InterviewSolution
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InterviewSolution
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Suppose that 90% of people areright-handed. What is the probability that at most 6 of a random sample of 10people are right-handed? |
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Answer» A person can be either right-handed or left-hended. It is given that `90%` of the people are right-handed. Therefore, `p=P("right-handed")=9/10` `q=P("left-handed")=1-9/10=1/10` Using binomial distribution, the probability that at most six people are right-handed is given by `underset(r=0)overset(6)sum""^(10)C_(r)p^(r)q^(n-r)underset(r=0)overset(6)sum""^(10)C_(r)((9)/(10))^(r)((1)/(10))^(10-r)` Therfore, the probability that at most 6 people are right-handed `underset(r=7)overset(10)sum""^(10)C_(r)(0.9)^(r)(0.1)^(10-r)` |
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