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Solution of Sheet 3
Probability Theory
Example 1
Consider the random experiment of tossing a coin. Find the sample space
for this experiment.
Solution:
The sample space can be written in the form: S={H, T}
Example 2
Consider the random experiment of tossing two different coins. Find the
sample space for this experiment.
Solution:
The sample space can be written in the form:
S={HH, HT, TH, TT}
First coin
Second coin
Example 3
Consider the random experiment of rolling a fair die. Find the sample space
for this experiment.
Solution:
The sample space can be written in the form: S={1, 2, 3, 4, 5, 6}.
Example 4
In a rolling a fair die.
i. What is the probability of obtaining at least 5
ii. The probability of Even numbers.
Solution:
i. Let A be the event represent “at least 5”.
i.e., A={5, 6}, has two elements.
P(A)=2/6=1/3.
ii. Let B be the event represent “Even Numbers”.
i.e., B={2, 4, 6}, has three elements.
P(B)=3/6=1/2.
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Example 5
A random experiment consists of tossing 2 fair coins (or tossing a fair coin
twice). Find
i. The probability of obtaining at exactly one head.
ii. The probability of obtaining at least one head.
Solution:
Since the sample space: S={HH, HT, TH, TT}
i. Let A be the event of having exactly one head, thus
A={HT, TH}, has two elements.
P(A)=n(A)/n(S)=2/4=1/2.
ii. Let B be the event of obtaining at least one head, thus
B={ HH, HT, TH}, has three elements.
P(B)= n(B)/n(S)=3/4.
Example 6
A random experiment consists of throwing two fair die.
i. Find the sample space S.
ii. Let A be the event that the sum of the result is even. Find P(A)
iii. Find the probability of obtaining a sum of at most 6.
Solution:
i. Sample space
Die 1
Die 2
1
2
3
4
5
6
1
(1, 1)
(1, 2)
(1, 3)
(1, 4)
(1, 5)
(1, 6)
2
(2, 1)
(2, 2)
(2, 3)
(2, 4)
(2, 5)
(2, 6)
3
(3, 1)
(3, 2)
(3, 3)
(3, 4)
(3, 5)
(3, 6)
4
(4, 1)
(4, 2)
(4, 3)
(4, 4)
(4, 5)
(4, 6)
5
(5, 1)
(5, 2)
(5, 3)
(5, 4)
(5, 5)
(5, 6)
6
(6, 1)
(6, 2)
(6, 3)
(6, 4)
(6, 5)
(6, 6)
From the figure, it is clear that:
n(s)=36, n(A)=18, n(B)=15, then
ii. P(A)=18/36=1/2
iii. P(B)=15/36=5/12.
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Example 7
A ball is drawn at random from a box containing 6 red balls, 4 white balls,
and 5 blue balls. Determine the probability that it is
a) Red
b) white
c) Blue
d) Not red
e) Red or White.
Solution:
Let: R, W, and B denote the event of drawing a red ball, white ball, and
blue ball, respectively.
Sample consists of n(S)=6+4+5=15 sample points.
a) P( R) 
number of red balls
6
6 2

  .
Total number of balls 6  4  5 15 5
number of white balls
4
4

 .
Total number of balls 6  4  5 15
number of blue balls
5
5 1

  .
c) P( B) 
Total number of balls 6  4  5 15 3
2 3
C
d) P( Not R)  P( R )  1  P( R)  1   .
5 5
b) P(W ) 
e) P( R W )  P( R)  P(W ) 
2 4 10 2
 
 .
5 15 15 5
Example 8
Three horses A, B, C are in race, A is twice as likely to win as B, and B is
twice as likely to win as C.
a) Find their respective probabilities of winning, i.e., P(A), P(B), P(C).
b) Find the probability that B or C wins.
Solution:
Let: P(C)=x.
Since B is twice as likely to win as C. Then P(B)=2x, and
Thus P(A)=2P(B)=2 (2x)=4x
Since P(S)=1, hence P(A)+ P(B)+P(C)=x+2x+4x=1. Then x=1/7.
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2
1
a) P( A)  , P( B)  , & P(C )  .
7
7
7
2 1 3
b) P( B C )  P ( B )  P (C )    .
7 7 7
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Example 9
Suppose a student is selected at random from 80 students where 30 are
taking mathematics, 20 are taking chemistry and 10 taking mathematics
and chemistry. Find the probability that the student’s taking mathematics
or chemistry.
Solution:
Let M be the event represent Mathematics,
C be the event represent Chemistry.
Since the space is equiprobable space, we have
P (M ) 
30 3
20 1
 , P (C ) 
 , & P (M
80 8
80 4
P (M or C )  P (M
C)
10 1
 .
80 8
C )  P (M )  P (C )  P  M
3 1 1 1
C    .
8 4 8 2
Example 10
The graduates of a certain college were classified according to sex and
employment. The result is given in the following table
Employed Unemployed Total
E
U
Male
M
460
40
500
Female F
140
260
400
If a graduate is selected at random from 900 graduates and was found to
be a male. What is the probability that he is unemployed?
Solution:
It is required to find: P (U / M )  ?.
P (U M ) 40 900
40
P (U / M ) 


 0.08.
P (M )
500 900 500
Example 11
Two dice are tossed twice. What is the probability of obtaining a score of 7
on the first toss and a score of 11 on the second toss?
Solution:
Let A be the event of obtaining 7 on the first toss and
B be the event of obtaining on the second toss.
Since the result of the 2nd toss does not depend on the result of the 1st toss,
then A and B are independent. Thus
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A  1,6 ,  2,53, 4 4,35, 2  6,1, & B  5,6  , 6,5 .
and P (A ) 
Then P (A
6
,
36
2
.
36
6 2
B )  P (A ) . P (B )  .
36 36
& P (B ) 
Example 12
In a certain college, 25 present of the students failed mathematics, 15
percent failed chemistry, and 10 percent failed both mathematics and
chemistry. A student is selected at random.
a. If the student failed chemistry, what is the probability that he or she
failed mathematics?
b. If the student failed mathematics, what is the probability that he or she
failed chemistry?
c. What is the probability that the student failed mathematics or
chemistry?
d. What is the probability that the student failed neither mathematics nor
chemistry?
Solution:
Let M be the event represent the student failed Mathematics,
C be the event represent the student failed Chemistry.
a. It is required to find: P (M / C )  ? , the probability that the student
failed Mathematics given that he or she failed Chemistry
P (M C ) 0.10 10 2
P (M / C ) 


 .
P (C )
0.15 15 3
b. It is required to find probability that the student failed Chemistry given
that he or she failed Mathematics
P (C / M ) 
c. Since
P (M
P (M C ) 0.10 10 2


 .
P (M )
0.25 25 5
C )  P (M )  P (C )  P M
C   0.25  0.15  0.10  0.30.
d. Students who failed neither Mathematics nor Chemistry from the
complement of the set MUC
P (M C )C  1  P  M C   1  0.3  0.70.
Example 13
5
Let A and B be events with
P  A   0.6, P  B   0.3, and P  A B   0.2.
Find:
b . P  A B  c . P  A C  and P  B C 
a. P  A / B  and P  B / A 

d . P AC / B C
 and P  B
C
/ AC

Solution
a. By the definition of conditional probability
P A / B  
P  A B  0.2 2

 ,&
P B 
0.3 3
P B / A  
P  A B  0.2 1


P A 
0.6 6
b. P A
B   P A   P B   P A
 
P  B   1  P  B   1  0.3  0.7
P A
B  1 P A
d . P A / B  

0.7
P B 
P A
B  0.3 3
P B / A  


0.4 4
P A 
B   0.6  0.3  0.2  0.7
c . P A C  1  P  A   1  0.6  0.4,&
C
C
C
B
C
C
C
C
C

1  0.7 0.3 3

 ,&
0.7
0.7 7
C
C
C
Example 14
Let A and B be events with
P  A   0.6, P  B   0.3, and P  A
Find the probability that:
a. A does not occur
b. B Does not occur
c. A or B occur
d. Neither A nor B occur.
B   0.2.
Solution
a. By the Complement rule
 
P  not A   P A C  1  P  A   1  0.6  0.4.
6
 
b . P  not B   P B C  1  P  B   1  0.3  0.7.
c . By Addition Rule
P A
B   P A   P B   P A
B   0.6  0.3  0.2  0.7
d . P  Neither A nor B   1  P  A
B   1  0.7  0.3
Example 15
Find
P B / A 
If
a) A is subset of B
b) A and B are mutually exclusive (disjoint) events.
Solution
a. P  B / A  
P A B  P A 

1
P A 
P A 
b. P B / A  
P  A B  P  
0


0
P A 
P A  P A 
Example 16
A fair die is tossed twice. Find the probability of getting a 4, 5 or 6 on the
first toss and a 1, 2, 3, or 4 on the second toss.
Solution:
Let A be the event of getting 4, 5, or 6 on the first toss and
B be the event of getting 1, 2, 3, or 4 on the second toss.
Since the result of the 2nd toss does not depend on the result of the 1st toss,
then A and B are independent. Thus
A  4,5,6, & B  1,2,3,4.
3
4
and P (A )  , & P (B )  .
6
6
3 4
6 6
1
3
Then P (A B )  P (A ) . P (B )  .  .
Example 17
Three different boxes contain colored balls. Box I contains 2 red, 3 white,
and 5 blue balls all numbered 1, Box II contains 4 red, 1 white, and 3 blue
balls all numbered 2, while Box III contains 3 red, 4 white, and 3 blue balls
all numbered 3. The ball in the three boxes is mixed together in one box
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and then a ball is drawn. If the ball drawn was white what is the probability
that it came from box II.
Solution:
Let B1 be the event represent of Box I
B2 be the event represent of Box II
B3 be the event represent of Box III
W be the event represent of White ball
It is required to find: P  B 2 /W   ?
  P  B 2  P W / B 2 
P W 
P  B 2  P W / B 2 

P  B 1  P W / B 1   P  B 2  P W / B 2   P  B 3  P W
P  B 2 /W  
P B2 W
P W 
/ B3 
 8  1
1
  . 
1
 28   8 
28


 .
 10   3   8   1   10   4  3  1  4 8
  .     .     .  28 28 28
 28   10   28   8   28   10 
White
Ball
Box I
Box II
Box III
2R
4R
3R
3W
1W
4W
5B
3B
3B
28 Ball
Example 18
Suppose the following three different boxes are given:
Box A contains 3 red and 5 white marbles.
Box B contains 2 red and 1 white marbles.
Box C contains 2 red and 3 white marbles.
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A box is selected at random, and a marbles is randomly drawn from the
box. If the marble is red. Find the probability that it came from box A.
Solution:
Let R be the event represent red marble
It is required to find: P  A / R   ?
By Bays Theorem
P A R  P A  P R / A 
P A / R  

P R 
P R 
There are three paths leading to a red marble R, hence, by the total
probability
R
3
W
A
5
R
2
Boxes
B
2
1
W
R
C
3
W
P  R   P  A  P  R / A   P  B  P  R / B   P C  P  R /C 
1 3 1 2 1 2 173
 .  .  . 
.
3 8 3 3 3 5 360
Thus, we have
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P A / R  
P A R  P A  P R / A 

P R 
P R 
1 3
.
45
3
 8
 0.26.
173 173
360
Example 19
Three machine A, B, and c produce respectively 60%, 30%, and 10% of the
total number of items of a factory. The percentage of defective output of
these machines is 2%, 3%, and 4%.
a. An item is selected at random and is found defective. Find the
probability that the item was produced by machine C.
b. If an item is selected at random, find the probability that the item is
defective.
Solution:
a. Let D be the event represent defective item
It is required to find: P C / D   ?
By Bays Theorem
P C D  P C  P  D / C 
P C / D  

P D 
P D 

P C  P  D / C 
P  A  P  D / A   P  B  P  D / B   P C  P  D / C 
4
 0.10  . 0.04 
 .
 0.60  . 0.02    0.30  . 0.03   0.10 .  0.04  25
P  D   P  A  P  D / A   P  B  P  D / B   P C  P  D /C 
  0.60  .  0.02    0.30  .  0.03    0.10 .  0.04   0.025.

b.
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