In a tennis match, on a given point, the player who is serving has two chances to hit the ball in play. The ball must fall in the correct marked box area on the opposite side of the net. A serve that misses the box is called a fault. Most players hit the first serve very hard, resulting in a fair chance of making a fault. If they do make a fault, they hit the second serve less hard and with some spin, making it more likely to be successful. Otherwise, with two misses - a “double fault”- they lose the point. The 2006 men’s champion in the Wimbledon tennis tournament was Roger Federer of Switzerland. During the tournament he made 56% of his first serves. So, hefaulted on the first serve 44% of the time. Given that he made a fault on his first serve, he made a fault on his second serve only 2% of the time. Assuming these are typical of his serving performance, when he serves, what is the probability that he makes a double fault?

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Asked Oct 1, 2019
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In a tennis match, on a given point, the player who is serving has two chances to hit the ball in play. The ball must fall in the correct marked box area on the opposite side of the net. A serve that misses the box is called a fault. Most players hit the first serve very hard, resulting in a fair chance of making a fault. If they do make a fault, they hit the second serve less hard and with some spin, making it more likely to be successful. Otherwise, with two misses - a “double fault”- they lose the point. The 2006 men’s champion in the Wimbledon tennis tournament was Roger Federer of Switzerland. During the tournament he made 56% of his first serves. So, he
faulted on the first serve 44% of the time. Given that he made a fault on his first serve, he made a fault on his second serve only 2% of the time. Assuming these are typical of his serving performance, when he serves, what is the probability that he makes a double fault?

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Expert Answer

Step 1

Let A be the event that there is fault in first serve

Therefore, probability of fault in first serve = P(A) = 0.44

Let B be the event that there is fault in second serve 

Therefore ,Probability that there is default in second serve =  P(B) 

Probability of double fault = Probability of fault in second serve and  fault in first serve = P(BnA)

Let C be the event that there is default in second serve given there is default in first serve

Therefore, P(C)=0.02 ...

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