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A problem in statistics is given to three students A, B, and C. Their chances of solving the problem are 13, 14, and 15 respectively. If all of them try independently, what is the probability that

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Question

A problem in statistics is given to three students A, B, and C. Their chances of solving the problem are `1/3`, `1/4`, and `1/5` respectively. If all of them try independently, what is the probability that, exactly two students solve the problem?

Sum
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Solution

Let A be the event that student A can solve the problem.
B be the event that student B can solve problem.
C be the event that student C can solve problem.

∴ P(A) = `1/3`, P(B) = `1/4` and P(C) = `1/5`

P(A') = 1 − P(A) = `1-1/3=2/3`

P(B') = 1 − P(B) = `1-1/4=3/4`

P(C') = 1 − P(C) = `1-1/5=4/5`
Since A, B, C are independent events
∴ A', B', C' are also independent events

Let Z be the event that exactly two students solve the problem.
∴ P(Z) = P(A ∩ B ∩ C') ∪ P(A ∩ B' ∩ C) ∪ P(A' ∩ B ∩ C)
= P(A) · P(B) · P(C') + P(A) · P(B') · P(C) + P(A') · P(B) · P(C)

`=(1/3xx1/4xx4/5) + (1/3xx3/4xx1/5) + (2/3xx1/4xx1/5)`

= `4/60+3/60+2/60`

= `9/60`

= `3/20`

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Chapter 7: Probability - Exercise 7.4 [Page 107]

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