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A pair of dice is thrown 4 times. If getting a doublet is considered a success, find the probability of two successes.
Solution:
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Verified Answer
$\mathrm{n}(\mathrm{S})=36, \mathrm{~A}=\{11,22,33,44,55,66\}$
$$
\mathrm{n}(\mathrm{A})=6, \mathrm{p}=\frac{n(A)}{n(S)}=\frac{6}{36}=\frac{1}{6}
$$
$$
\begin{aligned}
&\mathrm{q}=1-\mathrm{p}=1-\frac{1}{6}=\frac{5}{6}, \mathrm{n}=4, \mathrm{r}=2 \\
&\mathrm{P}(\mathrm{X}=\mathrm{r})=\mathrm{C}(\mathrm{n}, \mathrm{r}) \mathrm{p}^{\mathrm{r}} \mathrm{q}^{\mathrm{n}-\mathrm{r}} \\
&\mathrm{p}(\mathrm{X}=2)=\mathrm{C}(4,2)\left(\frac{1}{6}\right)^2\left(\frac{5}{6}\right)^2 \\
&=6 \times 25 \times\left(\frac{1}{6}\right)^4=\frac{25}{216}
\end{aligned}
$$
$$
\mathrm{n}(\mathrm{A})=6, \mathrm{p}=\frac{n(A)}{n(S)}=\frac{6}{36}=\frac{1}{6}
$$
$$
\begin{aligned}
&\mathrm{q}=1-\mathrm{p}=1-\frac{1}{6}=\frac{5}{6}, \mathrm{n}=4, \mathrm{r}=2 \\
&\mathrm{P}(\mathrm{X}=\mathrm{r})=\mathrm{C}(\mathrm{n}, \mathrm{r}) \mathrm{p}^{\mathrm{r}} \mathrm{q}^{\mathrm{n}-\mathrm{r}} \\
&\mathrm{p}(\mathrm{X}=2)=\mathrm{C}(4,2)\left(\frac{1}{6}\right)^2\left(\frac{5}{6}\right)^2 \\
&=6 \times 25 \times\left(\frac{1}{6}\right)^4=\frac{25}{216}
\end{aligned}
$$
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