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.
whats the probability that he missed his first serve?
(round to 4 decimal places. leave your answer in decimal form.)

Explanation:

Step1: Calculate the probability of making the first serve

The probability of making the first serve is \(P(\text{make first serve}) = 0.63\).

Step2: Use the law of total probability

Let \(A\) be the event of winning a point. Let \(M\) be the event of making the first serve and \(N\) be the event of missing the first serve.
We know that \(P(A|M)=0.78\) (probability of winning a point given first - serve made) and \(P(A|N) = 0.57\) (probability of winning a point given first - serve missed).
By the law of total probability \(P(A)=P(A|M)P(M)+P(A|N)P(N)\). Since \(P(M) = 0.63\), then \(P(N)=1 - P(M)=1 - 0.63 = 0.37\).
We want to find \(P(N)\) using Bayes' theorem. Bayes' theorem states that \(P(N|A)=\frac{P(A|N)P(N)}{P(A)}\). First, find \(P(A)\):
\(P(A)=0.78\times0.63 + 0.57\times0.37\)
\(P(A)=0.78\times0.63+0.57\times0.37=0.4914+0.2109 = 0.7023\)
Now, use Bayes' theorem:
\(P(N|A)=\frac{0.57\times0.37}{0.7023}\)
\(P(N|A)=\frac{0.2109}{0.7023}\approx0.299\)

Answer:

\(0.299\)