Question

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

Answer 1

Let p denote the probability of getting a doublet in a single throw of a pair of dice. Then,
$$\begin{gathered} p=\frac{6}{36}=\frac{1}{6} \\ q=1-p=1-\frac{1}{6}=\frac{5}{6} \end{gathered}$$

p= Let X be the number of getting doublets in 4 throws of a pair of dice. Then, X follows a binomial distribution with n =4,
$$\begin{gathered} \mathrm{p}=\frac{1}{6}\text{and}\mathrm{q}=\frac{5}{6} \\ \mathrm{P}(\mathrm{X}=\mathrm{r})=\mathrm{Probability}\mathrm{of}\mathrm{getting}r\mathrm{doublets} \\ \mathrm{P}(\mathrm{X}=\mathrm{r})={}^{4}C_r\left(\frac{1}{6}{)}^{\mathrm{r}}\right(\frac{5}{6}{)}^{4-\mathrm{r}};r=0,1,2,3,4 \\ \text{If}\mathrm{X}=0,\text{t}\mathrm{hen}P(X=0)={}^{4}C_0\left(\frac{1}{6}{)}^0\right(\frac{5}{6}{)}^{4-\mathrm{r}} \\ \Rightarrow \mathrm{P}={\left(\frac{5}{6}\right)}^4 \\ \text{If}\mathrm{X}=1,\text{then}\mathrm{P}={}^{4}C_1{\left(\frac{1}{6}\right)}^1{\left(\frac{5}{6}\right)}^{4-1}=\frac{2}{3}{\left(\frac{5}{6}\right)}^3 \\ \text{If}\mathrm{X}=2,\text{then}\mathrm{P}={}^{4}C_2{\left(\frac{1}{6}\right)}^2{\left(\frac{5}{6}\right)}^{4-2}=\frac{1}{6}{\left(\frac{5}{6}\right)}^2 \\ \text{If}\mathrm{X}=3,\text{then}\mathrm{P}={}^{4}C_3{\left(\frac{1}{6}\right)}^3{\left(\frac{5}{6}\right)}^{4-3}=\frac{10}{3}{\left(\frac{1}{6}\right)}^3 \end{gathered}$$

$\text{If}\mathrm{X}=4,\text{then}\mathrm{P}={}^{4}C_4\left(\frac{1}{6}{)}^4\right(\frac{5}{6}{)}^{4-4}={\left(\frac{1}{6}\right)}^4=\frac{1}{1296}$



$$\begin{gathered} \mathrm{X}01234 \\ \mathrm{P}\left(\mathrm{X}\right)(\frac{5}{6}{)}^4\frac{2}{3}(\frac{5}{6}{)}^3\frac{25}{216}\frac{5}{324}\frac{1}{1296} \end{gathered}$$