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Ex 13.5

Ex 13.5, 1
Deleted for CBSE Board 2023 Exams

Ex 13.5, 2 Deleted for CBSE Board 2023 Exams You are here

Ex 13.5, 3 Important Deleted for CBSE Board 2023 Exams

Ex 13.5, 4 Important Deleted for CBSE Board 2023 Exams

Ex 13.5, 5 Deleted for CBSE Board 2023 Exams

Ex 13.5, 6 Important Deleted for CBSE Board 2023 Exams

Ex 13.5, 7 Important Deleted for CBSE Board 2023 Exams

Ex 13.5, 8 Deleted for CBSE Board 2023 Exams

Ex 13.5, 9 Deleted for CBSE Board 2023 Exams

Ex 13.5, 10 Important Deleted for CBSE Board 2023 Exams

Ex 13.5, 11 Deleted for CBSE Board 2023 Exams

Ex 13.5, 12 Deleted for CBSE Board 2023 Exams

Ex 13.5, 13 Important Deleted for CBSE Board 2023 Exams

Ex 13.5, 14 (MCQ) Important Deleted for CBSE Board 2023 Exams

Ex 13.5, 15 (MCQ) Important Deleted for CBSE Board 2023 Exams

Chapter 13 Class 12 Probability

Serial order wise

Last updated at May 29, 2018 by Teachoo

Ex 13.5, 2 A pair of dice is thrown 4 times. If getting a doublet is considered a success, find the probability of two successes. Let X : be the number of doublets Throwing a pair of die is a Bernoulli trial So, X has binomial distribution P(X = x) = nCx 𝒒𝒏−𝒙 𝒑𝒙 Where n = number of times die is thrown = 4 Finding p, q If 2 dies are thrown, there are 6 × 6 = 36 outcomes Doublet: It means same number is obtained on both throws of die Number of doublets possible on 2 throws of die are (1, 1), (2, 2), (3, 3), (4, 4), (5, 5), (6, 6) P(getting a doublet) = p = 636 = 16 P(not getting a doublet) = q = 1 – 16 = 56 Hence, P(X = x) = 4Cx 𝟏𝟔𝒙 𝟓𝟔𝟒 − 𝒙 We need to find probability of two successes. P(getting two successes) = P(getting 2 doublets) = P(X = 2) = 4C2 162 564 −2 = 4! 4 − 2! 2! 162 562 = 4 × 3 × 2!2! × 2! 162 562 = 2 × 3 × 16 × 6 × 2536 = 𝟐𝟓𝟐𝟏𝟔