14, Pr(A ∩ B′) = 14 and Pr(B′) = 34.
12 Two hundred and eighty children were asked to indicate their preference for ice-cream flavours. It was
found that 160 of the children like chocolate flavour, 145 like strawberry and 50 like both flavours. Use
this information to complete a Karnaugh map.
13 WE20 An examination of 250 people showed that of those in the group who are less than or equal
to 20 years of age, 80 wear glasses and 55 do not. Also, 110 people over 20 years of age must wear
glasses.
a Represent the information as a probability table.
b Calculate the probability that a randomly selected person of the group:
i does not wear glasses and is over 20 years of age
ii is 20 years of age or younger.
14 For the probability table shown, A is the event ‘no more than
15 years of age’ and B is the event ‘smoker’.
a Complete the probability table.
b What is the probability that:
i a person older than 15 years of age does not smoke?
ii a person is a smoker and is older than 15 years of age?
iii the person is a smoker over the age of 15 or is a nonsmoker less than or equal to 15 years of age?
496
Maths Quest 11 Mathematical Methods CAS
B
A
B′
0.08
0.6
A′
0.67
15 A survey of a group of business people indicates that
42% of those surveyed read the Daily Times newspaper
only each day and 18% read both the Daily Times and the
Bugle. Additionally, 12% of those questioned stated that
they do not read either of these newspapers.
a Show this information as a Karnaugh map.
b What is the probability that a randomly selected
member of the group:
i does not read the Daily Times?
ii reads the Bugle only?
iii does not read the Bugle or does not read either
newspaper?
c If the group consists of 150 business people, determine
how many members read at least one newspaper.
16 A lucky dip box contains 80 marble tokens that can be
exchanged for prizes. Some of the marbles have a red
stripe, some have a blue stripe, some have both a red and
a blue stripe, and some marbles have no stripes at all. It is
known that 25% of the marbles have a red stripe on them,
20% of them have a blue stripe and 65% have no stripe.
a Present the information as a Karnaugh map.
b What is the probability of choosing a marble that has a red stripe only?
c Find the probability of choosing a marble that has a red and a blue stripe or no stripe.
11F
Conditional probability
Erin thinks of a number from 1 to 10 (say 8) and asks Peter to guess what it is. The probability that Peter
makes a correct guess on his first try is 1 . If, however, Erin first tells Peter that the number is greater
10
than 7, his chances are better because he now knows that the number must be one of the numbers 8, 9 or
1
10. His probability of success is now .
3
This problem may be stated as: What is the probability of Peter choosing the right number from 1 to
10, given that the number is greater than 7?
This is an example of conditional probability, where the probability of an event is conditional on (that
is, it depends on) another event occurring first. The effect in this case is to reduce the event space and
thus increase the probability of the desired outcome.
For two events A and B, the conditional probability of event A given that event B occurs is denoted by
Pr(A | B) and is given by:
Pr(A ∩ B)
Pr(A | B) =
, Pr (B) ≠ 0
Pr(B)
Event B is sometimes called the reduced event space.
For the example above, if we let B be the event ‘numbers greater than 7’ and A be the event
‘Erin’s secret number’, then we may write:
Pr(A ∩ B)
Pr(A | B) =
Pr(B)
=
1
10
3
10
1
3
= (10
÷ 10
)
= 13
The reduced event space can be illustrated by the Venn diagram below.
ξ
A
B
1
5
2
8
9
10 6
3
7
4
ChapTEr 11 • Introductory probability
497
WorkEd ExamplE 21
If Pr(A ∩ B) = 0.8 and Pr(B) = 0.9, find Pr(A | B).
Think
WriTE
Pr(A | B) =
Substitute the values given into the expression for
conditional probability.
Pr(A ∩ B)
Pr(B)
= 0.8
0.9
=8
9
WorkEd ExamplE 22
If Pr(A) = 0.3, Pr(B) = 0.5 and Pr(A ∪ B) = 0.6, calculate:
a Pr(A ∩ B)
b Pr(A | B).
Think
WriTE
Pr(A ∪ B) = Pr(A) + Pr(B) − Pr(A ∩ B)
0.6 = 0.3 + 0.5 − Pr(A ∩ B)
so Pr(A ∩ B) = 0.2
a Use the Addition Law for probabilities
a
b Use the formula for conditional probability to
b Pr(A | B) =
to find Pr(A ∩ B).
find Pr(A | B).
Pr(A ∩ B)
Pr(B)
= 0.2
0.5
= 25
WorkEd ExamplE 23
Of a group of 50 Year 11 students, 32 study Art and 30 study Graphics. Each student studies at
least one of these subjects.
a How many students study both?
b Illustrate the information as a Venn diagram.
c What is the probability that a randomly selected student studies Art only?
d Find the probability that a student selected at random from the group studies Graphics, given
that the student studies Art.
Think
a 1 Define relevant events.
2
Find the number who study both subjects
using set theory.
b Show all the information on a Venn diagram.
WriTE/draW
a Let A = students who study Art
G = students who study Graphics
A ∩ G = students who study both
n(A ∪ G) = n(A) + n(G) − n(A ∩ G)
50 = 32 + 30 − n(A ∩ G)
50 = 62 − n(A ∩ G)
n(A ∩ G) = 12
b ξ
Art
20
c The Venn diagram reveals that 20 of the 50 students
study Art only. Calculate the probability.
498
Maths Quest 11 Mathematical Methods CAS
Graphics
12
18
c Pr(Art only) = 20
=
50
2
5
d Use the conditional probability formula to find the d Pr(G | A) =
probability that a student studies Graphics, given
that the student studies Art.
Pr(G ∩ A)
Pr(A)
÷ 32
= 12
50 50
= 12
32
= 38
WorkEd ExamplE 24
Seated in a Ford Falcon are 4 males and 2 females. Seated in a Holden
Commodore are 2 males and 1 female. One of the cars is randomly stopped
by the police and one person from the vehicle is randomly selected.
Draw a tree diagram to illustrate the situation and calculate the
probability that:
a the person selected by the police is female
b if a female is selected by the police, she was sitting in the Ford.
Think
a 1 Calculate the
probabilities.
TUTorial
eles-1451
Worked example 24
WriTE/draW
a Pr(Ford) =
1
2
Pr(male from Ford) = 46
= 23
Pr(female from Ford) = 26
Pr(Holden) = 12
Pr(male from Holden) = 23
Pr(female from Holden) = 13
= 13
2
Draw the tree diagram.
Car
1–
2
Ford
1–
2
3
Use the tree diagram
to work out the
probability that the
person is female.
Consider all the ways
a female may be
selected.
b Use the tree diagram and
the formula for conditional
probability.
Holden
Person
2–
3
male
1–
3
female Ford and female
2–
3
male
1–
3
female Holden and female
Ford and male
Holden and male
Pr(female selected) = Pr(Ford and female or Holden and female)
= Pr(Ford and female) + Pr(Holden and female)
= 12 × 13 + 12 × 13
= 13
b Pr(person is from the Ford | female is selected)
=
=
=
Pr(Ford and female)
Pr(female)
1
1
×
2
3
1
2
1
3
ChapTEr 11 • Introductory probability
499
Once the tree diagram was drawn, the calculation for part a in worked example 24 was quite
intuitive. In order to calculate the probability of a female being selected, the occupants of both cars
needed to be considered. In fact, there is a rule of probability that formalises the calculation performed
in part a above. The law is known as the Law of Total Probability, and it states:
Pr(A) = Pr(A | B)Pr(B) + Pr(A | B′)Pr(B′)
To calculate the answer to part a of worked example 24, let Pr(A) = Pr(female) and
Pr(B) = Pr(Ford).
Pr(female) = Pr(female given the car is a Ford)Pr(Ford)
+ Pr(female given the car is a Holden)Pr(Holden)
= 13 × 12 + 13 × 12
= 13
Note that the Law of Total Probability simplifies to give the rule used in part a of worked example 24:
Pr(A) = Pr(A | B)Pr(B) + Pr(A | B′)Pr(B′)
Pr(A) = Pr(A ∩ B) + Pr(A ∩ B′)
Exercise 11F
Conditional probability
1 WE21 If Pr(A) = 0.8, Pr(B) = 0.5 and Pr(A ∩ B) = 0.4, find:
a Pr(A | B)
b Pr(B | A).
diGiTal doC
doc-9808
SkillSHEET 11.1
Conditional probability
2 If Pr(A) = 0.65, Pr(B) = 0.75 and Pr(A ∩ B) = 0.45, find:
a Pr(A | B)
b Pr(B | A).
3 If Pr(A ∩ B) = 0.4 and Pr(A) = 0.5, find Pr(B | A).
If Pr(A ∩ B) = 0.25 and Pr(B) = 0.6, find Pr(A | B).
4
5 If Pr(B | A) = 0.32 and Pr(A) = 0.45, find Pr(A ∩ B).
6 If Pr(A | B) = 0.21 and Pr(B) = 0.8, what is Pr(A ∩ B)?
7 Calculate Pr(A) if Pr(B | A) = 0.75 and Pr(A ∩ B) = 0.5.
Calculate Pr(B) if Pr(A | B) = 0.96 and Pr(A ∩ B) = 0.8.
8
9 WE22 If Pr(A) = 0.7, Pr(B) = 0.5 and Pr(A ∪ B) = 0.9, calculate:
a Pr(A ∩ B)
b Pr(B | A).
10 mC If Pr(B | A) = 0.8 and Pr(A ∩ B) = 0.6, then Pr(A) is:
a
d
4
5
3
4
B
E
3
5
2
3
C
1
4
C
2
5
11 mC If Pr(A) = 0.9 and 2 × Pr(A ∩ B) = Pr(A), then Pr(B | A) is:
a
d
1
2
1
9
B
E
5
9
4
9
12 Show that if Pr(A ∩ B) = Pr(A) × Pr(B), then Pr(B | A) = Pr(B).
13
If Pr(A) = 0.23, Pr(B) = 0.27 and Pr(A ∪ B) = 0.3, find:
b Pr(A | B).
a Pr(A ∩ B)
14 If Pr(A) = 0.45, Pr(B) = 0.52 and Pr(A ∪ B) = 0.67:
a find Pr(A ∩ B)
b find Pr(B | A)
c represent the information as a Venn diagram.
15 A box contains marbles numbered 1, 2, 3, . . . 50. One marble is randomly taken out of the box. What is
the probability that it is:
a a multiple of 3, given that it is less than 21?
b between 11 to 39 inclusive, given that it is greater than 20?
500
Maths Quest 11 Mathematical Methods CAS
16 mC A group of 80 females consists of 54 dancers and 35 singers. Each member of the group is either
a dancer or a singer or both. The probability that a randomly selected member of the group is a singer
given that she is a dancer is:
a 0.17
B 0.44
C 0.68
d 0.11
E 0.78
17 WE23 A group of 60 adventurers comprises
30 mountain climbers and 45 scuba divers. If each
adventurer does at least one of these activities:
a How many adventurers are both climbers and
divers?
b Illustrate the information on a Venn diagram.
c What is the probability that a randomly
selected group member is a scuba diver only?
d Find the probability that an adventurer
randomly selected from the group is a scuba
diver, given that the adventurer is a mountain
climber.
18 Of 200 families surveyed, 85% have a TV and 70% possess a CD player. Assuming each family has at
least one of these items, what is the probability that one family randomly selected has a TV, given that
they also own a CD player?
19 During the Christmas holidays 42 students from a group of 85 VCE students found vacation
employment while 73 students went away on holidays. Assuming that every student had
at least a job or went on a holiday, what is the probability that a randomly selected
student worked throughout the holidays (that is, did not go away on holidays), given that
he/she had a job?
20 WE24 The probability that a machine in a chocolate factory does not coat a SNAP chocolate bar
adequately, therefore producing a defective product, is 0.08. The probability that it does not coat a
BUZZ chocolate bar adequately is 0.11. On any day the machine coats 250 SNAP bars and 500 BUZZ
bars. A chocolate bar is chosen at random from the production line. Draw a tree diagram to illustrate
the situation and find the probability that the chocolate bar chosen at random is:
a a BUZZ chocolate bar
b a SNAP chocolate bar and is defective
c defective, given that a SNAP bar is chosen.
21 The staff at Happy Secondary College is made up of 43 females and 29 males. Also, 22% of the
females are under 40 years old, and 19% of the males are under 40. If a staff member is selected at
random, what is the probability that:
a a male is selected?
b a male 40 years or over is selected?
c a female under the age of 40 is selected?
d a person under 40 years of age is selected?
e the person is a female given that the person selected is under 40 years of age?
22 Two letters are randomly picked from the word INFINITESIMAL. If a letter can be used more
than once, calculate the probability that both letters selected are vowels, given that the first letter
is a vowel.
Transition matrices and
markov chains
11G
introduction
In chapter 7 we saw many uses for matrices, from displaying information in an organised manner to
solving simultaneous equations or representing transformations. Matrices are also very useful in certain
conditional probability problems.
inTEraCTiViTY
int-0270
Transition matrices
ChapTEr 11 • Introductory probability
501
Example
A jar contains six red balls and four green balls. A ball is selected at random and not replaced. A second
ball is then selected. Find the probability that the second ball is a red ball.
Solution
The tree diagram illustrates the situation.
We may express the answer (where R2 is the event ‘selecting
a red ball on the second selection’) in the form:
Pr(R2) = Pr(R2 | R1)Pr(R1) + Pr(R2 | G1 )Pr(G1)
6
—
10
6
+6× 4
= 59 × 10
9
10
=
=
R1
R2
4–
9
G2
6–
9
4
—
10
54
90
3
5
5–
9
G1
3–
9
R2
G2
This is an example of the law of total probability, which may be stated as:
Pr(A) = Pr(A | B ) × Pr(B ) + Pr(A | B′) × Pr(B′)
Applying the law to the complement of event A gives:
Pr(A′ ) = Pr(A′ | B) × Pr(B) + Pr(A′ | B′ ) × Pr(B′ )
These two equations may be written in matrix form:
 Pr( A)   Pr( A | B) Pr( A | B′)   Pr( B) 
×


=
 Pr( A′)   Pr( A′ | B) Pr( A′ | B′)   Pr( B′) 
The example can now be written in matrix form, where:
A = selecting a red ball on the second selection
A′ = selecting a green ball on the second selection
B = selecting a red ball on the first selection
B′ = selecting a green ball on the first selection.

 Pr( A)  

=
 Pr( A′)  



=


5
9
4
9
6
9
3
9
5
9
4
9
6
9
3
9

 
×
 

 
 
×
 
 
Pr( B) 

Pr( B′) 

6
10
4
10
 
 
=
 
 
3
5
2
5





The first element in the final column matrix is the same as the answer we obtained in the example
shown. The second element is the probability of selecting a green ball on the second selection.
Note that the columns of the matrix each add to one. This is equivalent to one of the properties of
probability, Pr(A) + Pr(A′ ) = 1. In this case it is actually Pr(A | B) + Pr(A′ | B) = 1 for the first column and
Pr(A | B′ ) + Pr(A′ | B′ ) = 1 for the second column.
The matrix of conditional probabilities is called a transition matrix, usually denoted T.
B
B′

Pr( A | B) Pr( A | B′)
T = A 
A′  Pr( A′ | B) Pr( A′ | B′)





The preceding example can be thought of as a transition from an initial state (selection of the first
ball, B or B′ being ‘red ball’ or ‘green ball’ respectively) to the next state (selection of the second ball,
A or A′, that is ‘red ball’ or ‘green ball’).
 Pr( A) 
 Pr( B) 
 Pr( B) 
The column matrices 
 and  Pr( B′)  are called state matrices, where  Pr( B′)  is the
Pr(
A
′
)






 Pr( A) 
initial state, which we might label as S0, and 
 is the next state, S1. The matrix equation
 Pr( A′) 
becomes S1 = T × S0.
502
Maths Quest 11 Mathematical Methods CAS
WorkEd ExamplE 25
Consider a simple model of the behaviour of a football team. If it wins a
game, then the probability that it wins the next game is 0.8. If it loses, then
the probability that it wins the next game is only 0.5. Write the transition
matrix that represents the transition from one game to the next.
Think
1
2
3
WriTE
Set up a table showing the information given.
B (Wins
first game)
B′ (Loses
first game)
A (Wins
second game)
(Pr(A | B))
0.8
(Pr(A | B′))
0.5
A′ (Loses
second game)
(Pr(A′ | B))
?
(Pr(A′ | B′))
?
B (Wins first
game)
B′ (Loses first
game)
A (Wins
second game)
(Pr(A | B) )
0.8
(Pr(A | B′) )
0.5
A′ (Loses
second game)
(Pr(A′ | B) )
0.2
(Pr(A′ | B′))
0.5
Complete the table using the knowledge that
the columns must add to one.
Write the answer.
TUTorial
eles-1452
Worked example 25
 0.8 0.5 
T=

 0.2 0.5 
The matrix equation S1 = T × S0 describes the transition from state S0 (the first game) to state S1 (the
second game). If the conditional probabilities remain the same, then a similar equation will express the
transition from any particular state to the next state.
In general, Sn + 1 = T × Sn may be used to determine any state from the previous one.
WorkEd ExamplE 26
Suppose the conditional probabilities expressed in worked example 25 remain constant
throughout the football season. Also, suppose the team wins the first game of the season.
Express the problem in matrix form and find the probability that the team loses the third game
of the season.
Think
WriTE
1
Write down the transition matrix. This is the
same as in the previous worked example.
 0.8 0.5 
T= 

 0.2 0.5 
2
Write down a suitable initial state matrix that
shows the probabilities of the first game. The
first element will be the probability that the
team wins. Since we know it has won, this
probability must be 1. The second element must
be 0 as the column sums to 1.
S0 =  0 
1 
ChapTEr 11 • Introductory probability
503
3
Enter the transition matrix, T, into a CAS
calculator.
 0.8 0.5 
 0.2 0.5  → t


4
Enter the initial state matrix, S0, into the CAS
calculator.
 1 
 0  → s0
 
5
Perform matrix multiplication to calculate
S1 = T × S0.
t × s0 → s1
6
Record the result.
 0.8 
 0.2 


7
Perform matrix multiplication to calculate
S2 = T × S1.
t × s1 → s2
8
Record the result.
 0.74 
 0.26 


9
The answer will be the second element of the
state matrix as we want the probability of a loss.
The probability of losing game 3 is 0.26.
This worked example shows the power of using matrices. If the conditional probabilities remain
constant and the outcomes of any particular state depend only on the previous state, then we have a (twostate) Markov Process. We can easily calculate the probabilities associated with any of the later states
without determining all the intermediate ones.
As S1 = T × S0 and S2 = T × S1,
S2 = T × T × S0 = T2 × S0
Continuing in this fashion, we see that:
and, in general, Sn =
Tn
S3 = T × S2 = T × (T × S1) = T × T × T × S0 = T3 × S0
× S0.
WorkEd ExamplE 27
Using the data of worked example 25, find the probability that the team wins the fifth game of
the season, assuming it loses the first game.
Think
504
WriTE
1
Write down the transition matrix.
 0.8 0.5 
T =  0.2 0.5 


2
Write down a suitable initial state matrix. We
know that the team loses the first game, so the
second element must have a probability of 1.
S0 =  0 
1 
 
3
Identify which state matrix is required.
Since S0 corresponds to game 1, game 5 must
correspond to state matrix S4.
4
Enter the transition matrix, T, into a CAS
calculator.
 0.8 0.5 
 0.2 0.5  → t


5
Enter the initial state matrix, S0, into the CAS
calculator.
 0 
 1  → s0
 
Maths Quest 11 Mathematical Methods CAS
6
Calculate the probabilities of the outcome of
game 5 using the rule S4 = T 4 × S0.
t 4 × s0
7
Record the result.
 0.7085 
 0.2915 


8
Write down the answer. As we are interested in
the probability of winning the fifth game, we
look at the first element of the matrix.
The probability of winning the fifth game, given
that the team lost the first game, is 0.7085.
alternative forms of the state matrix
It is not necessary to express a state matrix in terms of probabilities. We can use percentages or even raw
numbers, as the following example illustrates.
WorkEd ExamplE 28
Suppose there are 800 people in a town who watch the two current affairs shows on television:
Breakdown and News Roundup. Assume they all watch just one of the shows every week night.
Also assume that if a person were to watch Breakdown on a particular night, then there is a
probability of 0.35 that she will watch Breakdown the next night. If she were to watch News
Roundup on a particular night, there is a probability of 0.45 that she will watch Breakdown the
next night. These probabilities remain constant.
If 300 people watch Breakdown on Tuesday night of a particular week, how many will watch
each show on the next Thursday night?
Think
1
WriTE
Set up a table that shows the
information given.
Watches
Watches News
Breakdown Roundup on
on first night first night
Watches Breakdown
on next night
0.35
0.45
Watches News
Roundup on next
night
2
Complete the table using the
knowledge that the columns must add
to one.
Watches
Watches News
Breakdown Roundup on
on first night first night
Watches Breakdown
on next night
0.35
0.45
Watches News
Roundup on next
night
0.65
0.55
3
Write the transition matrix.
 0.35 0.45 
T=

 0.65 0.55 
4
The initial state matrix must show how
many watch each show on the Tuesday
night. If 300 watch Breakdown, then
500 (800 − 300) must watch News
Roundup.
 300 
S0 = 

 500 
ChapTEr 11 • Introductory probability
505
5
Idenfity which state matrix is required.
Since S0 corresponds to Tuesday night, then Thursday
night must correspond to state matrix S2.
6
Enter the transition matrix, T, into a
CAS calculator.
 0.35 0.45 
 0.65 0.55  → t


7
Enter the initial state matrix, S0, into
the CAS calculator.
 300 
 500  → s0


8
Calculate the number of people who
watch each show on Thursday night.
Use S2 = T 2 × S0.
t 2 × s0
9
Record the result.
 327 
 473 


Write the answer, remembering to
round off to the nearest whole number
if necessary.
327 watch Breakdown and 473 watch News Roundup
on Thursday night.
10
Exercise 11G
Transition matrices and markov chains
1 WE25 Consider a simple model of the behaviour of a netball team. If it wins a game, then
the probability that it wins the next game is 0.75. If it loses, then the probability that it wins
the next game is only 0.5. Write down the transition matrix that represents the transition from
one game to the next.
2 WE26 Suppose the conditional probabilities expressed in question 1 remain constant throughout the
netball season. Also, suppose the team wins the first game of the season. Express the problem in matrix
form and find the probability that the team loses the third game of the season.
 0.6 0.7 
 30 
 and an initial state matrix S0 =  70  , calculate:
0.4
0.3




b S2
c S3.
3 WE27 For a transition matrix T = 
a S1
 0.35 0.5 
 800 
and an initial state matrix S0 = 

, calculate, giving
 0.65 0.5 
 200 
answers to the nearest whole number:
a S1
b S2
c S4.
4 For a transition matrix T = 
 0.6 0.25 
3
, calculate T . Use the result to calculate S3 if the initial
 0.4 0.75 
5 For a transition matrix T = 

state matrix S0 =  1 .
0 
 0.85 0.5 
4
, calculate T . Use the result to calculate S4 if the initial
0.15
0.5


6 For a transition matrix T = 
state matrix S0 =  1 .
0 
7 WE28 A school canteen offers vegetable or tomato soup in a cup each day. It is noticed that 30%
of students who have vegetable soup on a given day select tomato soup the next day, and 60% who
have tomato soup choose vegetable soup on the next day. There are 500 students who use the canteen
each day, and they all have vegetable or tomato soup, but not both. On a particular Monday, 200 have
vegetable soup and 300 have tomato soup.
a Set up a transition matrix for this situation.
b Write down the initial state matrix.
506
Maths Quest 11 Mathematical Methods CAS
c How many students will have vegetable soup and how many will have tomato soup on the Friday
of that week? (Answers must be given to the nearest whole number.)
d How many will have each type of soup on the following Friday? (Note that the canteen does not
open on the weekend.)
8 Frank Viccuci is the goalkeeper for his soccer team. Assume that he always dives sideways when
defending penalty kicks. If he dives to the right on a given occasion, then the probability that he dives to
the right on the next occasion is 0.4. The transition matrix for the probabilities of Frank diving to
either side, given the side he dived to on the previous occasion, is  0.4 0.8 .
 0.6 0.2 


a If Frank dives to the left on a penalty he defends in today’s game, what is the probability that he
will dive to the right for the next penalty?
b Suppose there is a penalty shoot-out that consists of each team taking five penalty shots at goal.
The player taking the fifth penalty kicks to the right. If Frank had dived to the left on the first
penalty, what is the probability that he dives to the right on the fifth penalty? Give answers
accurate to 2 decimal places.
9 Assume that the probability of a particular football team winning its next game is 0.75 if it won its
previous game and 0.55 if it lost its previous game.
a If the team was successful in the opening game of the season, calculate the probability that it
will win:
i the second game of the season
ii the fifth game of the season
iii the tenth game of the season
iv the twelfth game of the season.
Give answers accurate to 4 decimal places.
b Repeat the calculations to find the corresponding probabilities if the team lost its opening game.
c Can you predict, without further calculation, what is likely to happen in the last game of the
season? (Assume the season lasts for twenty-four games.)
10 Tasha likes to vary her gym routines in a special way. If she does a cycling class one day, the
probability that she will do a Pilates session the next day is 0.6, and if she does a Pilates session one
day, the probability she does cycling next day is 0.75. Assume that Tasha goes to the gym every day
and does only cycling or Pilates.
a Write down the transition matrix for this situation.
b If Tasha cycles on Saturday, what is the probability that she also cycles the next Tuesday?
c What is the probability that she does Pilates on that Tuesday?
d What is the probability that she does Pilates on the next Tuesday?
11h
independent events
Two events A and B are independent if each event has no effect on the likelihood of the other.
Consider two independent events A and B, where event A follows event B. If the probability of event A
is unaffected by event B, then we can say that the probability of A, given B has happened, is the same as
the probability of A (whether or not B has happened), or, using symbols:
Pr(A | B) = Pr(A)
But Pr( A | B) =
we have:
[1]
P r (A ∩ B)
using the conditional probability formula. Rearranging the above equation
P r (B)
Pr(A ∩ B) = Pr(B) Pr(A | B)
[2]
Note: Equation [2] has wide application in probability. It may be extended and
interpreted as: ‘When calculating the probability of a chain of events, you may simply multiply by
the probability of the next event, as long as the effect of previous events is taken into account’.
Substituting [1] into [2] we have:
Pr(A ∩ B) = Pr(A) Pr(B)
ChapTEr 11 • Introductory probability
507
1. Pr(A ∩ B) means the probability of events A and B occurring.
2. If Pr(A ∩ B) = Pr(A) × Pr(B), then the events A and B are independent.
One of the 12 outcomes possible when a coin and a die are simultaneously tossed is a Head for the coin
and a 5 on the die. The number 5 obtained with the die does not come about because the coin comes up a
Head, and getting a Head with the coin is not a result of the number 5 appearing uppermost on the die.
We can verify the expression given for independent events by looking further at the example of the
coin and die.
What is the probability of getting a Tail and a number from 3 to 4 inclusive?
The event space is ξ = {H1, H2, H3, H4, H5, H6, T1, T2, T3, T4, T5, T6}, with n(ξ ) = 12.
Let A be the event ‘getting a Tail’ and B be the event ‘getting a number from 3 to 4 inclusive’.
n(A ∩ B) 2 1
Then A ∩ B = {T3, T4}, n(ξ) = 12, so Pr(A ∩ B) =
= = .
12
6
n(ξ )
Now Pr(A) = 1 and Pr(B) = 2 = 1, so Pr(A) × Pr(B) = 1 × 1 = 1.
2
6
3
2
3
6
So Pr(A ∩ B) = Pr(A) × Pr(B); thus the two events are independent.
WorkEd ExamplE 29
Given that events A and B are independent, find the value of x if Pr(A) = 0.55, Pr(B) = 0.6 and
Pr(A ∩ B) = x.
Think
WriTE
Pr(A ∩ B) = Pr(A) × Pr(B)
1
Write the formula for independent events.
2
Substitute the given information.
x = 0.55 × 0.6
3
Simplify.
x = 0.33
WorkEd ExamplE 30
Show that if Pr(A) = 0.5, Pr(B) = 0.8 and Pr(A ∪ B) = 0.9, then A and B are independent.
Think
WriTE
1
Use the Addition Law for probabilities to find
Pr(A ∩ B).
Pr(A ∪ B) = Pr(A) + Pr(B) − Pr(A ∩ B)
0.9 = 0.5 + 0.8 − Pr(A ∩ B)
Pr(A ∩ B) = 0.4
2
Calculate Pr(A) × Pr(B).
Pr(A) × Pr(B) = 0.5 × 0.8
= 0.4
Since Pr(A ∩ B) = Pr(A) × Pr(B), A and B are
independent events.
WorkEd ExamplE 31
Two spinners each labelled with the numbers 1, 2, 3 are spun.
A is the event ‘an odd number with the first spinner’.
B is the event ‘an even number with the second spinner’.
C is the event ‘an odd number from each spinner’.
a Calculate Pr(A), Pr(B) and Pr(C ).
b Decide which of the pairs of events AB, AC, BC is independent.
Think
a 1 List ξ, A, B and C.
2
508
Calculate Pr(A), Pr(B) and Pr(C ).
Maths Quest 11 Mathematical Methods CAS
1
2
1
3
Spinner 1
2
3
Spinner 2
WriTE
a ξ = {11, 12, 13, 21, 22, 23, 31, 32, 33}
A = {11, 12, 13, 31, 32, 33}, B = {12, 22, 32},
C = {11, 13, 31, 33}
Pr(A) = 6 = 2, Pr(B) = 3 = 1, Pr(C) = 4
9
3
9
3
9
b Check to see if
Pr(A ∩ B) = Pr(A) × Pr(B),
Pr(A ∩ C) = Pr(A) × Pr(C ),
Pr(B ∩ C) = Pr(B) × Pr(C ).
b A ∩ B = {12, 32}, Pr(A ∩ B) =
Pr(A) × Pr(B) = 23 × 13
2
9
= 29
Pr (A ∩ B) = Pr(A) × Pr(B), so A and B are
independent.
A ∩ C = {11, 13, 31, 33}, Pr(A ∩ C ) = 49
Pr(A) × Pr(C ) = 23 × 49
8
= 27
Pr(A ∩ C ) ≠ Pr(A) × Pr(C ), so A and C are not
independent.
B ∩ C = ϕ, Pr(B ∩ C ) = 0
4
Pr(B) × Pr(C ) = 13 × 49 = 27
Pr(B ∩ C ) ≠ Pr(B) × Pr(C ), so B and C are not
independent.
When the probabilities of all possible outcomes are not equally likely, the probability of
each outcome is placed on the corresponding branch of the tree diagram. When each branch is
representing an outcome from independent events, you can follow the branches and multiply the
probabilities together.
WorkEd ExamplE 32
A moneybox contains three $1 coins and two $2 coins. The moneybox is shaken; one coin falls out and
is put back in the box. This is repeated twice more. If each coin has an equal probability of falling out:
a represent this information on a tree diagram
b calculate the probability of getting three $1 coins
c calculate the probability of getting at least two $2 coins.
Think
a 1 There are three $1 coins and five coins
WriTE/draW
a
altogether. The probability of a $1 coin
falling out is 35 = 0.6.
2
There are two $2 coins and five coins
altogether. The probability of a $2 coin
falling out is 25 = 0.4.
3
Place the probability of each outcome on the
corresponding branch of the tree diagram.
0.6
$1
0.4
$2
b Multiply the probabilities obtained from the tree
diagram corresponding to three $1 coins.
c 1 Outcomes corresponding to ‘at least two $2
0.6
$1
0.4
$2
0.6
$1
0.4
$2
0.6
0.4
0.6
0.4
0.6
0.4
0.6
0.4
$1 $1, $1, $1
$2 $1, $1, $2
$1 $1, $2, $1
$2 $1, $2, $2
$1 $2, $1, $1
$2 $2, $1, $2
$1 $2, $2, $1
$2 $2, $2, $2
b Pr(three $1 coins) = 0.6 × 0.6 × 0.6
= 0.216
c
coins’ are ($1, $2, $2), ($2, $1, $2),
($2, $2, $1) and ($2, $2, $2).
ChapTEr 11 • Introductory probability
509
2
Calculate and add the probabilities.
Pr(at least two $2 coins)
= Pr($1, $2, $2 or $2, $1, $2 or $2, $2, $1
or $2, , $2$2)
= (0.6 × 0.4 × 0.4) + (0.4 × 0.6 × 0.4)
+ (0.4 × 0.4 × 0.6) + (0.4 × 0.4 × 0.4)
= 0.096 + 0.096 + 0.096 + 0.064
= 0.352
WorkEd ExamplE 33
Christos estimates his chances of passing Maths, Science and English as 0.75,
0.6 and 0.5 respectively.
a Represent this information on a tree diagram.
b Assuming the events are independent, calculate the probability that:
i he passes all three subjects
ii he passes at least Maths and English
iii he passes at least one subject.
Think
WriTE/draW
a 1 Name the three events.
a
Let M, S, E be the events ‘passing Maths’,
‘passing Science’ and ‘passing English’
respectively.
2
Calculate Pr(M ′), Pr(S ′) and Pr(E ′). M ′ is the
event of failing Maths.
Pr(M) = 0.75
Pr(S) = 0.6
Pr(E ) = 0.5
3
Use the information to draw the tree diagram.
Maths
0.75
M'
i Multiply the probabilities corresponding
to passes in all three subjects.
ii
1
We require that Christos pass both
Maths and English and either pass or fail
Science.
Science
0.6
S
0.4
S'
0.6
S
0.4
S'
English
0.5
0.5
0.5
0.5
0.5
0.5
0.5
0.5
E
E'
MSE
MSE'
E
E'
MS'E
MS'E'
E
E'
M'SE
M'SE'
E
E'
M'S'E
M'S'E'
b i Pr(MSE) = Pr(M) × Pr(S) × Pr(E)
= 0.75 × 0.6 × 0.5
= 0.225
ii Pr(MSE or MS ′E)
= Pr(MSE) + Pr(MS ′E)
2
These events are independent, so we may
multiply the individual probabilities.
= Pr(M) × Pr(S) × Pr(E)
+ Pr(M) × Pr(S ′) × Pr(E)
= 0.75 × 0.6 × 0.5 + 0.75 × 0.4 × 0.5
3
Simplify.
= 0.375
iii Passing at least one subject is the complement of
failing all three subjects.
510
Pr(M′) = 0.25
Pr(S′) = 0.4
Pr(E′) = 0.5
M
0.25
b
TUTorial
eles-1453
Worked example 33
Maths Quest 11 Mathematical Methods CAS
iii Pr(passes at least one subject)
= 1 – Pr( M′S′E′)
= 1 – 0.25 × 0.4 × 0.5
= 1 – 0.05
= 0.95
Exercise 11h
independent events
1 WE29 Given that events A and B are independent, find the value of x if:
a Pr(A) = 0.4, Pr(B) = 0.5 and Pr(A ∩ B) = x
b Pr(A) = 0.7, Pr(B) = x and Pr(A ∩ B) = 0.49
c Pr(A) = x, Pr(B) = 0.8 and Pr(A ∩ B) = 0.32
d Pr(A) = 0, Pr(B) = 0.5 and Pr(A ∩ B) = x
e Pr(A) = 0.375, Pr(B) = x and Pr(A ∩ B) = 0.225.
2 WE30 Show that if Pr(A) = 0.6, Pr(B) = 0.25 and Pr(A ∪ B) = 0.7, then A and B are independent.
3 Two coins are tossed.
a List the event space.
b Show that the two events ‘Heads with the first coin’ and ‘Tails with the second coin’ are
independent.
4 A coin is tossed twice. If A is the event ‘Heads with the first toss’ and B is the event ‘two Heads’,
decide if the two events are independent.
5 A coin is tossed and a die is rolled.
a What is the probability of getting Heads with the coin and a number greater than 2 with
the die?
b Establish if the events ‘Tails with the coin’ and ‘getting an even number with the die’ are
independent.
6 mC Pr(A) = 0.4 and Pr(B) = 0.5. If A and B are independent events, the value of Pr(A ∪ B) is:
a 0.5
B 0.7
C 0.4
d 0.9
E 0.8
A standard die coloured red and a standard die coloured blue are rolled. If A = ‘two odd
numbers’, B = ‘a 1 or a 5 with the first die’ and C = ‘the sum of the two numbers is less than 4’:
a calculate Pr(A), Pr(B) and Pr(C )
b decide whether each of AB, AC and BC are independent.
7 WE31
8 mC Two coins are tossed and a die is rolled. The probability that there are less than two Heads and
the number showing uppermost on the die is a 2 or a 5 is:
a
3
4
B
1
3
C
2
3
d
1
24
E
1
4
9 mC The probabilities of Anna, Bianca and Celia passing a Geography test are 0.75, 0.5 and 0.6
respectively. The probability that only two girls will pass the next Geography test is:
a 0.65
B 0.275
C 0.45
d 0.14
E 0.15
10 mC A die is biased so that the probability of rolling a 1, 2, 3, 4, 5 or 6 is 0.25, 0.2, 0.1, 0.1, 0.15
or 0.2 respectively. If the die is rolled twice, the probability that the sum of the two numbers rolled is
greater than 9 is:
a 0.2255
B 0.4355
C 0.5650
d 0.1625
E 0.1255
11 WE32 A box contains 6 red marbles and 4 blue marbles. One marble is randomly drawn, its colour
is noted and the marble is put back in the box. This procedure is done two more times. Represent the
information as a tree diagram and calculate the probability of getting:
a three red marbles
b two red marbles and one blue marble in any order
c three red marbles or three blue marbles.
12
One card is randomly drawn from a standard deck of 52 cards, then the card is replaced and a
second card randomly chosen. Determine the probability that:
a both cards are aces
b both cards are spades
c the two cards are different colours.
13 A die is biased so that the probability of obtaining the numbers 1, 2, 3, 4, 5 or 6 is 0.1, 0.3, 0.1, 0.2, 0.2
and 0.1 respectively. If the die is rolled twice, find the probability of rolling:
a two 6s
b an odd number followed by an even number
c two numbers that sum to 4
d two numbers whose sum is greater than 10.
ChapTEr 11 • Introductory probability
511
14 A Krisp-O cereal box contains an action card of a famous sports player selected from 10 cricket stars,
25 football identities and 15 tennis celebrities.
a What is the probability that a randomly selected box of Krisp-O contains a card of a cricket
player?
b What is the probability of finding cards of football players in each of two boxes randomly
selected from the supermarket shelf?
c What is the probability that three randomly selected packs of Krisp-O will provide cards of
different sports?
15 WE33 Three types of seedling (daisy, rose and orchid) have probabilities of surviving any one week as
0.9, 0.85 and 0.8 respectively.
a Represent this information on a tree diagram.
b Assuming the seedlings’ chance of survival are independent, calculate the probability that:
i after one week all three seedlings will have survived
ii more than one seedling will survive to the end of the first week.
16
Ibrahim estimates the probability of rain on Monday and Tuesday as 0.7 and 0.4 respectively.
Assuming that the events ‘rain on Monday’ and ‘rain on Tuesday’ are independent:
a represent this information on a tree diagram
b calculate the probability of rain on both days
c calculate the probability of no rain on at least one of the two days.
17 A school junior swim team has five Year 7 students and seven Year 8 students. A merit certificate is to
be awarded to one student from Year 7 and to one student from Year 8.
a If each student has an equal chance of selection, state the probability of a particular student from
Year 7 and a particular student from Year 8 receiving the award.
b If a special achievement certificate is also to be given to one of the students, what is the
probability that a particular student wins two awards, assuming that each student has the same
chance of being selected?
18 To open a combination lock involves using the correct sequence of
three digits selected from 0, 1, 2, 3, 4, 5, 6, 7, 8 and
9, where a digit may be used more than once. If each
digit is randomly selected:
a calculate the probability of success after
one try
b calculate the probability of success after
one try given that the three digits are known to
be odd
c calculate the probability of success after one try
given that the first and third digits are known to be the same.
19 A leather bag contains 4 black beads, 3 red beads and 3 white beads. Inside a plastic bag are
5 black beads, 2 red beads and 3 white beads. A nylon bag contains 6 black beads, 1 red bead and
3 white beads.
One bead is randomly withdrawn from each bag. What is the probability of getting:
a three black beads?
b a red bead from the leather bag but not a red bead from the plastic bag?
c at least two white beads?
diGiTal doC
doc-9809
SkillSHEET 11.2
Sampling without
replacement
20
One letter is randomly selected from each of the words HOORAY FOR MATHS. Find the
probability of getting:
a three vowels
b at least two consonants
c at least one vowel which is not the letter O.
21 At Greengate Secondary College, the probability of a VCE student proceeding to university studies is
50%, the probability of enrolling in TAFE courses is 20%, and there is a 30% probability of finding
employment immediately after completion of the VCE. For a group of four randomly selected VCE
students, what is the probability that:
a all of them will undertake university studies?
b all will seek employment or all will undertake TAFE courses?
512
Maths Quest 11 Mathematical Methods CAS
11i
Simulation
Simulation methods are used to model events when direct investigations may not be possible or practical
because of factors such as insufficient time, possible danger or the expense involved. The aim of such
methods is to obtain results comparable to the outcomes that would have been observed if the event had been
examined directly. For example, the possible effects of air resistance on the structure and stability of a newly
designed aeroplane may be investigated by performing wind-tunnel experiments using a scale model of the
aircraft. On a larger scale, the economic implications arising from interest rate changes may be investigated
using mathematical models that will take into account relevant variables such as unemployment and the cost
of living. Altering the value and type of variable will provide a range of predicted outcomes.
Many basic simulation techniques involve the generation of random numbers. Methods used include
coins, dice, cards, spinners, random number tables, calculators and computers. A CAS calculator can be
especially useful.
diGiTal doC
doc-9810
Extension
Sampling without
replacement
WorkEd ExamplE 34
Pina estimates her chance of passing an English test as 23, and must take 5 tests during the year.
a Explain how a CAS calculator may be used to simulate the 5 tests.
b Carry out a simulation to obtain an estimate of Pina’s performance on the 5 tests.
Think
a 1 Define the event as shown.
2
State the relevant probabilities.
3
Decide which numbers (to be generated on a
CAS calculator) will represent the event space
and which will represent favourable outcomes.
Use the random number generator feature
of the CAS calculator to generate 5 random
numbers from the set {1, 2, 3}.
Note that the results will vary every time.
b 1
2
3
Three of the 5 outcomes (1, 2 and 2) represent
passing the test.
WriTE/diSplaY
a Let P be the event ‘passing the test’.
Pr(P) = 2 and Pr(P′) = 1
3
3
Let the numbers 1, 2, 3 represent the event
space, 1 and 2 represent passing the test and
3 represent failing the test.
b randInt(1, 3, 5)
{3, 3, 1, 2, 2} is a possible set of 5 random
numbers.
The simulation predicts Pina will pass 3 of the
5 tests.
The results can be summarised as follows.
Trial
1
2
3
4
5
Random
number
3
3
1
2
2
Test
outcome
Fail
Fail
Pass
Pass
Pass
WorkEd ExamplE 35
The probability of a Jonathan apple tree producing fruit in any one season is 56 and the
probability of a Granny Smith apple tree bearing fruit in a season is 34.
a Assuming the two events are independent, calculate the probability that during a particular
season:
i both trees will produce fruit ii both trees will bear no fruit
iii only one of the trees will bear fruit.
b Devise a suitable simulation model consisting of 10 trials for each tree to obtain estimates
for the probabilities obtained in a.
ChapTEr 11 • Introductory probability
513
WriTE
Think
a 1 Define the events as shown.
2
a Let
J = Jonathan bears fruit
G = Granny Smith bears fruit
Pr(J ) = 5, Pr(G) = 3
State the given probabilities.
6
i Find Pr(Jonathan and Granny Smith bear
fruit). Recall ∩ means ‘and’.
Since the events are independent,
Pr(J ∩ G) = Pr(J) × Pr(G).
ii Find Pr(both trees bear no fruit).
4
=
1
6
1
24
3
4
ii Pr(J′ ∩ G′) = ×
Note:
Pr(J ′ ) = 1 − Pr(J ) = 1 − 5 = 1 and
6
5
6
5
8
i Pr(J ∩ G) = ×
=
1
4
6
Pr(G′ ) = 1 − Pr(G) = 1 − 3 = 1
4
4
iii Consider all situations in which only
one of the trees bears fruit and add the
probabilities (recall ‘or’ means + ).
iii Pr(one tree bears fruit)
= Pr(J ∩ G′) + Pr(J′ ∩ G)
= 56 × 14 + 16 × 34
= 13
b 1 Decide on a method of simulation and decide
2
3
which numbers will represent the event
space and which will represent favourable
outcomes. There are 6 possible outcomes
when a die is rolled: 1, 2, 3, 4, 5, 6.
There are 4 possible outcomes when two
coins are tossed: HH, TT, HT, TH.
Roll a die 10 times and record the results.
Possible results are shown in the table.
Alternatively, use a CAS calculator to
generate 10 random numbers between
1 and 6 by entering (1, 6, 10) in randInt.
b Let rolling a 1, 2, 3, 4 or 5 on a die represent the
Jonathan bearing fruit.
Let the outcomes HH, TT and HT when two
coins are tossed represent the Granny Smith
bearing fruit.
Trial Die
3
1
4
2
3
1
4
2
5
6
6
5
7
3
8
6
1
9
1
10
✓ = Fruit × = No fruit
Coin Jonathan Granny Smith
TH
✕
✓
HT
✓
✓
TT
✓
✓
HH
✓
✓
TH
✕
✕
HT
✓
✓
TH
✕
✓
HH
✓
✕
HT
✓
✓
HH
✓
✓
Toss two coins 10 times and record the results.
Possible results are shown in the table.
Alternatively, use a graphics calculator to
generate 10 random numbers between 1 and 4
by entering (1, 4, 10) in randInt.
i Trials 2, 3, 4, 6, 9 and 10 (6 trials)
correspond to both trees bearing fruit.
ii Only trial 5 corresponds to both trees
bearing no fruit.
iii Trials 1, 7 and 8 correspond to only one
tree bearing fruit.
i Estimated probability that both trees bear
6
fruit = 10
ii Estimated probability that both trees bear no
fruit =
1
10
iii Estimated probability that only one tree bears
3
fruit = 10
Compare the calculated and simulated probabilities in worked example 35. Repeat the simulation to
obtain a new set of results. How do they compare?
514
Maths Quest 11 Mathematical Methods CAS
Exercise 11i
Simulation
1 WE34 A student estimates the probability of stopping at a particular set of traffic lights when being
driven to school in the morning is 1.
5
a Explain how a graphics calculator can be used to simulate 5 trips to school.
b Carry out 5 simulations to obtain an estimate for the probability of stopping at the intersection.
2 The table below shows the number of bullseyes scored by 40 dart players after 5 throws each.
a
b
c
d
1
4
0
3
4
2
1
5
4
2
3
0
4
5
2
1
4
3
2
1
0
2
1
4
3
5
3
2
4
4
0
2
1
0
3
5
4
2
3
1
Explain how a die or a CAS calculator can be used to obtain the range of numbers given in the
table.
What proportion of players scored at least 3 bullseyes?
Using a die (or by some other means), conduct 20 trials and obtain a second value for b.
Analysis of the results of a particular tournament (at which each player threw 3 darts) gives
the probability of scoring more than 2 bullseyes to be twice the probability of scoring less than
3 bullseyes.
i Explain how a die or a CAS calculator can be used to simulate this situation.
ii Perform 40 trials and compare your results to the given probabilities.
diGiTal doC
doc-9811
random numbers
3 A student generated 30 three-digit random numbers using a calculator. The results are given in the table
below.
200
123
399
165
100
355
778
400
150
100
387
001
793
215
030
288
345
009
970
993
546
720
549
139
248
405
369
217
935
782
a Explain how you can obtain 30 random 2-digit numbers from the table to simulate the ages of 30
people aged from 10 years to 100 years.
b Using the method you suggested for a, obtain an estimate of the proportion of the group that is
younger than 50 years of age.
4 WE35 A car may be said to be safe if both of its airbags will operate properly in the event of a
collision. Suppose that the probability of the driver’s airbag failing is 1 and the probability of the front
4
passenger’s airbag not working is 1. Use one die and two coins to simulate the situation of the airbags
3
working/not working.
a Assuming the operation of the airbags to be
independent, find the probability that during a
collision:
i both airbags operate properly
ii neither airbag operates properly
iii only one airbag operates properly.
b Using a die and two coins, devise a simulation
model consisting of 20 trials to obtain an
estimate for the probabilities obtained in a.
5 A die and a coin are each tossed 20 times with the following results.
Die
1 3 2 4 6 5 4 3 2 3 1 6 4 6 2 1 4 2 2 1
Coin H H T H T T T H H T H T H T H T H T H H
H is Heads, T is Tails.
a Explain how the events of tossing a coin and rolling a die can be used to simulate the situation of
forming a particular dancing couple consisting of a man randomly chosen from a group of 2 men
and a woman randomly selected from a group of 3 women.
b Perform 20 simulations to obtain an estimate of the probability of a particular man and a
particular woman dancing together.
6 Each packet of Krisp potato chips contains 1 of 5 different fridge magnets. Use 20 trials for a simulation
analysis to determine how many packets of Krisps need to be purchased in order to obtain all 5 magnets.
ChapTEr 11 • Introductory probability
515
Summary
introduction to
experimental
probability
Sets and Venn diagrams:
• { } A set is a collection of objects (or elements).
• x ∈ A denotes that element x belongs to set A.
• x ∉ A denotes that element x does not belong to set A.
• A finite set can be listed; an infinite set cannot be listed.
• A null set (denoted ϕ) contains no elements.
• A unit set contains one element.
• n(A) The cardinal number of a set is the number of elements it contains.
• ξ The universal set is a set containing all elements being considered.
• A′ The complement of set A is all the elements of the universal set not contained in set A.
• A = B Two sets are equal if they contain the same elements.
• A ↔ B Two sets are equivalent if they have the same cardinal number.
• A ∩ B The intersection of two sets is the set of elements common to both sets.
• A ∩ B = ϕ Two sets are disjoint sets if they have no elements in common.
• A ∪ B The union of two sets A, B is the set that contains all elements belonging to A or B or to
both A and B.
• n(A ∪ B) = n(A) + n(B) − n(A ∩ B)
• A ⊂ B Set A is a subset of set B if all the elements of set A are contained in set B.
• A ⊃ B Set A contains set B if set A contains all the elements of set B.
• Properties of sets can be represented as Venn diagrams.
Using sets to solve practical problems:
• Use overlapping regions (for example, circles) to represent sets within a universal set (for example,
a rectangle).
• Label each set.
• Fill in any given information.
• Calculate required missing quantities.
Estimated probability and expected number of outcomes:
number of favourable outcomes observed
total number of trials
• Expected number of favourable outcomes
= experimental probability (long-run proportion) × number of trials
• Experimental probability =
Calculating
probabilities
• Pr(favourable outcome) =
• For event E, Pr(E ) =
•
•
•
•
•
•
number of favourable outcomes
total number of outcomes
n(E )
.
n(ξ )
Pr(E ) + Pr(E ′) = 1
Pr(E ′) = 1 − Pr(E )
If event E is impossible, Pr(E ) = 0.
If event E is certain to occur, Pr(E ) = 1.
0 ≤ Pr(E ) ≤ 1
Pr(ξ) = 1 and Pr(ϕ) = 0
Tree diagrams and
lattice diagrams
• Outcomes can be illustrated by a tree diagram. The order of events determines the structure of the ‘tree’.
• A lattice diagram is a grid used to show the possible outcomes when two events occur. It is
particularly useful when dealing with outcomes from rolling a die.
The addition law of
probabilities
•
•
•
•
516
For events A, B, Pr(A ∪ B) = Pr(A) + Pr(B) − Pr(A ∩ B).
∪ means ‘or’ and ∩ means ‘and’.
Two events are mutually exclusive if they cannot occur at the same time.
If A ∩ B = ϕ, then A, B are mutually exclusive and Pr(A ∩ B) = 0.
Maths Quest 11 Mathematical Methods CAS
karnaugh maps and
probability tables
• Karnaugh maps and probability tables summarise all combinations of two
events (for example, A and B) and their complements (A′ and B′ ).
• Use known values, subtotals and the fact that the sum of the probabilities
in the outer row and column is equal to 1 to fill in any missing values.
B
B'
A
A'
1
Conditional probability
• For two events A and B, the conditional probability of event A given event B occurs is
Pr(A ∩ B)
Pr(A | B) =
, where Pr(B) ≠ 0 and B is the reduced event space.
n(ξ )
• Venn diagrams, tree diagrams and Karnaugh maps are useful aids in conditional probability
problems.
• The Law of Total Probability: Pr(A) = Pr(A | B) × Pr(B) + Pr(A | B′) × Pr(B′)
Transition matrices
and markov chains
• In general, Sn + 1 = T × Sn may be used to determine any state from the previous one.
• In general, Sn = T n × S0.
independent events
• If Pr(A ∩ B) = Pr(A) × Pr(B), then events A and B are independent.
• Pr(A ∩ B) means the probability of events A and B.
Simulation
• Simulation techniques are used to model events.
• Outcomes of the event space are randomly obtained but based on certain probabilities.
• Coins, dice, random number tables and calculators can provide random numbers that can be
associated with outcomes of the event space.
• Performing more trials or repeating the simulation will provide better estimates of the probability.
ChapTEr 11 • Introductory probability
517
Chapter review
1 A random sampling of 80 ceramic tiles produced at a ceramics factory reveals 8 scratched, 3 chipped
2
$50
$100 Bo
o
Pri dy
ze
$200
se
Lo 00
$1
3
and 4 broken tiles. Estimate the probability that a tile produced at the factory will be:
a scratched
b chipped or broken
c damaged in some way.
A card is randomly selected from a deck, its suit is noted, and then the card is placed back in the deck.
The experiment is repeated to obtain a second card.
a List the possible outcomes for selecting 2 cards in this way.
b Are the outcomes all equally likely? Explain.
$1000 $
$10 Jackpot 60
A game show host spins the wheel shown at right. What is the probability that
the wheel ends on:
a the jackpot ($1000 prize)?
b a prize greater than $50?
A letter is chosen at random from each of the words GO BLUES. Represent all
possible outcomes on a tree diagram and find the probability that:
a G and B are chosen
b S is chosen
c G or S is chosen.
A standard die is thrown and the spinner shown at right is spun.
1
a Show all possible outcomes on a lattice diagram.
3
b Find the probability of getting a number greater than 4 on the die and
2
an odd number on the spinner.
A set of 20 uniformly sized cards numbered 1 to 20 is shuffled. What is the probability of drawing a
number less than 8 or an even number from this set?
A class of 30 students was asked if there was a pet dog at home and if the students were responsible for
pooper scooping before the backyard lawn was mown. Fourteen students had a dog but only 6 did the
pooper scooping.
a Draw a Karnaugh map showing this information.
b Complete a probability table.
c State the probability that a randomly selected student has a dog but avoids pooper scooping.
If Pr(A) = 0.3 and Pr(B | A) = 0.4, find Pr(A ∩ B).
Two identical, equally accessible cookie jars sit on a kitchen bench. Jar 1 contains 6 chocolate and
9 plain biscuits, and jar 2 contains 12 chocolate and 8 plain biscuits. One biscuit is selected randomly
from one of the jars. If a chocolate biscuit is selected, what is the probability that it came from jar 1?
Of 50 people surveyed, 35 played tennis and 26 played netball. Everyone surveyed played at least
one of these sports.
a How many people played both netball and tennis?
b If one person is selected at random, what is the probability that:
i he/she plays tennis only?
ii he/she plays netball?
iii he/she plays tennis, given that he/she also plays netball?
Lo
it a se
ll
S h orT
anS WEr
7
8
9
10
mUlT ip l E
Ch oiCE
$30
6
0 $500
5
$4
4
1 Twelve nuts are taken from a jar containing macadamias and cashews. If 3 macadamias are obtained,
the estimated probability of obtaining a cashew is:
a
1
12
B
1
4
C
1
3
d
3
4
E
3
1
2 From a normal pack of 52 playing cards, one card is randomly drawn and replaced. If this is done
208 times, the number of red or picture cards (J, Q or K) expected to turn up is:
a 150
B 130
C 120
d 160
E 128
3 A cubic die with faces numbered 2, 3, 4, 5, 6 and 6 is rolled. The probability of rolling an even number is:
a
518
1
3
B
2
3
Maths Quest 11 Mathematical Methods CAS
C
1
6
d
5
6
E
1
2
4 The probability of rolling an odd number or a multiple of 2 using the die in question 3 is:
a 1
B
1
3
1
4
C
d
3
4
E
2
3
5 The tree diagram that describes the outcomes when three coins are tossed is:
a
T
H
d
TH
TH
TH
T
H
H
T
TH
H
T
H
T
H
B
H
T
H
T
H
T
C
H
T
T
HT
E
TH
H
T
H
T
H
T
H
T
H
T
H
T
HT
6 Bag A contains 3 red and 4 blue marbles, and
Bag B contains 3 yellow and 2 green marbles, as
shown at right. A marble is drawn from Bag A,
then one is taken from Bag B. Which diagram
below best illustrates this situation?
Bag A
a
Yellow RY
Red
Blue
C
Red
Green
B
RG
Red YR
Yellow
Blue YB
Red GR
Yellow BY
Green
Bag B
Green
BG
Blue
RB
Green
RG
d
Blue GB
Red
Blue
Yellow
Yellow
E
Red
Blue
Yellow
Green
Blue
YB
Green
YG
Red
Blue
Yellow
Green
Red
Blue
Yellow
Green
Red
Blue
Yellow
Green
Red
Blue
Yellow
Green
Green
RR
RB
RY
RG
BR
BB
BY
BG
YR
YB
YY
YG
GR
GB
GY
GG
ChapTEr 11 • Introductory probability
519
7 If Pr(A) = 0.6, Pr(B) = 0.7 and Pr(A ∪ B) = 0.8, then Pr(A ∩ B) is:
a
B
C
d
E
0.1
0.5
0.9
0.2
0.6
8 Consider the Karnaugh map at right, where C is ‘people who like comedy
movies’ and A is ‘people who like action movies’.
The number of people who like only action movies is:
a 26
B 30
C 18
d 20
E 23
A
A′
19
C
25
C′
26
50
9 Which of the following alternatives gives the correct values of a and b in
the probability table at right?
a a = 0.2, b = 0.3
B a = 0.3, b = 0.2
C a = 0.3, b = 0.6
d a = 0.6, b = 0.3
E a = 0.7, b = 0.4
B
B′
A
0.4
0.7
A′
a
b
0.4
10 If Pr(B | A) = 0.45 and Pr(A ∩ B) = 0.35, then Pr(A) is:
a
7
9
B
3
8
C
4
9
d
5
8
E
8
9
11 A fair coin is tossed twice. If A = Tails on first toss, B = Heads on second toss and C = both tosses are
Heads, which of the following is true?
1
a Pr(A) =
4
B Pr(B) = Pr(C)
C A and B are independent.
d A and C are independent.
E B and C are independent.
3
12 An archer has a probability of approximately of hitting the bullseye from a particular distance. The
8
spinner at right is used to simulate 10 rounds of 4 shots at such a target, and the
results are as follows, where B = blue (bullseye) and R = red (non-bullseye).
B, R, R, R
B, R, B, R
R, B, R, R
R, B, R, R
R, B, B, R
R, R, B, B
R, B, R, R
R, R, R, R
R, B, B, R
B, R, B, B
Based on this simulation, the probability of getting 2 bullseyes in a round of 4 shots is:
a
E x TEndEd
r E S p onS E
520
1
4
B
3
8
C
2
5
d
5
8
E
3
4




1 For a transition matrix T =  0.22 0.33  and an initial state matrix S0 =  0.5  , calculate, accurate
0.78 0.67
0.5




to 3 decimal places:
a S1
b S4.
2 The manager of a snow resort has noticed that, if it snows on a given day, there is a 70% chance that
it will snow the following day. If it does not snow, there is only a 30% chance that it will snow the
following day. John arrived on a Saturday when the weather was sunny and clear.
a What is the probability that he will have fresh snow the following Tuesday?
b What is the probability that he will have fine weather for the drive home on the following
Saturday? (Give answers to 3 decimal places.)
3 A factory has a machine in poor working condition that often produces faulty components. If it
produces a faulty component, there is a probability of 0.25 that it will follow this up with another faulty
component. However, each time it produces a good component, there is a probability of only 0.05 that
Maths Quest 11 Mathematical Methods CAS
it will next produce a faulty component. If there is a 20% chance that the first component of the day is
faulty, set up the initial state matrix and find the probability that:
a the second component is faulty
b the fifth component is faulty.
4 One student is selected at random from each of Years 7, 8 and 9. If there are 148 girls and 114 boys in
Year 7, 126 girls and 97 boys in Year 8, and 115 girls and 122 boys in Year 9, find the probability that
all students chosen are boys. Give your answer to 3 decimal places.
5
9 7
that he arrives at work to find a free parking space in his company’s car park is 10. Assuming these
5 The probability that the newspaper is delivered to Geoff’s house before 8.00 am is , and the probability
events are independent:
a use a CAS calculator to conduct a simulation of 28 days’ duration
b find the probability, based on the simulation, that on one day Geoff misses out on a company car
park space and his paper arrives late
c calculate the theoretical probability of the combination of events described in part b.
6 Consider 3 fair coins being tossed.
If A = Heads with the first coin, B = Tails with the
second coin and C = Tails with the third coin:
a list the event space (for example, use HTH for
‘Head then Tail then Head’)
b find:
i Pr(A)
ii Pr(B)
iii Pr(C)
iv Pr(A ∩ B)
v Pr(A ∩ C)
vi Pr(B ∩ C)
vii Pr(A ∩ B ∩ C)
viii Pr(A) × Pr(B) × Pr(C)
c propose how you might define independence for 3 events.
ChapTEr 11 • Introductory probability
521
7 A large number of asthma sufferers were asked to volunteer for the testing of a new drug. Only some
of the volunteers were given the drug, but all of the volunteers were observed to see if they developed
asthma on a smoggy day. The results are shown in the table below.
Developed asthma
Did not develop
asthma
Given drug
148
Not given drug
59
566
184
a How many people were selected to take part in the study?
b If a person was randomly selected from the volunteers, what is the probability that they were
given the drug?
c What is the probability that a randomly selected person developed asthma on this day?
d Given that a volunteer was given the drug, what is the probability that they developed asthma?
e From this information, what can you conclude about the effectiveness of the drug in preventing
diGiTal doC
doc-9812
Test Yourself
Chapter 11
522
asthma?
It was decided that conclusive observations about the effectiveness of the drug could not be made
after one day, so the same volunteers continued with the study for three months. (Assume that the
number of people given the drug is unchanged.) The results were:
395 people were given the drug and did not have an asthma attack
143 people were given the drug but had exactly one episode of asthma per month
97 people were not given the drug and developed asthma more than once a month
84 people were not given the drug and had exactly one episode of asthma each month.
f Represent this information using a Venn diagram.
g How many people were given the drug and had more than one episode of asthma per month?
h How many people in the study were not given the drug and did not have an episode of asthma?
i Given that a volunteer had been given the drug, what is the probability that they have more than
one episode of asthma per month?
j Given that a volunteer had more than one episode of asthma per month, what is the probability
they had taken the drug?
Maths Quest 11 Mathematical Methods CAS
ICT activities
Chapter opener
diGiTal doC
• 10 Quick Questions doc-9801: Warm-up with ten quick questions on
introductory probability (page 475)
11a
introduction to experimental probability
diGiTal doCS
• doc-9802: Investigate long-run proportion using a spreadsheet
(page 475)
• doc-9803: Simulate the rolling of a die multiple times using a
spreadsheet (page 475)
• doc-9804: Simulate the rolling of two dice multiple times using a
spreadsheet (page 475)
• WorkSHEET 11.1 doc-9805: Calculate probabilities for everyday
events and random selection (page 478)
11B
Calculating probabilities
diGiTal doC
• Extension doc-9810: Investigate sets and Venn diagrams (page 478)
11C
Tree diagrams and lattice diagrams
TUTorial
• WE 9 eles-1448: Assuming replacement, use a tree diagram to
calculate the probabilities of drawing two specific cards (page 483)
diGiTal doC
• doc-9806: Investigate Stirling’s formula using a spreadsheet
(page 485)
11d
The addition law of probabilities
TUTorial
• WE 15 eles-1449: Use the addition law of probabilities to
calculate probabilities and determine whether two events are
mutually exclusive (page 488)
diGiTal doC
• WorkSHEET 11.2 doc-9807: Use Venn diagrams to help calculate
probabilities (page 492)
11E
11F
Conditional probability
TUTorial
• WE 24 eles-1451: Use a tree diagram to calculate a conditional
probability (page 499)
diGiTal doC
• SkillSHEET 11.1 doc-9808: Practise conditional probability
(page 500)
11G
Transition matrices and markov chains
inTEraCTiViTY
• Transition matrices int-0270: Consolidate your understanding of
transition matrices (page 501)
TUTorial
• WE 25 eles-1452: Represent conditional probabilities of two
events in a table (page 503)
11h
independent events
TUTorial
• WE 33 eles-1453: Using a tree diagram, calculate probabilities
assuming three events are independent (page 510)
diGiTal doC
• SkillSHEET 11.2 doc-9809: Practise sampling without replacement
(page 512)
11i
Simulation
diGiTal doCS
• Extension doc-9810: Investigate sampling without replacement
(page 513)
• doc-9811: Investigate random numbers using a spreadsheet
(page 515)
Chapter review
diGiTal doC
• Test Yourself doc-9812: take the end-of-chapter test to test your
progress (page 522)
To access eBookPLUS activities, log on to www.jacplus.com.au
karnaugh maps and probability tables
TUTorial
• WE 18 eles-1450: Complete a probability table and use it to
represent the information in a Venn diagram (page 493)
ChapTEr 11 • Introductory probability
523
Answers CHAPTER 11
2 a
Red
Head
White HW
Exercise 11a
Red
Tail
3 a
A
T
5 a
2
3
4
5
6
Calculating probabilities
a ξ = {Red, Blue, Yellow, Green}
b n(ξ) = 4
ξ = {A, B, C, D, E, F, G, . . . , X, Y, Z}
n(ξ) = 52
a ξ = {spades H, spades T, clubs H,
clubs T, hearts H, hearts T, diamonds H,
diamonds T}
b S = {spades H, spades T}
ξ = {HH, HT, TH, TT}
a ξ = {boy Time Out, boy Mars Bar, boy
Violet Crumble, girl Time Out, girl Mars
Bar, girl Violet Crumble}
b M = {boy Mars Bar, girl Mars Bar}
1
6
7 a 7
1
1
13
10
1
b
1
154
b
1
4
c
73
154
c
3
13
1
b 6
1
T
C
AT
TC
A
TA
H
HH
T
HT
H
T
TH
TT
Red
Blue
Red
Blue
Red
Blue
Red
Blue
Red
Blue
2
4
6
8
10
8
4
5
6
4
5
6
2
3
1
b 5
5
1
c 4
d 2
24
25
26
34
35
36
b 11
T1
G
B
3
R
3
7
8
16 a 17
14
12
b 19
c 19
1
c 0
b 85
d 19
T2
diagrams
B
B
R
G
BB
BR
BG
RB
RR
RG
GB
GR
GG
b
n(ξ ) = 9
i
1
9
ii
1
9
iii
2
9
G
B
Tree diagrams and lattice
B
R
G
B
R
G
B
R
G
G
R
T3
Exercise 11C
V
C
L
C
S
C
B
C
13
Bjorn
1
J2
T1
T2
T3
T4
J3
T1
T2
T3
T4
R
T4
G
B
L
F
L
F
L
F
L
F
L
F
L
F
L
F
L
F
L
F
L
F
L
F
L
F
S
T
S
H
S
S
S
1
1
b 4
MSF
MSN
TSF
TSN
HSF
HSN
SSF
SSN
1
3
1
c 4
17 D
18 a
Outcomes
n(ξ) = 12
J1 is the
first jacket,
T1 is the
first tie.
H
H
T
T
H
T
H
T
H
T
H
T
H
T
H
T
H
T
H
T
H
T
H
T
H
T
H
T
H
T
1
6
n(ξ ) = 6
b 4
16 8
1
12
n(ξ) = 12
1
3
F
N
F
N
F
N
F
N
15 a 4
1
2
n(ξ ) = 24
T1 is the first small triangle.
R is red, G is green, B is blue.
L is low sheen.
F is full gloss.
Maths Quest 11 Mathematical Methods CAS
M
1
b 6
c
T1
T2
T3
T4
Tail BT
Head CH
Tail CT
Carl
14 a
Outcomes
S VCS
L VCL
B VCB
V LCV
S LCS
B LCB
V SCV
L SCL
B SCB
V BCV
S BCS
L BCL
Head AH
Tail AT
Head BH
Alan
1
4
3
10
n(ξ ) = 6
=8
c 2
1
= 10
HHH
HHT
HTH
HTT n(ξ )
THH
THT
TTH
TTT
1
b 8
b 2
Outcomes
R
Tail
12
c
Tail
1
b 6
1
10
14 21 = 7
15 a 19
1
Head
Tail
Head
Tail
Head
Tail
Head
Tail
Head
Head
Tail
a 3
9
4
13 a 11
n(ξ ) = 4
1
4
2R
2B
4R
4B
6R n(ξ )
6B
8R
8B
10R
10B
d 1
c 2
4
12 a 2
524
CT
AC
J1
1
11 a 2
1 a
T
C
b
1
40 000
9
CA
e 1
d 13
10
6 B
7
Head
TR
A
1
4
b 7
8a
9 a
H
T
Exercise 11B
11 a
White TW
C
4 a
4
19 a
HHHH
HHHT
HHTH
HHTT
HTHH
HTHT
HTTH
HTTT
THHH
THHT
THTH
THTT
TTHH
TTHT
TTTH
TTTT
3
b 16
Die 2
introduction to
experimental probability
1 0.6
2 0.7
3 a 0.66
b 0.34
4 48
5 250
6 200
7 A
8 C
9 18 squares, 24 circles
10 a 120
b 140
11 E
12 249 (Hot-Shot), 401 (Zap Inc)
13 10
14 a i 72 ii 24 iii 12
b 4 losses or draws
15 A
16 a 0.7597
b 0.2403
c i 1200 ii 380
1
b 1
HR
1
c 8
d 2
1
b 4
6
5
4
3
2
1
c
1
9
1 2 34 5 6
Die 1
20 a
Coin
inTrodUCTorY proBaBiliTY
T
H
1 2 3 4 5 6
Die
1
1
b 12
1
21 a 36
1
d 9
c 4
1
b 18
1
e 4
5
c 6
Exercise 11E karnaugh maps and
probability tables
22 1
6
23
E1
E2
E3
D1
D2
D1
D2
D1
D2
D1
D2
D1
D2
D1
D2
M1
M2
M1
M2
M1
M2
1
6
1 a
E1 is the first entrée.
M1 is the first main meal.
D1 is the first dessert.
n(ξ ) = 12
Exercise 11d
10 a i
B′
A
8
17
25
A′
7
6
13
15
23
38
A
ii
iii
A′
17
8
3
52
3
2
ii 5
12 a 10
b 2
1
13 a
1
1
2
3
b
A
69
33
102
A′
45
27
72
114
60
174
2
33
69
2
27
d 52
3
27
17 a 80
b 8
1
14
1
7
18 a
c
1
b
2
B′
A
0.27
0.3
A′
0.4
0.03 0.43
12
0.67 0.33
c 65
19 a
b
13
A
B
3
4
29
40
0.27
d
B
B′
A
0.61 0.03 0.64
b
23
40
A′
0.14 0.22 0.36
0.75 0.25
30
A
32
10
20
2
ii 5
9
iv 10
b The areas are stated to be equal to
ensure that each area has an equal
chance of being hit, that is, to ensure
equally likely outcomes.
B′
A
7
15
22
A′
18
5
23
25
20
45
B
B′
0.27 0.37 0.64
0.09 0.27 0.36
A′
B
1
B′
A
0.15 0.35
0.5
A′
0.25 0.25
0.5
0.6
1
7 E
8 a ξ
F
S
24
1
6
0.03
b
0.61
20
2 C
3
S
S′
F
6
24
30
F′
20
10
30
26
34
60
0.14
0.22
1
10
A
B
3
iii 2
4
10
24 10
1
16
0
ξ
or Clotto is 100 , 100 and 100 respectively, so
Alotto is easier to win.
23 0.75
25 a i 10
10
0.4
d
d 0
33
6
0.4
4
5
22 The probability of winning Alotto, Blotto
B′
B
1
1
4
0.03
b 5
c 1
21 a
0.3
5
8
2
7
0.12
A
b
d 24
20 a 10
0
A′
6 a
B
11
c 24
A′
ξ
d 14
19
24
0.61 0.27 0.88
0.36 0.64
9
c 7
0.57
B′
0.12
d
45
B
1
A
c
27
b 4
c 13
A′
0.41
0.11
0.73 0.27
2
1
0.59
0.3
B
13
20
15 5
16 a 13
c
A
B′
0.45 0.14
B
ξ
1
c
14 3
100
164
0.75 0.25
B′
d 2
3
10
44
73
7
B
11
26
iv 1
c 20
56
A
6
b
64
91
b
iv
iii 0
B′
29
B
b No
11 a i 5
b Yes
B
35
5a
B
The addition law of
3
13
A
ξ
probabilities
1 0.7
2 0.68
3 0.89
4 0.3
5 0.45
6 0.41
7 0.78, 0.39
8 A and B are mutually exclusive.
9 A⊂B
1
4
B
4
9
B
B′
A
87
63
150
A′
13
55
68
100
118
218
F
F′
S
S′
0.1
0.4
0.5
0.33 0.17
0.5
0.43 0.57
1
F is football, S is soccer.
ChapTEr 11 • Introductory probability
525
10 A
11 a
Exercise 11F
A
B
B′
0.3
0.3
0.6
0.2
0.2
0.4
0.5
0.5
1
B
B′
A
0.1
0.1
0.2
A′
0.7
0.1
0.8
0.8
0.2
1
A′
b
c
4
35
2
6 0.168
7 3
9 a 0.3
b 7
11 A
10 D
Pr( B ∩ A) Pr( A) × Pr( B)
=
Pr( A)
Pr( A)
= Pr( B)
12 Pr( B | A) =
20
27
b
0.8
14 a 0.3
0.1
0.2
c
b 3
0.6
0.4
1
B
B′
A
0
1
4
1
4
A′
1
4
1
2
3
4
1
4
3
4
1
B′
A
0.5
0.3
A′
0.1
12
S′
C
50
110
160
C′
95
25
120
145
135
280
13 a
G
0.32 0.22 0.54
A′
0.44 0.02 0.46
3
1
0.4
A′
0.25 0.35
0.6
0.33 0.67
1
ii 0.25
B
iii 0.57
D is Daily Times,
B is Bugle.
ii 0.28
iii 0.54
0.46 0.54 1
are not independent. B and C are not
independent.
8 E
9 C
10 D
30
1
11
d 2
2
18 14
19 7
2
2
2
20 a 3
b 75
c 25
21 a 0.40
d 0.21
b 0.32
e 0.63
c 0.13
 0.75 0.5 
 0.25 0.5 
67 
3a  
33 
63.67 
c 
36.33 
1
380 
4a 
620 
R′ 10% 65% 75%
20% 80% 100%
2 0.31
63.3 
b 
36.7 
 443 
b 
557 
 0.411 0.368 125   0.411 
 0.589 0.631875   0.589 
 0.772 694 0.757 688   0.772 694 
0.242 313   0.227 306 
6
 0.227 306
 0.70 0.60 
0.40 
Maths Quest 11 Mathematical Methods CAS
b 0.432
1
 435 
7a 
 0.30
a 0.216
1
12 a 169
c 
565 
5
Second draw Third draw
0.6 Red
First draw
0.6 Red
0.4 Blue
Red
0.6 Red
0.6
0.4 Blue
0.4 Blue
0.6 Red
0.6 Red
0.4
0.4 Blue
Blue
0.6 Red
0.4 Blue
Blue
0.4
1
c 2
1
4
b A and B are not independent. A and C
Scuba
15
independent events
and Pr(A) × Pr(B) = 0.15, so A and B are
independent.
a {H1H2, H1T2, T1H2, T2T2}
1
b Pr(H1 ∩ T2) = and Pr(H1) × Pr(T2) =
4
Not independent
1
a
b Independent
3
B
1 1
1
a 4 , 3 , 12
11
B′
R 10% 15% 25%
b 0.15
c 0.75
6
7
Transition matrices and
markov chains
B′
D′ 0.28 0.12 0.4
B
5
19
b 30
Exercise 11G
D 0.18 0.42 0.6
b i 0.4
c 132
16 a
4
6
0.08 0.32
b i 0.35
3
22 13
B′
A
15 a
A is people
less than 20
years of age,
G is people
who wear
glasses.
ii 0.54
B
0.22
C is chocolate,
S is strawberry.
A
b i 0.02
14 a
0.3
15 a 10
16 A
17 a 15
b ξ Mountain
15
Exercise 11h
B
0.33
G′
0.76 0.24
ξ
A
Vegetable 333, tomato 167
Same answer as c
0.80
b 0.58
i 0.7500
ii 0.6880
iii 0.6875
iv 0.6875
b i 0.5500
ii 0.6864
iii 0.6875
iv 0.6875
c 0.6875
0.40 0.75 
10 a 
b 0.5365
 0.60 0.25 
c 0.4635
d 0.4444
c
d
8 a
9 a
1 a 0.2
b 0.7
c 0.4
d 0
e 0.6
2 Using the Addition Law, Pr(A ∩ B) = 0.15
2
0.15
S
5
8 6
3
13 a 0.2
B
d
526
Conditional probability
b 0.5
b 0.69
5
4 12
5 0.144
1 a 0.8
2 a 0.6
 200 
b 
300 
13 a 0.01
c 0.11
1
1
9
b 4
0.85
Survive
0.15
Not
survive
b i 0.612
16 a
Monday
0.7
0.3
b 0.28
c 0.72
c 50
Orchid
Rose
Daisy
0.1
c 2
b 0.24
d 0.05
14 a 5
15 a
0.9
c 0.28
1
b 16
0.85
0.15
Survive
0.8
0.2
0.8
Not
survive 0.2
0.8
Survive
0.2
0.8
Not
survive 0.2
ii 0.941
Tuesday
0.4 Rain
Rain
0.6 No rain
No 0.4 Rain
rain
No rain
0.6
Survive
Not survive
Survive
Not survive
Survive
Not survive
Survive
Not survive
1
1
18 a 1000
c
b 125
3
5 a, b Answers will vary.
1
6
1
3
21 a 0.0625
b 0.0097
P′
D
6
8
14
v 4
D′
0
16
16
vii 8
6
24
30
b
P
P′
D
3
15
4
15
7
15
D′
0
8
15
8
15
3
15
12
15
1
ChapTEr rEViEW
ShorT anSWEr
b
7
80
c
3
16
2 a LetH = hearts, D = diamonds,
S = spades and C = clubs.
Possible outcomes are:
HH DH SH CH
HD DD SD CD
HS DS SS CS
HC DC SC CC
b All outcomes are equally likely as there
is an equal number of cards in each suit
during each selection.
3 a
5
1
12
b 12
4
G
O
1
a 10
B
L
U
E
S
B
L
U
E
S
GB
GL
GU
GE
GS
OB
OL
OU
OE
OS
1
b 5
c
iii 2
P
Exercise 11i
Simulation
Answers will vary.
i 2
1
7a
11
1
b
7
6 10
b 15
1
10
TTH, TTT}
2
b 9
27
125
1
1 a
c 35
6 a {HHH, HHT, HTH, HTT, THH, THT,
1 2 3 4 5 6
Die
b 25
20 a 30
c
4 0.097
3
2
1
1
100
19 a 25
c
5 a
1
b 35
Spinner
1
17 a 35
1
1
b 238 ≈ 0.746
319
c 69 ≈ 0.216
319
74
d 357 ≈ 0.207
e Answers will vary. Make reference to
2
10 a 11
13
ii 25
11
iii 26
mUlTiplE ChoiCE
2E
5C
8D
11 C
the percentage who developed asthma
given the drug, compared with those not
given the drug (20.7% compared with
24.3%).
f ξ
Drug
395
3B
6A
9A
12 C
143
Asthma = 1
84
ExTEndEd rESponSE
0.275 
1 a 
3
c 5
 0.725 
2 a 0.468
b 0.501
0.2 
3 S0 
 0.8 
1
viii 8
7 a 957
9 5
1D
4A
7B
10 A
1
vi 4
Pr(C ) and A, B and C are ‘piecewise
independent’ (that is, AB, AC and BC
are all independent pairs of events), then
A, B and C are independent.
8 0.12
12
1
iv 4
c If Pr(A ∩ B ∩ C ) = Pr(A) × Pr(B) ×
4
15
b i 25
1
ii 2
176
Asthma > 1
97
62
 0.297 
b 
 0.703 
g 176
h 62
88
i 357 ≈ 0.246
a 0.09
b 0.063
176
j 273 ≈ 0.645
ChapTEr 11 • Introductory probability
527