Ex 13.5, 11 Find the probability of getting 5 exactly twice in 7 throws of a die. Let X : Number of times getting 5 Die thrown is a Bernoulli trial So, X has a binomial distribution P(X = x) = nCx n = number of times die is thrown = 7 p = Probability of getting 5 = 1 6 q = 1 p = 1 1 6 = 5 6 Hence, P(X = x) = 7Cx We need to find Probability of getting 5 twice i.e. P(X = 2) Putting x = 2 in (1) P(X = 2) = 7C2 1 6 2 5 6 7 2 = 7! 7 2 ! 2 ! 1 6 2 5 6 5 = 7 ! 5 ! 2 ! 1 6 6 5 6 5 = 7 6 5 ! 5 ! 2 ! . 6 . 6 5 6 5 = 7 2 1 6 5 6 2 = 7 12 5 6 5 Required Probability =

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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.

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