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Last updated at May 29, 2018 by Teachoo
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Misc 4 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? Let X : be the number of right handed people Picking people is a Bernoulli trial So, X has binomial distribution P(X = x) = nCx ๐^(๐โ๐) ๐^๐ n = number of people = 10 p = Probability of getting right handed people = 90% = 90/100 = 9/10 q = 1 โ p = 1 โ 9/10 = 1/10 Hence, P(X = x) = 10Cx (๐/๐๐)^๐ (๐/๐๐)^(๐๐ โ ๐) We need to find probability that at most 6 of a random sample of 10 people are right-handed P(at most 6 are right handed) = P(X โค 6) = 1 โ P(X โฅ 7) = 1 โ ( "10C7" (9/10)^7 (1/10)^(10โ7)+"10C8" (1/10)^8 (9/10)^(10โ8)+"10C9" (1/10)^9 (9/10)^(10โ9) +"10C10" (1/10)^10 (9/10)^(10โ10)) = 1 โ ("10C7" (9/10)^7 (1/10)^3+"10C8" (1/10)^8 (9/10)^2+"10C9" (1/10)^9 (9/10)^1+"10C10" (1/10)^10 (9/10)^0)โ_(๐ = ๐)^๐๐โ๐๐๐ช๐ (๐.๐)^๐ (๐.๐)^(๐๐โ๐)
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