Sum

A pair of dice is thrown 4 times. If getting a doublet is considered a success, find the probability of two successes.

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#### Solution

Let X = number of doublets.

p = probability of getting a doublet when a pair of dice is thrown

∴ p = `6/36 = 1/6` and

q = 1 - p = `1 - 1/6 = 5/6`

Given: n = 4

∴ X ~ B`(4, 1/6)`

The p.m.f. of X is given by

P(X = x) = `"^nC_x p^x q^(n - x)`

i.e. p(x) = `"^nC_x (1/6)^x (5/6)^(4 - x)`, x = 0, 1, 2, 3, 4

∴ P(2 successes) = P(X = 2)

= p(2) = `"^4C_2 (1/6)^2 (5/6)^(4 - 2)`

`= (4!)/(2!*2!) (1/6)^2 (5/6)^2`

`= (4 * 3 * 2!)/(2!.2.1) xx 1/36 xx 25/36`

`= 25/216`

Hence, the probability of two successes is `25/216`.

Concept: Binomial Distribution

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