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Unit 7: Modern Approach to Probability

is drawn from U . The event W can occur with any one of the mutually exclusive events A, B Notes
2
and C.

( ) (W A
( ) 
P W P A .P / ) P B P+ ( ) (W /B ) P C P+ ( ) (W /C )
9 C 5 4 C 3 9 4 4 57
´
 2 ´ + 2 ´ + ´ 
13 13 13
C 11 C 11 C 11 143
2 2 2
Example 42: A bag contains tickets numbered as 112, 121, 211 and 222. One ticket is
drawn at random from the bag. Let E (i = 1, 2, 3) be the event that i th digit on the ticket is 2.
i
Discuss the independence of E , E and E .
1 2 3
Solution.

1
P E
The event E occurs if the number on the drawn ticket 211 or 222, therefore, ( )  . Similarly
1 1
2
1 1
( ) 
P E 2 and ( )  .
P E
2 3 2
i d 1
Now P E  E ji  (i, j = 1, 2, 3 and i  j).
4
Since (E  E j ) P E P E ( ) ( ) for i  j, therefore E , E and E are pair-wise independent.
P
i
i
j
2
1
3
1 b 1 P E b g b g b g
Further, P E  E  g   1 . P E . P E 3 , therefore, E , E and E are not mutually
E
2
3
2
1
2
3
independent. 4
Example 43: Probability that an electric bulb will last for 150 days or more is 0.7 and that
it will last at the most 160 days is 0.8. Find the probability that it will last between 150 to 160
days.
Solution.
Let A be the event that the bulb will last for 150 days or more and B be the event that it will last
at the most 160 days. It is given that P(A) = 0.7 and P(B) = 0.8.
The event A B is a certain event because at least one of A or B is bound to occur. Thus,
 b
 b
P A Bg  1. We have to find P A Bg . This probability is given by
P Ag b
P A Bg b + P Bg b  . + .  .  05
 b

P A Bg  07 08 10
.

Example 44: The odds that A speaks the truth are 2 : 3 and the odds that B speaks the truth
are 4 : 5. In what percentage of cases they are likely to contradict each other on an identical point?
Solution.
Let A and B denote the respective events that A and B speak truth. It is given that
2 4
P A and P B .
( )
( ) 
5 9
The event that they contradict each other on an identical point is given by A B d  i A  d Bi ,
where A B d i and A  d Bi are mutually exclusive. Also A and B are independent events. Thus,
we have
P é (A  B ) (A  B ù P (A  B ) ( A+ P  B ) P A ( ) ( ) ( ) ( )
P A
.P B
) 
.P B +
ë û
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