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Exercise 12.3 · Q12

Q.Suppose the chances of hitting a target by a person XX are 3 times in 4 shots, by YY are 4 times in 5 shots, and by ZZ are 2 times in 3 shots. They fire simultaneously exactly one time each. What is the probability that the target is damaged by exactly 2 hits?

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Step 1. Given. P(X hits)=34P(X\text{ hits})=\dfrac34, P(Y hits)=45P(Y\text{ hits})=\dfrac45, P(Z hits)=23P(Z\text{ hits})=\dfrac23, independent shots. So P(X misses)=14P(X\text{ misses})=\dfrac14, P(Y misses)=15P(Y\text{ misses})=\dfrac15, P(Z misses)=13P(Z\text{ misses})=\dfrac13.

Step 2. List the three ways to get exactly 2 hits. (X,Y hit,Z misses)(X,Y\text{ hit},Z\text{ misses}), (X,Z hit,Y misses)(X,Z\text{ hit},Y\text{ misses}), (Y,Z hit,X misses)(Y,Z\text{ hit},X\text{ misses}).

Step 3. Compute each term.

X,YX,Y hit, ZZ misses: 34×45×13=1260=15\dfrac34\times\dfrac45\times\dfrac13=\dfrac{12}{60}=\dfrac15. …

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