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Social ScienceA basketball player has a free throw percentage of 93%. What is the probability she misses at least one of his next 27 free throws?
A basketball player has a free throw percentage of 93%. What is the probability she misses at least one of his next 27 free throws?
Expert Answer
The probability of she misses at least one of his next 27 free throws is obtained below:
Let X be the number of throws until she the first missed, 93% of player has a free throw, that is q = 0.93. 7% of does not make her free throw, that is p = 0.07 (1–0.93).
Therefore, the distribution is binomial distribution with sample size 27 and probability of success 0.07. That is, X~ Binomial (27, 0.07).
The formula for the probability of binomial distribution is,
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Aк -те-" "Вove-pт Р(х - х)- р)"" here x %3D0,1,2,....,п for 0sp<1 . ,00)-е
The required proba...
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P(X21)-1-P(X0) 27 =1- -Jaorya-o.0 07(1-0.0774 27-0 1-(1)(1)(0.93) 27 1-0.1409 0.8591
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