# Ex 13.5, 10 - Chapter 13 Class 12 Probability (Term 2)

Last updated at Feb. 15, 2020 by Teachoo

Last updated at Feb. 15, 2020 by Teachoo

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Ex 13.5, 10 A person buys a lottery ticket in 50 lotteries, in each of which his chance of winning a prize is 1/100 . What is the probability that he will win a prize (a) at least once (b) exactly once (c) at least twice?Let X : Number of times he wins a prize Winning a prize on lottery is a Bernoulli trial So, X has a binomial distribution P(X = x) = nCx ๐^(๐โ๐) ๐^๐ Here, n = number of lotteries = 50 Let X : Number of times he wins a prize Winning a prize on lottery is a Bernoulli trial So, X has a binomial distribution P(X = x) = nCx ๐^(๐โ๐) ๐^๐ Here, n = number of lotteries = 50 p = Probability of winning a prize = 1/100 q = 1 โ p = 1 โ 1/100 = 99/100 Hence, P(X = x) = 50Cx (๐/๐๐๐)^๐ (๐๐/๐๐๐)^(๐๐โ๐) (a) Probability that he wins the lottery atleast once P (at least once) = P(X โฅ 1) = 1 โ P (0) = 1 โ 50C0 (1/100)^0 (99/100)^(50โ0) = 1 โ 1 ร 1 ร (99/100)^50 = 1 โ (99/100)^50 (b) Probability that he wins the lottery exactly once P (exactly once) = P(X = 1) = 50C1 (1/100)^1 (99/100)^(50โ1) = 50 ร 1/100 ร (99/100)^49 = 1/2 (99/100)^49 (c) Probability that he wins the lottery atleast twice P (atleast twice) = P(X โฅ 2) = 1 โ [P(X = 0) + P(X = 1)] = 1 โ ["50C0 " (1/100)^0 (99/100)^(50โ0) "+ 50C1 " (1/100)^1 (99/100)^(50โ1) ] = 1 โ [(99/100)^50 "+" 1/2 (99/100)^49 ] = 1 โ (99/100)^49 [99/100 "+" 1/2] = 1 โ (99/100)^49 [(99 + 50)/100] = 1 โ 149/100 (99/100)^49

Ex 13.5

Ex 13.5, 1
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Ex 13.5, 3 Important Deleted for CBSE Board 2022 Exams

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Ex 13.5, 10 Important Deleted for CBSE Board 2022 Exams You are here

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Chapter 13 Class 12 Probability (Term 2)

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Davneet Singh

Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 10 years. He provides courses for Maths and Science at Teachoo.