anonymous
  • anonymous
Kim has a strong first serve; whenever it is good (that is, in) she wins the point 70% of the time. Whenever her second serve is good, she wins the point 57% of the time. Seventy percent of her first serves and 76% of her second serves are good. What is the probability that Kim wins the point when she serves?
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
You have the following: P(winning the point) = P(winning point on first serve) + P(winning point on second serve) - P(winning on first and winning on second) = P(winning point on first serve) + P(winning point on second serve) (the probability of the intersection of events is zero since she can't have both a successful first and second serve when she's trying to win a point). Now, P(winning pt on first serve) = 0.7 x 0.7 = 0.49 (she wins serves successfully on the first serve 70% of the time, AND in these cases, wins it 70% of the time) P(winning pt on second serve) = 0.57 x 0.76 = 0.43332 So, subbing into the first equation, P(winning the point) = 0.49 + 0.43332 = 0.9232. Check that this makes sense/agrees with others...the day's over and my head's fried.

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