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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?
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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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