# Question 10 - Bernoulli Trial - Chapter 13 Class 12 Probability

Last updated at April 16, 2024 by Teachoo

Chapter 13 Class 12 Probability

Serial order wise

Last updated at April 16, 2024 by Teachoo

Question 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 𝒒^(𝒏−𝒙) 𝒑^𝒙 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 𝒒^(𝒏−𝒙) 𝒑^𝒙 Here, n = number of lotteries = 50 p = Probability of winning a prize = 1/100 q = 1 – p = 1 − 1/100 = 99/100 Hence, P(X = x) = 50Cx (𝟏/𝟏𝟎𝟎)^𝒙 (𝟗𝟗/𝟏𝟎𝟎)^(𝟓𝟎−𝒙) (a) Probability that he wins the lottery atleast once P (at least once) = P(X ≥ 1) = 1 − P (0) = 1 − 50C0 (1/100)^0 (99/100)^(50−0) = 1 − 1 × 1 × (99/100)^50 = 1 − (99/100)^50 (b) Probability that he wins the lottery exactly once P (exactly once) = P(X = 1) = 50C1 (1/100)^1 (99/100)^(50−1) = 50 × 1/100 × (99/100)^49 = 1/2 (99/100)^49 (c) Probability that he wins the lottery atleast twice P (atleast twice) = P(X ≥ 2) = 1 – [P(X = 0) + P(X = 1)] = 1 – ["50C0 " (1/100)^0 (99/100)^(50−0) "+ 50C1 " (1/100)^1 (99/100)^(50−1) ] = 1 – [(99/100)^50 "+" 1/2 (99/100)^49 ] = 1 – (99/100)^49 [99/100 "+" 1/2] = 1 – (99/100)^49 [(99 + 50)/100] = 1 − 149/100 (99/100)^49