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PROBABILITY
Introduction
John and Peter each have two pennies. John tosses each of his two coins once and gives to Peter
those which fall heads. Peter now tosses once each of the coins in his possession, giving to John
those which fall heads. Using a probability tree or otherwise, determine the probability that,
a) Peter now has all the coins
b) John now has all the coins
c) John and Peter each now have two coins
Consider Johns tosses first of course.
John
HH
HT
TH
TT
The results above will affect the number of coins Peter will toss.
HHHH
HHH
HHH
HH
________________________________________________________
HHHT
HHT
HHT
HT
HHTH
HTH
HTH
HTHH
THH
THH
THHH
________________________________________________________
HHTT
HTT
HTT
TH
HTTH
THT
THT
TTHH
TTH
TTH
THTH
HTHT
THHT
TTTH
TTHT
THTT
HTTT
_________________________________________________________
TTTT
TTT
TTT
TT
1 1
1
a) P(Peter now has all coins) = J(HH)  P(TTTT) =

=
4 16
64
b) P(John now has all the coins) = J(HH)  P(HHHH) + 2[J(HT)  P(HHH)] + J(TT)  P(HH)

= ¼
=
1
16
+2[
1 1

4 8
]
+ ¼

¼
9
64
c) P(Each has now two coins) = J(HH)  6[P(HHTT)] + 2[J(HT)  3[P(HHT)]]+ J(TT)  P(TT)
1
= ¼ 
6( )
+ 2 [ ¼  3(1/8) ] + ¼
 ¼
16
11
=
32
PROBABILITY
Mutually exclusive events
If an event A can occur or an event B can occur, but not both A and B can occur, then the two events
are said to be mutually exclusive. Sometimes known as the “or” rule.
Prob (A or B) = P (A) + P (B)
Example 1
A card is drawn at random from a pack. The probability of a King or a Queen is
P(King or Queen) = P(King) + P(Queen)
=
=
4/52 + 4/52
2/13
Example 2
Two ordinary dice are thrown. Find the probability that the sum of the scores obtained is
a) a multiple of 5
b) a prime number
c) a multiple of 5 or a prime number
a) P(multiple of 5) =
7
36
b) P(prime number) =
11
36
c) P(a multiple of 5 or a prime number) =
7
11 18
+
=
36 36
36
Example 3
The probability that student A ‘comes top’ in P1 is 0.3 . The probability that student B ‘comes top’
is 0.25 and the probability that student C ‘comes top’ is 0.4. Assume there is no tie for first place.
Find the probability that
a) Student A or B comes top.
b) Neither student A or C comes top
Since there can only be one person coming ‘top’ the events are mutually exclusive.
a)
P(A or B )
= 0.3 + 0.25
= 0.55
P (student A or B comes top) = 0.55
b)
P(neither A or C)
=
1  (A or C ‘coming top’)
=
1  [0.3 + 0.4]
=
0.3
P(neither student A or C coming top) = 0.3
PROBABILITY
Independent Events
If either of the two events A and B can occur without being affected by the other, then the two events
are said to be independent. This is sometimes known as the “and” rule.
Prob (A and B) = P (A)  P (B)
Example 1
A die is thrown and a coin is tossed.
P( a six and a head) = P (six)  P (head)
= 1/6  1/2
1
=
12
Example 2
The probability that it will rain on any day in May is ¼ . Find the probability that
a) it will rain on both May 1st and May 5th
b) it will not rain on May 5th
c) it will rain on may 1st but not May 5th
a) P(rain and rain) = ¼  ¼
=
b) P(not rain) = 1  P(rain)
1
16
=1¼
= ¾
c) P(rain and not rain) = ¼  ¾
=
3
16
Example 3
A box of 20 counters; 1 red, 5 blue, 10 green and the rest white. Two counters are chosen at random,
the first being replaced before the second is chosen. Find the probability that ;
a) the first drawn is white
b) the first is white the second is blue
c) one of the counters is white and one is blue
a) P(first is white) =
c) P(one White, one Blue)
b) P(W then B) =
= P(WB) + P(BW)
=
=
Probability trees
Probability trees are a useful way of illustrating probability and information can be easily extracted
from them.
Example 1
George has a 0.6 chance of passing his Maths exam and a 0.4 chance of passing his History exam.
Draw a tree diagram to show illustrate this data and hence calculate the probability of
a) passing both exams
b) passing exactly one exam
c) failing both exams
a) P(passing both exams) =
b) P(passing exactly one exam) =
c) P(failing both exams) =
Example 2
A biassed coin is such that the probability of getting a head is 2/3 and a tail is 1/3 . The coin is tossed
3 times. Draw a tree diagram showing this information. Hence calculate the probability of throwing
a) three heads
b) no heads
c) one head and two tails in any order.
a) P(3 heads) =
b) P(No heads) =
c) P(One head and two tails) =
CONDITIONAL PROBABILITY
Look at the following information in this two way table;
7X
7Y
BOYS
12
8
GIRLS
10
4
If a pupil is selected at random, what is the probability that it is a boy?
P(boy) =
20
(which we could simply!)
34
What is the probability that from the year, the pupil was from 7X?
P(7X) =
22
(which again can be simplified)
34
However, suppose that a child is chosen from 7X. The chance that it will be a boy is
12
22
P(Boy given that they are from 7X) =
even though P(Boy) =
20
34
So having already been told the pupil is from 7X affects the chance of choosing a boy. This is called
CONDITIONAL probability.
We write this conditional notation as :
P(Boy / 7X)
This probability, like all probabilities can be shown as a fraction
P (choosing a boy AND a pupil from 7X)
P (choosing a pupils from 7X)
So
P(boy/7X) = P(boy and 7X)
P(7X)
P(boy/7X) =
 12 
 
 34  =
 22 
 
 34 
12
22
as above.
More generally,
P( B / A) 
P( AandB)
P( A)
Rewriting this equation leads to the multiplication law of probability
P( AandB)  P( A)  P( B / A)
Example 1
A card is chosen at random from a pack without replacement. A second card is
picked. Given that the first card is a King what is the probability of the second being a
King ?
Answer: The probability of the 2nd card being a King has been affected by the first
card’s result.
P(K2 /K1) =
3
51
Therefore the probability of choosing 2 Kings :P (K1 and K2) = P(K1  K2) = P(K1)  P(K2/K1)
= 4/52  3/51
1
=
221
Example 2
In a game of basketball, if a foul is committed on a player in the act of shooting then 2
“free throws” are awarded to the shooter. The probability of scoring a basket is 0.7,
and independence is assumed. Using a probability tree or otherwise calculate the
probability that exactly one point (1 goal) is scored from the two shots.
0.7
0.7
SCORES
P(SS)
DOES NOT
SCORE
P(S,NS)
SCORES
P(NS, S)
DOES NOT
SCORE
P(NS, NS)
SCORES
0.3
0.3
0.7
DOES NOT
SCORE
0.3
P(Exactly one basket is scored) = P(S, NS) + P(NS, S)
= [0.7  0.3] + [0.3  0.7]
=
2  0.21
=
0.42
The probability that exactly one basket is scored is 0.42
Example 3
Consider the basketball example again. In an alternative model, the probability of the
shooter scoring on the second shot is reduced to 0.6 if he misses the first shot.
Calculate the probability that
a) exactly one basket is scored from the two shots
b) the second shot is scored
c) the second shot is scored given that only one basket is scored
0.7
SCORES
0.3
DOES NOT
SCORE
SCORES
0.7
0.3
0.6
SCORES
DOES NOT
SCORE
0.4
DOES NOT
SCORE
P(Exactly one basket is scored)
= P(S, NS) + P(NS, S)
= [0.7  0.3] + [0.3  0.6]
=
0.21 +
0.18
=
0.39
P(the second shot is scored)
=
=
=
=
P(NS, S) + P(S,S)
[0.3  0.6] + [0.7  0.7]
0.18 +
0.49
0.67
P(the second shot is scored given exactly one basket is scored)
= P(2nd shot scored
exactly1 basket)
P(exactly 1 basket)
From P( A / B) 
P( AandB)
P( B)
=
P(NS, S)
0.39
=
0.3  0.6
0.39
= 0.462 (3 sig fig)
Example 4
Three coins are such that one of them is biased and the other two are fair. The
probability that the biased coin lands heads up is 0.7. If one of the coins is tossed twice
and lands heads up on both occasions, what is the probability that it is the biased coin?
The events here are “choosing a coin” and “Tossing a coin twice”
0.49
2H
BIASED
1/3
2H
2/3
0.25
FAIR
P(biased / 2heads) 
P(biased  2heads)
P(2heads)
P(biased  2heads)  0.16333...
P(2heads)  0.1633.3..  0.1666...  0.33
P(biased / 2heads) 
0.1633..
 0.495(3sf )
0.33
CONDITIONAL PROBABILITY
Give your answers correct to 3 significant figures.
1. One box contains 50 red pens and 25 blue pens. A second box contains 35 red pens
and 40 blue pens. One box is chosen at random and then one pen is chosen at random
from that box. If it is a red pen, what is the probability that it came from the first box?
2. Three coins are tossed. Two of the coins are fair and one is biased so that a head is
three times as likely as a tail. What is the probability that
(a) exactly one head shows (b) when only one coin lands head up, it is the biased coin?
3. There are 50 straws in a pot. The hidden end of one straw is painted red and if that
straw is chosen, a prize is won. Competitors are each allowed to remove up to 4
straws, one at a time and without replacement. What is the probability that a
competitor will win a prize?
4. A company offers a repair service for washing machines. On any one day, the
probability that no repairs are requested is 0.1, the probability that one repair is
requested is 0.3 and the probability that two repairs are requested is 0.5.
(a) On any one day, what is the probability that more than two repairs are requested?
(b) What is the probability that there are no requests for repairs on three consecutive
days?
5. Two dice are thrown. One die is fair and the other is biased so that the probability of
scoring a six on it is 0.12.
(a)What is the probability of scoring two sixes?
(b) If one six is scored, what is the probability that it is on the biased die?
6. A survey by a school catering company asks schools to give a questionnaire to each
child in the school whose birthday is on the 10th day of any month.
(a) What is the probability that a randomly chosen child will be asked to take part in
the survey? What assumptions do you need to make?
(b) A sixth form college has 200 students in the first year. What is the probability that
none of these students will be asked to take part?
7. A company buys paper from two suppliers A and B and places twice as many orders
with A as with B. The probability that supplier A will deliver an order within one week
is 0.8 and the probability that supplier B will deliver an order within one week is 0.7.
(a) What is the probability that a randomly selected order takes more than a week to be
delivered?
(b) Given that an order has taken more than a week to be delivered, what is the
probability that it was placed with supplier B ?