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Question
A problem in Mathematics is given to three students whose chances of solving it are `1/3, 1/4` and `1/5`. What is the probability that exactly one of them will solve it?
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Solution
`"P"(bar"A"_1) = 1 - "P"("A"_1) = 1 - 1/3 = 2/3`
`"P"(bar"A"_1) = 1 - "P"("A"_2) = 1 - 1/4 = 3/4`
`"P"(bar"A"_1) = 1 - "P"("A"_3) = 1 - 1/5 = 4/5`
Probability of Exactly one student solving the problem
= `"P" [("A"_1 ∩ bar"A"_2 ∩ bar"A"_3) ∪ (bar"A"_1 ∩ "A"_2 ∩ bar"A"_3) ∪ (bar"A"_1 ∩ bar"A"_2 ∩ "A"_3)]`
= `"P"("A"_1 ∩ bar"A"_2 ∩ bar"A"_3) + "P"(bar"A"_1 ∩ "A"_2 ∩ bar"A"_3) + "P"(bar"A"_1 ∩ bar"A"_2 ∩ "A"_3)`
= `"P"("A"_1) "P"(bar"A"_2) "P"(bar"A"_3) + "P"(bar"A"_1) "P"("A"_2) "P"(bar"A"_3) + "P"(bar"A"_1) "P"(bar"A"_2) "P"("A"_3)`
= `1/3 xx 3/4 xx 4/5 + 2/3 xx 1/4 xx 4/5 + 2/3 xx 3/ xx 1/5`
= `(12 + 8 + 6)/60`
= `26/16`
= `13/30`
[Probability of exactly one student solving the problem = Probability [(A1 solving the problem and A2, A3 non solving the problem) or (A1, A3 non solving and A2 solving) or (A1, A2 solving and A3 non solving)]
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