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Question
Suppose that 90% of people are right-handed. What is the probability that at most 6 of a random sample of 10 people are right-handed?
Solution
Let p be the probability of success that people are right-handed
⇒ `p = 90/100 = 9/10`
and `q = 1 - p = 1 - 9/10 = 1/10`
∴ X has a binomial distribution with
`n = 10, p = 9/10, q = 1/10`
∴ P (X = r) = nCr (q)n-r pr
Required probability = P(at most 6 of 10 people are right-handed)
= P (X ≤ 6) = 1 - P (7 ≤ X ≤ 10)
`= 1 - sum_(r =17)^10 ""^10C_r (9/10)^r (1/10)^(10-r)`
`= 1- sum_(r=7)^10 ""^10C_r (0.9)^r (0.1)^(10-r)`
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