Getting a doublet is considered a success, find the probability of two successes.

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

Text Solution

Solution : Given, experiment is to toss a pair of dice 4 times. When a pair of dice is thrown once, the number of possible outcomes `= 6×6=36.` The number of possible doublets in this case is : `{(1,1), (2,2), (3,3), (4,4), (5,5), (6,6)}` P(getting a doublet on a roll of two dice) `= p=6/36=1/6` P(not getting a doublet ) = `q=1−p=5/6` Since X has a binomial distribution, the probability of x successes in n-Bernoulli trials, `P(X=x)=((n-x)!)/((n!)(x!))p^xq^(n−x)` =>`P(X=2)=((4-2)!)/((4!)(2!))(1/6)^2(5/6)^(2)` `=6(1/6)^2×(5/6)^2` `=6×1/36×25/36=25/216`

Last updated at May 29, 2018 by

Solve all your doubts with Teachoo Black (new monthly pack available now!)

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 𝒒﷮𝒏−𝒙﷯ 𝒑﷮𝒙﷯ 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 = 6﷮36﷯ = 1﷮6﷯ P(not getting a doublet) = q = 1 – 1﷮6﷯ = 5﷮6﷯ Hence, P(X = x) = 4Cx 𝟏﷮𝟔﷯﷯﷮𝒙﷯ 𝟓﷮𝟔﷯﷯﷮𝟒 − 𝒙﷯ We need to find probability of two successes. P(getting two successes) = P(getting 2 doublets) = P(X = 2) = 4C2 1﷮6﷯﷯﷮2﷯ 5﷮6﷯﷯﷮4 −2﷯ = 4!﷮ 4 − 2﷯! 2!﷯ 1﷮6﷯﷯﷮2﷯ 5﷮6﷯﷯﷮2﷯ = 4 × 3 × 2!﷮2! × 2!﷯ 1﷮6﷯﷯﷮2﷯ 5﷮6﷯﷯﷮2﷯ = 2 × 3 × 1﷮6 × 6﷯ × 25﷮36﷯ = 𝟐𝟓﷮𝟐𝟏𝟔﷯

Let X denote the number of times of getting doublet.

If a pair of dice is thrown, then there are total 36 possible outcomes, out of which 6 [i.e. (1, 1), (2, 2), …,(6, 6)] are doublets.

∴ P(getting a doublet) = p = `(6)/(36) = (1)/(6)`

∴ q = 1 – p = `1 - (1)/(6) = (5)/(6)`

Given, n = 3

∴ X ∼ B `(3, 1/6)`

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

P(X = x) = `""^3"C"_x (1/6)^x (5/6)^(3 - x) ,x ` = 0, 1, 2, 3

∴ P(two successes) = P(X = 2)

= `""^3"C"_2(1/6)^2(5/6)`

= `3 xx (1)/(36) xx (5)/(6)`

= `(5)/(72)`.

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

The repeated tosses of a pair of dice are Bernoulli trials. Let X denote the number of times of getting doublets in an experiment of throwing two dice simultaneously four times.

Probability of getting doublets in a single throw of the pair of dice is

Concept: Bernoulli Trials and Binomial Distribution

Is there an error in this question or solution?