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Question
Three persons A, B and C shoot to hit a target. If A hits the target four times in five trials, B hits it three times in four trials and C hits it two times in three trials, find the probability that:
1) Exactly two persons hit the target.
2) At least two persons hit the target.
3) None hit the target.
Solution
`P(A) = 4/5; P(B) = 3/4; P(C) = 2/3`
`P(A') = 1/5; P(B') = 1/4; P(C') = 1/3 `
1) P (Exactly two person hit the target)
= 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)
`= 4/5 xx 3/4 xx 1/3 + 4/5 xx 1/4 xx 2/3 + 1/5 xx 3/4 xx 2/3`
`= 12/60 + 8/60 + 6/60`
`= 26/60 = 13/30`
2) P (At least two person hit the target)
= P (Two person hit the target) + P (All three hit the target)
`= 26/60 + 4/5 xx 3/4 xx 2/3`
`= 26/60 + 24/60`
`= 50/60 = 5/6`
3) P (None hit the target)
= P(A' ∩ B' ∩ C')
`= P(A'). P(B').P(C')`
`= 1/5 xx 1/4 xx 1/3 = 1/60`
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