# Example 25

Last updated at March 11, 2017 by Teachoo

Last updated at March 11, 2017 by Teachoo

Transcript

Example 25 (Method 1) 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) = 636 = 16 P(not getting a doublet) = 1 – 16 = 56 We need to find probability distribution of number of doublets in three throws of a pair of dice. Let X : Number of doublets Throwing a die is a Bernoulli trial So, X has a binomial distribution P(X = x) = nCx 𝒒𝒏−𝒙 𝒑𝒙 n = number of times pair of die is thrown = 3 p = Probability of getting doublet = 16 q = 1 – p = 1 – 16 = 56 Hence, P(X = x) = 3Cx 𝟏𝟔𝒙 𝟓𝟔𝟑 − 𝒙 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 Now, P(X = 0) = 3C0 160 563−0= 1 × 1 × 563= 125216 P(X = 1) = 3C1 161 563−1= 3 × 16 × 562= 75216 P(X = 2) = 3C2 162 563−2= 3 × 162× 56 = 5216 P(X = 3) = 3C3 163 560= 1 × 163 × 1 = 1216 So, probability distribution is Example 25 (Method 2) 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) = 636 = 16 P(not getting a doublet) = 1 – 16 = 56 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) = 56 × 56 × 56 = 125216 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) = 16 × 56 × 56 + 56 × 16 × 56 + 56 × 56 × 16 = 3 × 56 × 56 × 16 = 75216 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) = 16 × 16 × 56 + 16 × 56 × 16 + 56 × 16 × 16 = 3 × 16 × 16 × 56 = 15216 P(X = 3) P(X = 3) = P(three doublets on three throws) = P(one doublet) × P(one doublet) × P(one doublet) = 16 × 16 × 16 = 1216 So, probability distribution is

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Chapter 13 Class 12 Probability

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Davneet Singh

Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He provides courses for Mathematics from Class 9 to 12. You can ask questions here.