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I've been working on a test review for class, and this question popped up.

Fred is serving in tennis. 60% of the time his first serve goes in. If it doesn’t he serves again and 90% of the time his second serve goes in. If his first serve goes in then he wins 70% of the time, while he wins 40% of the time if his second serve goes in, and he loses if neither serve goes in. Find P(his first serve went in | he won).

Now, I know how to solve normal conditional problems (by dividing the intersect of A and B by the probability of B), but in this case, I'm not sure how I would go about finding the intersect of A and B. Should I just multiply the probability that his first serve went in and multiply it by the probability that he won? Is it possible this a problem that requires Baye's Theorem?

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    $P(A\cap B)=P(A)P(B\mid A)$ is the "*multiplication principle*" of probability, which you should recognize is just a rearrangement of terms from the definition of conditional probability. – 2017-02-26

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Bayes' Theorem helps, but you can do it from first principles.

How can Fred win? Well, two ways. Either his first goes in and he wins or his first serve misses, second serve goes in, and he wins. Thus the total probability that he wins is $$.6\times .7 +.4\times .9\times .4 = .564$$

That is your denominator. The numerator is that portion of that probability which is explained by Fred getting his first serve in. Thus your answer is $$\frac {.6\times .7}{.6\times .7 +.4\times .9\times .4}\approx \boxed {.745}$$