Misc 4 - Suppose 90% of people are right handed. Probability - Binomial Distribution

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  1. Chapter 13 Class 12 Probability
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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’’^(๐’โˆ’๐’™) 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Ÿ—/๐Ÿ๐ŸŽ)^๐’™ (๐Ÿ/๐Ÿ๐ŸŽ)^(๐Ÿ๐ŸŽ โˆ’ 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 โ‰ค 6) = 1 โ€“ P(X โ‰ฅ 7) = 1 โ€“ ( "10C7" (9/10)^7 (1/10)^(10โˆ’7)+"10C8" (1/10)^8 (9/10)^(10โˆ’8)+"10C9" (1/10)^9 (9/10)^(10โˆ’9) +"10C10" (1/10)^10 (9/10)^(10โˆ’10)) = 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)โˆ‘_(๐’“ = ๐Ÿ•)^๐Ÿ๐ŸŽโ–’๐Ÿ๐ŸŽ๐‘ช๐’“ (๐ŸŽ.๐Ÿ—)^๐’“ (๐ŸŽ.๐Ÿ)^(๐Ÿ๐ŸŽโˆ’๐’“)

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Davneet Singh
Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 8 years. He provides courses for Maths and Science at Teachoo. You can check his NCERT Solutions from Class 6 to 12.