Ex 13.5, 10 - A person buys a lottery ticket in 50 lotteries

Ex 13.5, 10 - Chapter 13 Class 12 Probability - Part 2
Ex 13.5, 10 - Chapter 13 Class 12 Probability - Part 3 Ex 13.5, 10 - Chapter 13 Class 12 Probability - Part 4

Learn in your speed, with individual attention - Teachoo Maths 1-on-1 Class


Transcript

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

Ask a doubt
Davneet Singh's photo - Co-founder, Teachoo

Made by

Davneet Singh

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 13 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.