


Binomial Distribution
Ex 13.5, 12 Deleted for CBSE Board 2022 Exams
Ex 13.5, 2 Deleted for CBSE Board 2022 Exams
Ex 13.5, 4 Important Deleted for CBSE Board 2022 Exams
Ex 13.5, 9 Deleted for CBSE Board 2022 Exams
Ex 13.5, 6 Important Deleted for CBSE Board 2022 Exams
Ex 13.5, 11 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
Example 32 Important Deleted for CBSE Board 2022 Exams
Ex 13.5, 13 Important Deleted for CBSE Board 2022 Exams
Ex 13.5, 3 Important Deleted for CBSE Board 2022 Exams
Misc 6 Important Deleted for CBSE Board 2022 Exams
Misc 7 Important
Ex 13.5, 7 Important Deleted for CBSE Board 2022 Exams
Example 31 Important Deleted for CBSE Board 2022 Exams
Ex 13.5, 5 Deleted for CBSE Board 2022 Exams
Ex 13.5, 10 Important Deleted for CBSE Board 2022 Exams You are here
Misc 5 Important Deleted for CBSE Board 2022 Exams
Misc 9 Deleted for CBSE Board 2022 Exams
Misc 4 Deleted for CBSE Board 2022 Exams
Misc 10 Important
Example 35 Deleted for CBSE Board 2022 Exams
Ex 13.5, 8 Deleted for CBSE Board 2022 Exams
Example 34 Deleted for CBSE Board 2022 Exams
Binomial Distribution
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 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