Question

Bill and George go target shooting together. Both shoot at atarget at the same time. Suppose Bill hits the target withprobability 0.7, whereas George independently, hits the target withprobability 0.4.(a) Given that exactly one shot hit the target, what is theprobability that it was George's shot?(b) Given that the target is hit, what is the probability thatGeorge hit it?

Answer 1

Answer: (a) $\frac{2}{9}$       (b) $\frac{6}{41}$

Step-by-step explanation:

(a) P( Bill hitting the target) = 0.7        P( Bill not hitting the target) = 0.3

    P( George hitting the target) = 0.4     P(George not hitting the target) = 0.6

Now the chances that exactly one shot hit the target is = 0.7 x 0.6 + 0.4 x 0.3

                                                                                            = 0.54

Chances that George hit the target is = 0.4 x 0.3 = 0.12

So given that exactly one shot hit the target, probability that it was George's shot = $\frac{0.12}{0.54}$ = $\frac{2}{9}$ .

(b) The numerator in the second part would be the same as of (a) part which is 0.12.

The change in the denominator will be that now we know that the target is hit so now in denominator we include the chance of both hitting the target at same time that is 0.4 x 0.7 and the rest of the equation is same as above i.e.

Given that the target is hit,probability that George hit it =                                                 $\frac{0.4\times 0.3}{0.7\times 0.6 +0.4\times 0.3+0.4\times 0.7}$  = = $\frac{6}{41}$