Question

Tennis great Roger Federer made 63% of his first serves in a recent season. When Federer made his first serve, he won 78% of the points. When Federer missed his first serve and had to serve again, he won only 57% of the points. Suppose you randomly choose a point on which Federer served. You get distracted before seeing his first serve but look up in time to see Federer win the point. What's the probability that he missed his first serve?

Answer 1

0.3

Explanation:

We need to reverse this because we only know the outcome. P(A|B) = P(B|A) * P(A) / P(B), so P(B|A) is B given that A happens, P(A) in this case is the probability that he missed the first serve. P(B) is the probability of winning the point. P(B) = P(B|A') * P(A') + P(B|A) * P(A), which means P(B) is the probability of the point scored given that the serve was made multiplied by the probability of the serve being made plus the probability of B given that serve is missed multiplied by the probability that the serve is missed. This is 57% * 37% + 78% * 63%=0.21+0.49=0.7. So P(B)=0.7. P(B|A)*P(A)/P(B) would be 57% * 37% / 70% = 0.3. The probability that Federer missed his first serve given that he scored the point is 0.3.

Answer 2

Sounds like a Bayes Theorem problem.

Events:

F: Roger makes his first serve. We'll write ~F for "not F".

P(F) = .63

P(~F) = 1 - .63 = .37

W: Roger wins the point. We don't know P(W) but we are given

P(W | F) = .78

P(W | ~F) = .57

We're asked for

P(~F | W)

The basic conditional probability theorem is

P(~F and W) = P(~F | W) P(W) = P(W | ~F) P(~F)

P(~F | W) = ( P(W | ~F) P(~F) ) / P(W)

We write

P(W) = P(W | F) P(F) + P(W | ~F) P(~F)

Substituting gives Bayes' Theorem:

P(~F | W) = ( P(W | ~F) P(~F) ) / ( P(W | F) P(F) + P(W | ~F) P(~F) )

We know all the parts so we substitute,

P(~F|W) = ( .57(.37) ) / (.78(.63) + .57(.37) ) = 0.30029901751388294

Let's call that 30%

Answer: 30%