Asked by mark
                Data shows that 93% of the people in a certain population are right-handed. A group of 9 people from this population are selected at random. What is the probability that exactly 7 of the people in the group are right-handed?
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        2
Answers
                    Answered by
            Reiny
            
    prob of right-handed = .93
prob of NOT right-handed = .07
prob (7 of 9 are right-handed)
= C(9,7)(.93)^7(.07)^2
= appr. 0.10614
    
prob of NOT right-handed = .07
prob (7 of 9 are right-handed)
= C(9,7)(.93)^7(.07)^2
= appr. 0.10614
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