Question

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

Answer 1

The experiment is to toss a pair of dice (n=4) times. Getting a doublet is considered a success. It is a case of Bernoulli trials.

When a pair of dice is thrown once, the number of possible outcomes = $6\times 6=36$.

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) = $p=\frac{6}{36}=\frac{1}{6}$

P(not getting a doublet ) = $q=1-p=\frac{5}{6}$

X is a binomial distribution.

Here, $n=4,p=\frac{1}{6},q=\frac{5}{6}$

Since X has a binomial distribution, the probability of x successes in n-Bernoulli trials,

$P(X=x)={n_C}_x.p^x.q^{n-x}$

$P(X=2)={4_C}_{2}times(\frac{1}{6}{)}^2\times (\frac{5}{6}{)}^2$

$=6\times \frac{1}{36}\times \frac{25}{36}=\frac{25}{216}$