


Learn in your speed, with individual attention - Teachoo Maths 1-on-1 Class
Bernoulli Trial
Question 2 Deleted for CBSE Board 2024 Exams
Question 3 Important Deleted for CBSE Board 2024 Exams
Question 4 Important Deleted for CBSE Board 2024 Exams
Question 5 Deleted for CBSE Board 2024 Exams
Question 6 Important Deleted for CBSE Board 2024 Exams
Question 7 Important Deleted for CBSE Board 2024 Exams
Question 8 Deleted for CBSE Board 2024 Exams
Question 9 Deleted for CBSE Board 2024 Exams
Question 10 Important Deleted for CBSE Board 2024 Exams You are here
Question 11 Deleted for CBSE Board 2024 Exams
Question 12 Deleted for CBSE Board 2024 Exams
Question 13 Important Deleted for CBSE Board 2024 Exams
Question 14 (MCQ) Important Deleted for CBSE Board 2024 Exams
Question 15 (MCQ) Important Deleted for CBSE Board 2024 Exams
Last updated at May 29, 2023 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