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Question
A, B, and C try to hit a target simultaneously but independently. Their respective probabilities of hitting the target are `3/4, 1/2` and `5/8`. Find the probability that the target
- is hit exactly by one of them
- is not hit by any one of them
- is hit
- is exactly hit by two of them
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Solution
Let event A: A can hit the target,
event B: B can hit the target,
event C: C can hit the target.
∴ P(A) = `3/4`, P(B) = `1/2`, P(C) = `5/8`
∴ P(A') = 1 – P(A) = `1 - 3/4 = 1/4`
P(B') = 1 – P(B) = `1 - 1/2 = 1/2`
P(C') = 1 – P(C) = `1 - 5/8 = 3/8`
Since A, B, C are independent events,
A', B', C' are also independent events.
(a) Let event W: Target is hit exactly by one of them.
P(W) = 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)
= `(3/4 xx 1/2 xx 3/8) + (1/4 xx 1/2 xx 3/8) + (1/4 xx 1/2 xx 5/8)`
= `9/64 + 3/64 + 5/64`
= `17/64`
(b) Let event X: Target is not hit by any one of them.
P(X) = P(A' ∩ B' ∩ C')
= P(A') · P(B') · P(C')
`= 1/4 xx 1/2 xx 3/8`
`= 3/64`
(c) Let event Y: Target is hit.
P(Y) = 1 - P (target is not hit by any one of them)
`= 1 - 3/64`
`= 61/64`
(d) Let event Z: Target is hit by exactly two of them.
∴ 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)
`= (3/4 xx 1/2 xx 3/8) + (3/4 xx 1/2 xx 5/8) + (1/4 xx 1/2 xx 5/8)`
`= 9/64 + 15/64 + 5/64`
`= 29/64`
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