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Transcript

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 𝒒^(π’βˆ’π’™) 𝒑^𝒙 n = number of people = 10 p = Probability of getting right handed people = 90% = 90/100 = 9/10 q = 1 – p = 1 – 9/10 = 1/10 Hence, P(X = x) = 10Cx (πŸ—/𝟏𝟎)^𝒙 (𝟏/𝟏𝟎)^(𝟏𝟎 βˆ’ 𝒙) 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 ≀ 6) = 1 – P(X β‰₯ 7) = 1 – ( "10C7" (9/10)^7 (1/10)^(10βˆ’7)+"10C8" (1/10)^8 (9/10)^(10βˆ’8) + "10C9" (1/10)^9 (9/10)^(10βˆ’9) +"10C10" (1/10)^10 (9/10)^(10βˆ’10)) = 1 – ("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) = 1 – ("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) = 1 – βˆ‘_(π‘Ÿ = 7)^10β–’10πΆπ‘Ÿ (9/10)^π‘Ÿ (1/10)^(10βˆ’π‘Ÿ) = 1 – βˆ‘_(𝒓 = πŸ•)^πŸπŸŽβ–’πŸπŸŽπ‘ͺ𝒓 (𝟎.πŸ—)^𝒓 (𝟎.𝟏)^(πŸπŸŽβˆ’π’“)

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Davneet Singh

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 14 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.