


Ex 13.5
Ex 13.5, 2 Deleted for CBSE Board 2022 Exams
Ex 13.5, 3 Important Deleted for CBSE Board 2022 Exams
Ex 13.5, 4 Important Deleted for CBSE Board 2022 Exams
Ex 13.5, 5 Deleted for CBSE Board 2022 Exams
Ex 13.5, 6 Important Deleted for CBSE Board 2022 Exams
Ex 13.5, 7 Important Deleted for CBSE Board 2022 Exams
Ex 13.5, 8 Deleted for CBSE Board 2022 Exams
Ex 13.5, 9 Deleted for CBSE Board 2022 Exams
Ex 13.5, 10 Important Deleted for CBSE Board 2022 Exams You are here
Ex 13.5, 11 Deleted for CBSE Board 2022 Exams
Ex 13.5, 12 Deleted for CBSE Board 2022 Exams
Ex 13.5, 13 Important Deleted for CBSE Board 2022 Exams
Ex 13.5, 14 (MCQ) Important Deleted for CBSE Board 2022 Exams
Ex 13.5, 15 (MCQ) Important Deleted for CBSE Board 2022 Exams
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 π^(πβπ) π^π 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 π^(πβπ) π^π Here, n = number of lotteries = 50 p = Probability of winning a prize = 1/100 q = 1 β p = 1 β 1/100 = 99/100 Hence, P(X = x) = 50Cx (π/πππ)^π (ππ/πππ)^(ππβπ) (a) Probability that he wins the lottery atleast once P (at least once) = P(X β₯ 1) = 1 β P (0) = 1 β 50C0 (1/100)^0 (99/100)^(50β0) = 1 β 1 Γ 1 Γ (99/100)^50 = 1 β (99/100)^50 (b) Probability that he wins the lottery exactly once P (exactly once) = P(X = 1) = 50C1 (1/100)^1 (99/100)^(50β1) = 50 Γ 1/100 Γ (99/100)^49 = 1/2 (99/100)^49 (c) Probability that he wins the lottery atleast twice P (atleast twice) = P(X β₯ 2) = 1 β [P(X = 0) + P(X = 1)] = 1 β ["50C0 " (1/100)^0 (99/100)^(50β0) "+ 50C1 " (1/100)^1 (99/100)^(50β1) ] = 1 β [(99/100)^50 "+" 1/2 (99/100)^49 ] = 1 β (99/100)^49 [99/100 "+" 1/2] = 1 β (99/100)^49 [(99 + 50)/100] = 1 β 149/100 (99/100)^49