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Question
The probability that a student X solves a problem in dynamics is `2/5` and the probability that student Y solves the same problem is `1/4`. What is the probability that
- the problem is not solved
- the problem is solved
- the problem is solved exactly by one of them
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Solution
Let event A: Student X solves the problem in dynamics,
event B: Student Y solves the problem in dynamics.
∴ P(A) = `2/5`, P(B) = `1/4`
∴ P(A') = 1 – P(A) = `1 - 2/5 = 3/5`
P(B') = 1 – P(B) = `1 - 1/4 = 3/4`
A and B are independent events,
A' and B' are also independent events
(i) Let event C: Problem is not solved.
∴ P(C) = P(A' ∩ B')
= P(A') · P(B')
= `3/5 xx 3/4`
= `9/20`
(ii) Let event D: Problem is solved.
Problem can be solved if at least one of the two students solves the problem.
∴ P(C) = P(at least one student solves the problem)
= 1 - P (no student solves the problem)
= 1 - P(A' ∩ B')
= 1 - P (A') · P (B')
`= 1 - 3/5 xx 3/4`
`= 1 - 9/20`
`= 11/20`
(iii) Let event E: The problem is solved exactly by one of them.
∴ P(E) = P(A' ∩ B) ∪ P(A ∩ B')
= P(A') · P (B) + · P (A) · P (B')
`= (3/5 xx 1/4) + (2/5 xx 3/4)`
`= 3/20 + 6/20`
`= 9/20`
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