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Question
Suppose the chances of hitting a target by a person X is 3 times in 4 shots, by Y is 4 times in 5 shots, and by Z is 2 times in 3 shots. They fire simultaneously exactly one time. What is the probability that the target is damaged by exactly 2 hits?
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Solution
Given
Probability X hitting the target P(X) = `3/4`
Probability Y hitting the target P(Y) = `4/5`
Probability Z hitting the target P(Z) = `2/3`
`"P"(bar"X")` = 1 – P(X) = `1 - 3/4 = 1/4`
`"P"(bar"Y")` = 1 – P(Y) = `1 - 4/5 = 1/5`
`"P"(bar"Z")` = 1 – P(Z) = `1 - 2/3 = 1/3`
Probability hitting the target exactly by 2 hits
= `"P"[("X" ∩ "Y" ∩ bar"Z") ∪ (bar"X" ∩ "Y" ∩ "Z") ∪ ("X" ∩bar"Y" ∩ "Z")]`
= `"P"("X" ∩ "Y" ∩ bar"Z") + "P"(bar"X" ∩ "Y" ∩ "Z") + "P"("X" ∩bar"Y" ∩ "Z")`
= `"P"("X") "P"("Y") "P"(bar"Z") + "P"(bar"X") "P"("Y") "P"("Z") + "P"("X") "P"(bar"Y") "P"("Z")`
= `3/4 xx 4/5 xx 1/3 + 1/4 xx 4/5 xx 2/3 + 3/4 xx 1/5 xx 2/3`
= `(12 + 8 + 6)/60`
= `26/60`
= `13/30`
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