
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 You are here
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
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, 9 On a multiple choice examination with three possible answers for each of the five questions, what is the probability that a candidate would get four or more correct answers just by guessing ? Let X : be the number of correct answers by guessing Guessing an answer is a Bernoulli trial So, X has binomial distribution P(X = x) = nCx 𝒒𝒏−𝒙 𝒑𝒙 n = number of questions = 5 p = Probability of getting correct answer = 13 q = 1 – p = 1 – 13= 23 ∴ P(X = x) = 5Cx 𝟏𝟑𝒙 𝟐𝟑𝟓−𝒙 Probability that he would get four or more correct answers = P(X ≥ 4) = P(4) + P(5) = 5𝐶4 134 231+ 5𝐶5 135 230 = 5 !4 ! 2 35 + 1 35 = 5 × 2 + 1 35 = 10 + 1 35 = 11 35 = 𝟏𝟏𝟐𝟒𝟑