Question

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

Answer 1

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

$\therefore q=1-p=1-\frac{1}{6}=\frac{5}{6}$
Clearly, X has the binomial distribution with n = 4, $p=\frac{1}{6}$ and $q=\frac{5}{6}$

$\therefore (X=x)= ^nC_xq^{n-x}p^x$ where x = 0,1,2,3….n

= $^4C_x\bigg(\frac{5}{6}\bigg)^{4-x}.\bigg(\frac{1}{6}\bigg)^x$

= $^4C_x.\frac{5^{4-x}}{6^{4}}$

∴ P (2 successes) = P (X = 2)
= $^4C_2.\frac{5^{4-2}}{6^{4}}$

= 6.$\frac{25}{1296}$

= $\frac{25}{216}$