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Unit 7: Modern Approach to Probability
(c) exactly two of them are able to solve the problem. Notes
(d) exactly one of them is able to solve the problem.
Solution.
Let A be the event that student A solves the problem. Similarly, we can define the events B and
C. Further, A, B and C are given to be independent.
(a) The problem is solved if at least one of them is able to solve it. This probability is given by
1 2 3 3
P (A C ) 1 P ( ) ( ) ( ) 1A .P B .P C ´ ´
B
2 3 4 4
Here we have to find P A B b g b B C g b A Cg
(b)
)
P é ë (A B ) (B C ) (A C ù û P A P B ( ) ( ) P A P C+ ( ) ( )
( ) ( ) P B P C+
( ) ( ) ( )
2P A P B P C
1 1 1 1 1 1 1 1 1 7
´ + ´ + ´ 2. ´ ´
2 3 3 4 2 4 2 3 4 24
C d
The required probability is given by P A B C i d A B i B Ci
d
(c) A
( ) ( ) ( )
( ) ( ) B +
( ) ( ) P+
P A .P P B .P C ( ) ( ) 3A .P C P A .P B .P C
1 1 1 1 1
+ + .
6 12 8 8 4
The required probability is given by P A B d C i d A B C i d A B Ci
(d)
( ) ( )
( ) P B+
P A ( ) P+ ( ) 2C P ( ) ( ) 2A .P B P B .P C
( ) ( ) ( )
2P ( ) ( ) 3A .P C + P A .P B .P C
1 1 1 1 1 1 1 11
+ + + .
2 3 4 3 6 4 8 24
Note that the formulae used in (a), (b), (c) and (d) above are the modified forms of corollaries
(following theorem 4) 3, 4, 5 and 6 respectively.
Example 32: A bag contains 2 red and 1 black ball and another bag contains 2 red and 2
black balls. One ball is selected at random from each bag. Find the probability of drawing (a) at
least a red ball, (b) a black ball from the second bag given that ball from the first is red; (c) show
that the event of drawing a red ball from the first bag and the event of drawing a red ball from
the second bag are independent.
Solution.
Let A be the event of drawing a red ball from the first bag and A be the event of drawing a red
1 2
ball from the second bag. Thus, we can write:
1 b 1 d 2i
´
A
n A g 2 2 4, n A A 2 2 4,
´
2
1 d
2i
1 d
2i
n A A 1 2 2, n A A 1 2 2
´
´
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