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Essential Maths Skills
Subject Knowledge and Understanding for
QTS Students
Handling Data
Book C
Contents:
Chapter 11: Easy Probability
Chapter 12: Harder Probability
Pages
1 - 16
17 - 35
1
Chapter 11
EASY PROBABILITY
Introduction
I toss a coin ten times. It lands heads up ten times.
Is this result typical of what would usually happen when this coin is tossed ten times?
Has the coin landing heads up ten times been an accident or a freak occurrence? Has it
happened by chance?
How likely is this result if the coin was fair - that is the chance of it landing heads up is the
same as the chance of it landing tails up?
Probability allows us to give this likelihood a number .
When writing about a survey or experiment it is important to explain what the chance is that
the results you have obtained happened by accident.
In everyday life words and phrases like certainly, almost certainly, might, perhaps, almost
certainly not, certainly not, describe how likely an event is to happen. Sometimes numbers
are put to events.
You will have noticed that there are several ways of describing a probability
a) in words e.g. quite likely
b) using numbers
1) ratio e.g. one in four
2) fraction 1/4
3) percentage 25%
4) decimal 0.25
A probability of 1 or 100% is absolute certainty, the event is bound to happen. That it will
happen is inevitable.
When I toss a coin in the air, what are the chances of it coming down heads?
You will probably answer one of the following :
1;
50/50;
1 in 2;
evens;
0.5.
All of these answers are correct - they are different ways of saying the same thing.
In mathematics, probability is expressed as either a vulgar fraction or as a decimal fraction,
in this case either as
1
or
0.5
2
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We know that a fair coin is equally as likely to come down heads as it is tails. There are two
possible OUTCOMES (head or tail) to this TRIAL (tossing a coin). The EVENT we are
interested in is "landing heads".
A definition of probability
PROBABILITY = NUMBER OF WAYS THE EVENT CAN HAPPEN
TOTAL NUMBER OF POSSIBLE OUTCOMES
So, when a coin is tossed
P (head)
=
NUMBER OF WAYS OF GETTING A HEAD
=
TOTAL NUMBER OF POSSIBLE OUTCOMES
1
2
N.B. P (head) is a short hand way of writing the probability of getting a head.
Example
The probability of picking a chocolate with a particular centre.
A box of six chocolates containing 3 caramels, 2 strawberry creams and one apricot cream.
Strawberry
Caramel
Caramel
Apricot
Cream
Caramel
Cream
Strawberry
Cream
What is the probability of choosing at random a caramel?
P(Caramel) = Number of Caramels
Number of chocolates
=
3=1
6 2
What is the probability of choosing at random a strawberry cream?
P( Strawberry cream)
=
=
Number of strawberry creams
Number of chocolates
2 = 1
6
3
What is the probability of choosing at random an apricot cream?
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P( Apricot cream)
=
=
Number of apricot creams
Number of chocolates
1
6
Probability of not getting apricot cream = Number of chocs not apricot creams
Number of chocolates
=5
6
Notice that
Probability of an apricot cream + probability of not an apricot cream
= 1/6 + 5/6 = 1
CARDS AND DICE
A pack of cards is made up of 52 cards, divided into four suits - Diamonds, Hearts,
Clubs and Spades. Diamonds and Hearts are red cards. Clubs and Spades are
black cards. Each suit has 13 cards. These carry the numbers from 2 - 10 PLUS
Jack, Queen, Kings and Ace (called the 'picture' cards).
So, each suit has the following cards:Ace, 2, 3, 4, 5, 6, 7, 8, 9,10, Jack, Queen, King.
A die (or dice if there are more than one die) is a six-faced cube, with numbers from
1 to 6 on each face.
Look at these examples which will include dice and cards:
Example 1
a) A die is rolled. What is the probability of getting a four?
The die may land with any of the following numbers on top 1,2,3,4,5,6.
There are six numbers so the number of possible outcomes is 6.
One of these numbers (outcomes) is 4, so the number of ways of getting a 4 is 1.
P(4) =
b)
Number of ways of getting a 4
=
Total number of possible outcomes
1
6
A card is drawn from a pack. What is the probability that it is a king?
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There are 52 cards in a pack so the total number of possible outcomes is 52. There is the
king of hearts, the king of spades, the king of clubs and the king of diamonds, so there are
4 kings altogether.
P(King) =
c)
=
4
52
=
13
52
=
1
13
What is the probability that the card drawn is a heart?
P(Heart)
d)
Number of ways "king" can happen
Total number of possible outcomes
= Number of ways of getting "heart" =
Total number of possible outcomes
1
4
What is the probability that the card drawn is the Ace of Spades?
P(Ace of Spades) =
1
52
Exercise 1
1)
There are twenty buttons in a box. Five are red, three are blue and twelve are green.
One is drawn at random. What is the probability that the button
a)
is blue?
b)
is green?
c)
is red or blue?
2)
A die is rolled. Calculate the probability that it will give
a)
a1
b)
an odd number
c)
a number greater than four.
3) A letter is chosen from the word STRANGER . What is the probability that it is:a)
a vowel ( a e i o u are the vowels)
b)
an R.
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SAMPLE SPACE
A list of everything that may happen. For example in the above example about a box of six
chocolates the sample space would be
apricot cream, strawberry cream, strawberry cream, caramel, caramel, caramel.
When the situation is complicated in any way eg we have 2 coins or 2 dice it is sometimes
helpful to make a list of possible outcomes. Such a list can be called a 'sample space'.
Example 1
If I toss two coins list the possible outcomes.
Find the probability of getting i) 2 heads and ii) one head and one tail in any order.
Possible results are HH, HT, TH, TT.
Notice that HT and TH are both included first coin head and second tail and the other way
round.
There are 4 possible results.
P (2 heads) =
1
4
P (head & tail) =
2
1
=
4
2
Example 2
If I roll 2 dice list the set of possible outcomes.
Find the probability of getting
i)
double six
ii)
same number on both dice
iii)
total score of ten.
It is easiest here to show the results in a table, showing 1st score, 2nd score.
Fill in the missing spaces.
1,1
2,1
3,1
4,1
5,1
6,1
1,2
2,2
1,3
6,2
6,3
1,4
1,5
1,6
Total number of possible outcomes = 36
P (double six) =
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1
36
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P (same number on both dice) = P ( (6,6), (5,5), (4,4), (3,3), (2,2), (1,1) )

6 1

36 6
P (Total score of 10) = P ( (5,5), (6,4), (4,6) )

3
1

36 12
The PROBABILITY SCALE
If an event is certain to happen the probability that it will happen is 1(one).
P (a tossed coin will come down again) = 1
P (one day I shall die) = 1
If an event can never happen the probability that it will happen is 0 (zero).
P (I will live for ever) = 0
P (I can fly without mechanized help) = 0
All probabilities must lie between 0 and 1. Probabilities can be expressed as decimal
fractions or vulgar fractions.
Both of the following are acceptable as answers:
P (a die will show '2') = 1
6
or 0.167 (correct to 3 dec.places)
TOTAL PROBABILITY
For most events, there is an opposite.
For example the opposite of throwing a six is not throwing a six.
When a die is thrown either a six is thrown or it isn't.
So we are certain to throw a six or not throw a six. The probability is 1.
P(throwing a six) + P(not throwing a six) = 1
P(not throwing a six) = 1 - P(throwing a six)
using the usual rules of rearranging equations.
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P(not throwing a six) = 1 - 1/6 = 5/6
If the event we are interested in happens, this is often thought of as a "success".
If the event does not happen, this is called a "failure"
Total probability covering all possible outcomes of an event must be 1. Therefore,
P (success) + P (failure) = 1
What is the probability of obtaining a 5 when throwing a die? What is the probability of not
getting a 5?
P (5) = 1
6
P (not get t i ng a 5 )
= 1 -
1 5
=
6 6
This shows us a very useful method for many probability examples.
Example
What is the probability that, when a card is drawn from a pack it is NOT an ace?
Solution
P (Ace) =
4
52
P ( not Ace ) = 1 -
=
1
13
1
12
=
13
13
Exercise 2
1) A box contains 100 coloured balls. 25 balls are red; 30 are blue. The rest are yellow or
green. One ball is drawn at random.
What is the probability that:a) it is red?
b) it is red or blue?
c) it is not red?
d) it is black?
2) Five red balls are now removed from the box, and then one ball is drawn at random.
What is the probability that:a) it is red?
b) it is blue?
HOW TO UNDERSTAND PROBABILITY
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When a coin is tossed, the probability of getting a head is 0.5. This does not mean that if we
tossed a coin ten times we would get five heads and five tails.
P=
1
2
would tell us what to expect in the long run.
If we tossed a coin 100 times, it would be very unlikely that we would get exactly 50 heads,
but if we repeated this experiment a large number of times, we would EXPECT that the
average results would be 50 heads.
If we tossed a coin 100 times and got 95 heads, we would suspect that someone had given
us an unfair coin (in this case, weighted to favour heads).
Remember that probability can never tell us what WILL happen, but it can tell us what is
LIKELY to happen.
Probability =
99 means VERY LIKELY TO HAPPEN
100
Probability =
1 means VERY UNLIKELY TO HAPPEN
100
and so on.
PROBABILITY SCALE
[___________[___________[___________[_________[
Impossible Unlikely
Likely
Certain
0 ------------------------------------0.5- ------------------------------- 1
EMPIRICAL (EXPERIMENTAL) PROBABILITY
In the examples we have done so far, we have been able to work out the probabilities from
our knowledge of the situation.
We know that all coins have two faces; that packs of cards are made up from 52 cards and
four of them are kings. In many situations, this kind of information is not available, and in
practice, the probabilities are worked out from information gathered from samples that have
been taken or experiments that have been done.
If we are told that the probability of rain on a June day is 13/30 this will have been arrived
at by studying the rainfall records for June, over, let us say, the past 20 years.
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Obviously, information about every June day that ever was or ever shall be is not available,
so the probability has been estimated from a sample.
If a manufacturer tells us that he is 99% certain ( P = 0.99) that an electric light bulb will last
for 200 hours, this figure will have been arrived at from experiments which have tested
samples of light bulbs.
THINK ABOUT - how you would give an estimate of the probability of a 25-year old
passing the driving test at his first attempt?
INDEPENDENT EVENTS
I toss a fair coin.
Four times in a row, it comes down heads.
What would you call for the next toss?
It might be tempting to think that if we have had four heads in a row, the next toss is more
likely to be a tail!
But, no, if the coin is a fair coin, the probability of getting a head on the next go, is still 0.5,
just as it was on all the other goes.
This is an example of independent events, because what has happened on one trial has
no effect on what happens or any other trial.
Look at another example.
I draw a card from a pack. It is the king of hearts. I put the card back and draw again. What
is the probability of drawing a heart this time? The answer is 1/4.
The two events are independent. The outcome of the first trial cannot affect the outcome of
the second trial.
Now suppose after I drew the first card, I did not put it back into the pack. What is the
probability of drawing a heart this time?
The answer is not 1/4 because the events are not independent. The answer depends on
what happened on the first draw.
There are two possibilities -
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a) If the first card drawn was a heart, then on the SECOND draw, I have a pack of 51
cards, containing 12 hearts.
Therefore, P (heart) = 12
51
b) If the first card drawn was not a heart, then on the second draw, I have a pack of 51
cards containing 13 hearts.
Therefore, P (heart) =
13
51
Exercise 3
1)
A card is drawn from a pack.
a) What is the probability that it is an ace?
b) The card is replaced and another one drawn. What is the probability that this is a
diamond?
2) A card is drawn from a pack. It is not replaced. A second card is drawn. What is the
probability that this card is an ace if
a)
b)
the first card drawn was an ace?
the first card drawn was not an ace?
3)
On throwing a die, what is the probability of turning up
a)
b)
c)
d)
a1
a number greater than 3
a number more than 6
an even number.
4)
In a pack of playing cards, what is the chance of drawing
a)
b)
c)
d)
a Jack
a spade
the Ace of Diamonds?
an 8, 9 or 10.
5) In a box containing 100 oranges, 5 are bad. What is the probability of picking out a bad
one.
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6) Two dice are thrown. What are the probabilities that the TOTAL SCORE is
a)
4
b)
1
c)
11
d)
12
e)
5
f)
less than 5
g)
more than 5.
7)
A box contains 40 coloured pencils, 15 red, 10 blue and the rest yellow. What is the
probability of choosing
a)
b)
c)
d)
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a red pencil?
a yellow pencil?
a red or a yellow pencil?
a white pencil?
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ANSWERS
Exercise 1
1
20 buttons
a)
P (blue) =
b)
P (green)
c)
P (red or blue)
2
A die
a
P (1) = 1
6
c
P (numbers greater than 4) =
3
STRANGER
a
P (vowel) = 2 = 1
8
4
P (R)
= 2 = 1
8
4
b
5 red, 3 blue, 12 green.
3
20
=
1
12
20
=
3
5
5+3 = 8
20
20
2 3
4
5
= 2
5
6
b P (odd numbers)
2
6
= 3
6
=
=
1
2
1
3
(vowels are A
and E)
Exercise 2
1)
100 balls
a)
P (red)
b)
P ( red or blue)
c)
P ( not red)
d)
P (black)
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25 red, 30 blue
=
25
100
=
=
1
4
55
100
=
= (100 - 25)
100
=
0
100
100 - 55 = 45 yellow or green.
=
11
20
=
75 =
100
3
4
0
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2)
95 balls
20 red, 30 blue
a)
P (red)
b)
P (blue) =
30
95
1)
a)
P (ace) =
4 = 1
52
13
b)
P (diamond)
2)
a)
51 cards
= 20
95
95 - 50 = 45 yellow or green.
= 4
19
=
6
19
Exercise 3
P (ace)
b)
3)
c)
13 =
52
1
4
3 aces left
=
51 cards
3 =
51
1
17
4 aces left
P (ace)
=
4
51
Die 1
2
3
a)
b)
=
4 5 6
P (1) = 1
6
P (number greater than 3) = P (4,5 or 6) = 3 = 1
6
2
P (number more than 6) = 0
d)
P (even number) =
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3 =1
6
2
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4)
a)
P (Jack)
b)
P (spade)
=
=
13
52
=
4
52
=
1
4
c)
P (ace of diamonds)
d)
P (8, 9 or 10)
5)
P (bad)
=
1
13
=
=
1
52
12
52
=
3
13
5 = 1
100
20
6)
2nd die
1
2
3
4
5
6
1st die
1
2
2
3
3
4
4
5
5
6
6
7
7
8
3
4
5
6
7
8
9
4
5
6
7
8
9
10
5
6
7
8
9
10
11
6
7
8
9
10
11
12
n = 36
a)
P (4)
b)
P (1) = 0
c)
P (11) =
d)
P (12)
= 1
36
e)
P (5)
= 4 = 1
36
9
f)
P ( less than 5)
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=
3 = 1
36
12
2 = 1
36
18
=
6 = 1
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g)
P ( more than 5)
Note : P ( more than 5)
= 1 - ( 4
(36
7
+
=
36
6
26
36
=
13
18
= 1 - P (not more than 5)
6 ) =
36 )
1
-
10
36
= 26
36
= 13
18
40 pencils 15 red, 10 blue, yellow pencils = 40 - (15+10) =15
a)
P (red)
b)
P (yellow)
c)
P ( red or yellow)
d)
P (white) =
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= 15 = 3
40 8
=
15
40
= 3
8
= 30 = 3
40
4
0
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Chapter 12
HARDER PROBABILITY
MUTUALLY EXCLUSIVE EVENTS AND THE ADDITION LAW OF PROBABILITY
Most of the last chapter was about the chance of one thing happening. The end of it briefly
introduced us to two things happening.
This chapter is going to explain the laws of chance for two or more things happening.
Always start from basic rules.
The next section is going to deal with things that cannot happen at the same time.
Earlier we talked about the probability of throwing a six and the probability of not throwing a
six.
These events cannot both happen. Either a six is thrown or it is not.
Events like this are called mutually exclusive.
Example
Throwing a 3 with a single die excludes the possibility of throwing a 4. Therefore, throwing
a 3 and a 4 are mutually exclusive.
Example
It is impossible to cut both a King and a Jack with a single cut of a pack of cards.
Therefore, the one excludes the other.
Example
Choosing an odd number excludes the choice of an even one.
Example
A coin cannot show both heads and tails, simultaneously.
Example
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Find the probability of throwing an even number with a single fair die. This event consists of
three separate mutually exclusive events :
throwing a 6
throwing a 4
throwing a 2.
The probability of throwing an even score P is
the probability of throwing a 2 P1
plus the probability of throwing a 4 P2
plus the probability of throwing a 6 P3
P (throwing an even score) = P (throwing a 2) + P (throwing a 4) + P (throwing a 6)
P =
1
1
1
1
+
+
=
6
6
6
2
If P1 P2 P3... are the separate probabilities of a set of mutually exclusive events,
then the probability of one of the events happening is
P = P1 + P2 + P3 + ...
Therefore, P (A or B) = P(A) + P(B)
The addition law is sometimes called the 'OR' LAW, because we require the probability that
one event OR another event will happen.
Example
What is the probability of throwing a 5 or a 6 with a single die?
P( 5 ) =
1
6
1
6
P( 6 ) =
Mutually exclusive events, therefore ADDING gives
P(5 OR 6) = P(5)  P(6) =
1
6

1
6
=
2
6
=
1
3
Example
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A pack of cards is cut once. Find the probability that the card will be the 7 of Hearts, the
King of Diamonds, or the Ace of Spades.
P(7 Hear t s OR Ki ng Di am onds OR A ce Spades)
P(7 Hear t s)  P( Ki ng Di amonds)  P( A ce Spades)
=
1
52

1
52

1
52
=
3
52
Mut ual l y excl usi ve event s, t her ef or e ADD, t o gi ve,
3
52
Example
In a partly used box of tissues, there are 12 purple, 15 green, 17 white, 10 yellow, and 6
blue tissues. Assuming the box is filled with colours at random what is the probability
that the last one is:
a) either blue or yellow
b) not white?
P(l ast one bl ue OR yel l ow ) = P(l ast one bl ue)  P(l ast one yel l ow )
=
6
60

10
60
=
16
60
=
4
15
b) The events white and not white are mutually exclusive.
P(white) + P(not white) = 1
because a tissue is either white or not. We have covered all the possibilities so the sum
of the probabilities is 1.
P(last one not white) = 1  P(white)
P(l ast one w hi t e) =
17
60
P(l ast one not w hi t e) = 1 -
17
60
=
43
60
INDEPENDENT EVENTS: THE MULTIPLICATION LAW OF PROBABILITY
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Two events are said to be independent if the outcome of one does not affect the other.
Examples
1. Drawing a card from a pack and tossing a coin
2. Drawing from a pack, replacing it, and then drawing another card.
3. When a die is thrown twice, what happens in the first throw does not affect what
happens on the second. The two throws are therefore independent events.
4. The probability of developing cancer is not independent of the probability of smoking.
5. The probability of having naturally brown eyes and naturally dark hair are not
independent. Brown eyes occur more often with dark hair than fair hair.
6. Now consider a set of eight identical electrical components numbered from 1 to 8.
Fr om a pr evi ous exper i ment i t has been di scover ed t hat t he pr obabi l i t y
of a f ai l ur e i n t he f i r st hour i s
1
100
P(component 1 f ai l s) =
1
100
P(component 2 f ai l s) =
1
100
What is the probability that both components will fail?
These events are not mutually exclusive because they can both happen at the same time.
Component 1 failing does not prevent component 2 failing or vice versa.
However component 1 failing does not change the probability of component 2 failing.
Component 1 not failing does not change the probability of component 2 failing either.
When one event happening does not affect the probability of another event happening the
events are said to be independent.
The probability of both events happening is the product of the two probabilities; this is often
called the multiplicative law or the 'AND' law.
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P(component 1 fails AND component 2 fails)
= P(component 1 fails) x P(component 2 fails)
=
1
100
x
1
100
=
1
1000
Now you have two rules :
1) Where you add the probabilities.
2) Where you multiply the probabilities.
It is important to think carefully about which rule should be applied.
If P1, P2, P3... are the separate probabilities of a set of independent events, then the
probability that ALL the events will occur is
P = P1 x P2 x P3...
P(A and B) = P(A) x P(B)
The multiplication law is sometimes called the 'AND' LAW because we require the
probability that one event AND another will happen.
Example
What is the probability of throwing a 4 and then a 5, with 2 throws of a die?
These are independent events :
P(4) =
1
6
P(5) =
1
6
Independent events, therefore MULTIPLY to give:
P( 4 t hen 5 ) =
1
1
1
x
=
6
6
36
Example
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What is the Probability of tossing a coin and obtaining a head and throwing a die and
obtaining a 6?
Again, these are independent events :
P(H) =
1
2
P(6) =
1
6
Independent events, therefore MULTIPLY, giving
P(H and 6 ) =
1 1
1
x
=
2 6 12
Look at these further examples
1)
From records held by the Registrar of Births, Marriages and Deaths it is worked out that
0.51 of all babies born are boys and 0.49 are female.
Mrs Jones has one baby, so does Mrs Khan. What is the probability that they are both
boys?
Solution
P(Mrs Jones has baby boy and Mrs Khan has baby boy)
= P(Mrs Jones has baby boy) x P(Mrs Khan has baby boy)
= 0.51 x 0.51 = 0.2601
2)
Assume t hat t he pr obabi l i t y of Fr ed's bi r t hday bei ng on t he 1st June
is
1
365
and t he pr obabi l i t y of Jaw ai d's bi r t hday i s t he 1st June i s
1
.
365
What is the probability of that neither of them have a birthday on the first of June?
Solution
P(Fred's birthday not 1st June)
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SUBJECT KNOWLEDGE AND UNDERSTANDING: HANDLING DATA
23
= 1  P(Fr ed's bi r t hday i s 1st June) = 1 
1
365
=
364
365
=
364
365
P(Jawaids birthday not 1st June)
= 1  P(Jaw ai d's bi r t hday i s 1st June) = 1 -
1
365
P(Jawaid's birthday not 1st June and Fred's birthday not 1st June)
= P(Jawaid's birthday not 1st June) X P(Fred's birthday not 1st June)
=
364
365
X
364
365
= 0.995 (t o 3 si g. f i g)
Another Example
I t i s Gr andad's bi r t hday. The pr obabi l i t y t hat Gr andad w i l l be gi ven a bot t l e
1
1
. The pr obabi l i t y t hat Gr andad w i l l be gi ven a t i e i s .
12
4
The pr obabi l i t y t hat Gr andad w i l l be gi ven a pai r of socks i s 2 .
3
of w hi sky i s
What is the probability that Grandad will receive
a) a tie or socks
b) a bottle of whisky, or a tie or socks
Solution
a)
P(tie OR socks) = P(tie) + P(socks)
=
1 2 3 + 8 11
+ =
=
4 3
12
12
b) P(whisky OR tie OR socks) = P(whisky) + P(tie) + P(socks)
=
1 1 2
+ + =1
12 4 3
TREE DIAGRAMS
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SUBJECT KNOWLEDGE AND UNDERSTANDING: HANDLING DATA
24
A tree diagram is useful when more than one thing happens. It allows a complete ordered
list of all results to be made.
The multiplication law can be illustrated by a "probability tree".
This method helps you see what is taking place.
Study the examples below, because they show you both methods in operation for the same
problem.
All possible outcomes, together with their probabilities, can be shown on a tree diagram.
Example
A bag containing 10 marbles, has 7 red and 3 black marbles. A marble is drawn at random
and then replaced. A second marble is then drawn.
What is the probability that both marbles are red?
P(f i r st mar bl e i s r ed) =
7
10
The marble is returned to the bag making 10 marbles.
P(second mar bl e i s r ed) =
7
10
So P(bot h mar bl es ar e r ed) =
7
7
49
x
=
10
10
100
Si mi l ar l y, P(bot h mar bl es bl ack) =
3
3
9
x
=
10 1 0
100
P (one r ed mar bl e t hen one bl ack mar bl e) =
7
3
21
x
=
10
10
100
But if one black then one red is equally acceptable, that is the black appears first instead
of the red then
P ( one black marble then one red ) =
3
7
21
x
=
10
10
100
So, provided the order of drawing the marbles is not important,
St Martin's College
SUBJECT KNOWLEDGE AND UNDERSTANDING: HANDLING DATA
25
P ( one red, one black, any order ) =
21
21
42
21
+
=
=
100 100 100
50
Pr obabi l i t y (f i r st mar bl e i s r ed) =
7
10
Red
7
10
P (Se con d ma rbl e i s re d)
7
10
7
10
Red
Red
P
red AND
( 1st2ndmarble
)
marble red
= P(1st marble red) X P(2nd marble red)
=
7
10
7
10
7
10
Red
7
10
X
7
10
=
49
1 00
Red
3
10
Bl ack
(2nd marble black )
P 1st marble red AND
= P(1st marble red) X P
=
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7
1 0
X
3
1 0
=
(2nd marble black)
2 1
1 00
SUBJECT KNOWLEDGE AND UNDERSTANDING: HANDLING DATA
26
7
10
7
10
Red
Red
3
1 0
Bl ack
3
1 0
P(fi rst m arbl e b l ack)
Bl ack
7
10
7
10
Red
=
3
1 0
Red
3
1 0
Black
3
1 0
7
10
Black
Red
P(black then red) = P(black) X P(red)
=
7
10
7
10
Red
3
X
1 0
7
=
10
21
100
Red
3
1 0
Bl ack
3
1 0
7
10
Red
Bl ack
Bl ack
St Martin's College
P(bo th bl a ck)
=
9
3
3
X
=
10
10
100
SUBJECT KNOWLEDGE AND UNDERSTANDING: HANDLING DATA
27
7
10
7
10
Red
Red
3
1 0
Bla ck
3
1 0
7
10
Red
}
T he se rou tes bo th gi ve on e
b la ck an d o ne re d, no t
n ece ssari l y in th at ord er, so
a dd th e pro ba bi l iti es.
Bl ack
Bl ack
By now you should have worked out that you multiply along the branches. If you want a
selection of end points then you add those probabilities.
P (one r ed, one bl ack, any or der ) =
(
) (
7
3
X
10 10

3
7
X
10 10
)
This can easily be illustrated by a tree diagram as shown :
St Martin's College
SUBJECT KNOWLEDGE AND UNDERSTANDING: HANDLING DATA
28
An alternative method
Since the sum of all the probabilities is 1
P(one red one black) = 1 - P ( all red ) - P ( all black )
49
9
100
100
=1 -
58
100
=1 -
P ( one r ed , one bl ack ) =
42
21
=
100
50
Example
A coin is tossed 3 times. Find the probabilities of :
a)
3 heads b) only 1 head.
Solution
a) P(3H) = P(Head AND Head AND Head) = P(H) X P(H) X P(H)
=
1
2
X
1
2
X
1
2
=
1
8
b) P(HTT) = P(Head AND Tails AND Tails) = P(H) X P(T) X P(T)
=
1
2
X
1
2
X
1
2
=
1
8
P(THT) = P(Tails AND Head AND Tails) = P(T) X P(H) X P(T)
=
1
2
X
1
2
X
1
2
=
1
8
P(TTH) = P(Tails AND Tails AND Heads) = P(T) X P(T) X P(H)
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SUBJECT KNOWLEDGE AND UNDERSTANDING: HANDLING DATA
29
1
2
=
1
2
X
1
2
X
=
1
8
P(exactly one Head) = P(Head Tails Tails OR Tails Head Tails OR Tails
Tails Head)
= P(HTT) + P(THT) + P(TTH)
=
1
8
+
1
8
+
1
8
=
3
8
To draw a tree diagram start at the left of page. On the first toss of the coin the result is
either a Head or a Tail.
1
2
H top branch
(I threw a Head
)
pr obabi l i t y of f ol l ow i ng t op br anch i s
1
2
pr obabi l i t y of f ol l ow i ng bot t om br anch i s
1
2
T bottom branch
1
2
(I threw a Tail)
I have chosen to go on the up branch for Heads. Notice how I have written the probability
of following a branch on the branch itself.
Let us follow the top branch for a while.
A Head was thrown with the first throw.
The probability of throwing a Head with the second throw is still a half.
So we might get a Head or a Tail; - the chances are equal.
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SUBJECT KNOWLEDGE AND UNDERSTANDING: HANDLING DATA
30
1
2
H
1
T
H
amhere
here II must
must have
have thrown
IfIfI Iam
throwna
head on the first throw and a head
a Head on the first throw and a
on the second throw.
head
on the second
The
The probability
that Ithrow.
am here
is
probability
1 x 1 = 1 that I am her e is
12 21 41
X =
2
1
2
2
4
IIthrew
andthen
thena aTail.
Tail.
threwaa Head
Head and
The probability
of being
Probability
of being
herehereisis
11 x 11 = 1 1
2 X 2 =4
2
1
2
2
2
4
T
1
2
H
1
2
1
2
1
2
T
H
1
2
T
H
1
2
T
}
Threw a Tail first
throw, either Head
or Tail Second
throw.
H
Threw HHH
H
T
H
Threw HHT
T
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SUBJECT KNOWLEDGE AND UNDERSTANDING: HANDLING DATA
31
H
1
Three Heads HHH P( HHH) =
2
H
1
2
H
1
1
2
1
2
1
2
2
T
H
T
1
2
1
T
H
1
8
1
HTH P( HTH) =
8
HHT P( HHT) =
HTT P( HTT) =
1
8
THH P( THH) =
1
8
THT P( THT) =
1
8
2
1
2
T
1
2
H
1
2
1
1
2
T
T
H
2
1
2
T
1
8
TTH P( TTH) =
TTT P( TTT) =
1
8
One Way t o
( get
one Head)
(
(
)
)
2 nd w ay t o
get one Head
Last w ay t o
get one Head.
1
8
Try these
1.
P, Q and R are three mutually exclusive events, with
P(P) = 1
6
P(Q) = 1
3
P(R) = 1
2
Calculate:
a) the probability of either P or R occurring
b) the probability of either P or Q occurring.
2.
What is the probability of throwing 6 heads in 6 tosses of a coin?
3.
A bag contains ten marbles, six are blue and four are yellow. One marble is drawn
at random. Its colour is noted and then it is replaced. This is repeated once more.
Find the probability that the following will occur:
a) two blue marbles are drawn one after the other
b) their colours are different.
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SUBJECT KNOWLEDGE AND UNDERSTANDING: HANDLING DATA
32
4. A bag contains twenty sweets, eight toffees and twelve chocolates. A
sweet is drawn at random and replaced. A second and third sweet are
also chosen, and replaced. What is the probability that the following
sweets are chosen?
a) three toffees
b) one toffee, following by two chocolates.
5. The probability that Anne and Jack will pass their driving tests are
2 and 1 respectively
3
4
Find the probability that:
a) Anne and Jack pass
b) they both fail
c) only Anne passes.
6. Find the probability of drawing 3 Kings in succession from a pack of cards, if the cards
are not replaced after each draw.
7. Find the probability of drawing either a Diamond or a Heart from a pack of cards at the
first draw.
8. Find the probability of drawing a King or a Queen or a Jack from a pack of cards at the
first attempt.
9. It is only possible for three beauty queens to win a major prize. The probabilities are as
follows:
Miss A 3 ,
8
Miss B 1 , Miss C x
3
Which girl is most likely to win?
What is the probability that it is Miss A or Miss B that wins the prize?
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SUBJECT KNOWLEDGE AND UNDERSTANDING: HANDLING DATA
33
ANSWERS
1.
a) P(P or R) = P(P) + P(R)

1 1 1 3 4 2
    
6 2 6 6 6 3
b) P(P or Q) = P(P) + P(Q)

2.
1 1 1 2 3 1
    
6 3 6 6 6 2
P(6 heads) = P(H) x P(H) x P(H) x P(H) x P(H) x P(H)
1 1 1 1 1 1 1
      
2 2 2 2 2 2 64
3. a) P(blue followed by blue) = P(blue) x P(blue)

6 6
9
 cancel ! 
10 10
25
b) P(blue followed by yellow) = P(blue) x P(yellow)

6 4
6
 
10 10 25
P(yellow followed by blue) = P(yellow) x P(blue)

4 6
6
 
10 10 25
P(blue followed by yellow or yellow followed blue)
= P(blue) P(yellow) + P(yellow) P(blue)

6
6 12


25 25 25
St Martin's College
SUBJECT KNOWLEDGE AND UNDERSTANDING: HANDLING DATA
34
4. a)
P(three toffees) = P(toffee) x P(toffee) x P(toffee)

8 8 8
8
 

20 20 20 125
b) P(toffee followed by 2 chocolates)
= P(toffee) x P(chocolate) x P(chocolate)
=
8 12 12 18
x
x
=
20 20 20 125
5.
a) P(Anne and Jack) = P(Anne) x P(Jack)
=
2
1
1
x
=
3
4
6
b) P(Anne fail and Jack fail) = P(Anne fail) x P(Jack fail)
=
c)
=
1 3 1
x =
3 4 4
P(Anne pass and Jack fail) = P(Anne pass) x P(Jack fail)
2 3 1
x =
3 4 2
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SUBJECT KNOWLEDGE AND UNDERSTANDING: HANDLING DATA
35
6.
P(3 Kings when cards not replaced) = P(King) x P(King) x P(King)
=
7.
4
3
2
1
x
x
=
52 51 50 5525
P(Heart or Diamond) = P(Heart) + P(Diamond)
1 1 1
= + =
4 4 2
8.
P(King or Queen or Jack) = P(King) + P(Queen) + P(Jack)
=
4
4
4
3
+
+
=
52 52 52 13
P  A 
9.
3
1
P  B 
8
3
P C   x
P(A) + P(B) + P(C) = 1
3 1 9 + 8 17
+ =
=
8 3
24
24
Ther ef or e, P(C) = 1 -
17
7
=
24 2 4
Miss A is more likely to win
P(Miss A or Miss B) = P(Miss A) + P(Miss B)
=
3 1 17
 =
8 3 24
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SUBJECT KNOWLEDGE AND UNDERSTANDING: HANDLING DATA