# Example 25 - Chapter 13 Class 12 Probability (Term 2)

Last updated at Feb. 17, 2020 by Teachoo

Last updated at Feb. 17, 2020 by Teachoo

Transcript

Example 25 Find the probability distribution of number of doublets in three throws of a pair of dice. 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) = 6/36 = 1/6 P(not getting a doublet) = 1 – 1/6 = 5/6 We need to find probability distribution of number of doublets in three throws of a pair of dice. Since Two dies are thrown thrice. We can get, 0 doublet or 1 doublet or 2 doublets or 3 doublets So, value of X can be 0, 1, 2, 3 P(X = 0) P(X = 0) = P(0 doublet on three throws) = P(0 doublet) × P(0 doublet) × P(0 doublet) = 5/6 × 5/6 × 5/6 = 125/216 P(X = 1) P(X = 1) = P(one doublet on three throws) = P(one doublet) × P(0 doublet) × P(0 doublet) + P(0 doublet) × P(one doublet) × P(0 doublet) + P(0 doublet) × P(0 doublet) × P(one doublet) = 1/6 × 5/6 × 5/6 + 5/6 × 1/6 × 5/6 + 5/6 × 5/6 × 1/6 = 3 × 5/6 × 5/6 × 1/6 = 75/216 P(X = 2) P(X = 2) = P(two doublet on three throws) = P(one doublet) × P(one doublet) × P(0 doublet) + P(one doublet) × P(0 doublet) × P(one doublet) + P(0 doublet) × P(one doublet) × P(one doublet) = 1/6 × 1/6 × 5/6 + 1/6 × 5/6 × 1/6 + 5/6 × 1/6 × 1/6 = 3 × 1/6 × 1/6 × 5/6 = 15/216 P(X = 3) P(X = 3) = P(three doublets on three throws) = P(one doublet) × P(one doublet) × P(one doublet) = 1/6 × 1/6 × 1/6 = 1/216 So, probability distribution is

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Chapter 13 Class 12 Probability (Term 2)

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About the Author

Davneet Singh

Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 10 years. He provides courses for Maths and Science at Teachoo.