Two men, A and B, independently shot at the same target once, with a hit rate of 0.6 and 0.7 respectively. Suppose the target is hit, what is its probability of being hit?

The probability that the target is not hit is (1-0.6) * (1-0.7) = 0.12.

The probability that the target is hit by both sides is 0.6*0.7=0.42.

The probability that the target is only hit by A is 0.6*( 1-0.7)=0. 18.

The probability that the target is only hit by B is 0.7*( 1-0.6)=0.28.

The probability that the target is hit by A is 0.6/(1-0.12) =15/22. The method in the university is relatively simple: let event A be hit by A, event B be hit by B, and event C be hit by the target. This problem is solved by P(A|C).

Then P(A)=0.6 and P(B)=0.7.

p(c)= 1-p(cNo)= 1-p(aNo)p(bNo)= 1-0.4 * 0.3 = 0.88。

So p (a | c) = p (AC)/p (c) = p (a)/p (c) = 0.6/0.88 = 0.68.