# 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

Example 7
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Example 6 Important

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Example 11 Important

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Example 25 Important You are here

Example 26 Important

Example 27 Important

Example 28 Important

Example 29 Important

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Example 31 Important

Example 32 Important

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Class 12

Important Question for exams Class 12

- Chapter 1 Class 12 Relation and Functions
- Chapter 2 Class 12 Inverse Trigonometric Functions
- Chapter 3 Class 12 Matrices
- Chapter 4 Class 12 Determinants
- Chapter 5 Class 12 Continuity and Differentiability
- Chapter 6 Class 12 Application of Derivatives
- Chapter 7 Class 12 Integrals
- Chapter 8 Class 12 Application of Integrals
- Chapter 9 Class 12 Differential Equations
- Chapter 10 Class 12 Vector Algebra
- Chapter 11 Class 12 Three Dimensional Geometry
- Chapter 12 Class 12 Linear Programming
- Chapter 13 Class 12 Probability

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CA Maninder Singh

CA Maninder Singh is a Chartered Accountant for the past 8 years. He provides courses for Practical Accounts, Taxation and Efiling at teachoo.com .