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

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

$\therefore q=1-p=1-\frac{1}{6}=\frac{5}{6}$

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

$\therefore \mathrm{P}(\mathrm{X}=x)={ }^{*} \mathrm{C}_{3} q^{p-x} p^{x}$, where $x=0,1,2,3 \ldots n$

$={ }^{+} \mathrm{C}_{x}\left(\frac{5}{6}\right)^{4-x} \cdot\left(\frac{1}{6}\right)^{x}$

$={ }^{4} \mathrm{C}_{x} \cdot \frac{5^{4-x}}{6^{4}}$

$\therefore \mathrm{P}(2$ successes $)=\mathrm{P}(\mathrm{X}=2)$

$={ }^{4} \mathrm{C}_{2} \cdot \frac{5^{4-2}}{6^{4}}$

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

$=\frac{25}{216}$