A die is tossed thrice. A success is 1 or 6 on a toss. Find the mean and variance of successes.

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A die is tossed thrice. A success is 1 or 6 on a toss. Find the mean and variance of successes.

Using Bernoulli’s Trial we have,
$P($ Success $=x)={ }^{n} C_{x} \cdot p^{x} \cdot q^{(n-x)}$ $x=0,1,2, \ldots \ldots . n$ and $q=(1-p), n=3$
$\begin{array}{l} p=2 / 6=1 / 3, q=4 / 6=2 / 3 \\ P(x=0)=P(\text { no success })=P \text { (all failures) }=\left(\frac{2}{3}\right)\left(\frac{2}{3}\right)\left(\frac{2}{3}\right)=\frac{8}{27} \end{array}$
$\mathrm{P}(\mathrm{x}=1)=\mathrm{P}(1$ success and 2 failures $)={ }^{3} \mathrm{C}_{1}\left(\frac{1}{3}\right)^{1}\left(\frac{2}{3}\right)^{2}=\frac{12}{27}$
$P(x=2)=P(2$ success and 1 failure $)={ }^{3} C_{2}\left(\frac{1}{3}\right)^{2}\left(\frac{2}{3}\right)^{1}=\frac{6}{27}$ $\mathrm{P}(\mathrm{x}=3)=\mathrm{P}($ all 3 success $)={ }^{3} \mathrm{C}_{3}\left(\frac{1}{3}\right)^{3}\left(\frac{2}{3}\right)^{0}=\frac{1}{27}$
$\therefore$ The probability distribution of the random variable $x$ is –
$\begin{array}{l} x: 0123 \\ P(x): \frac{8}{27} \frac{12}{27} \frac{6}{27} \frac{1}{27} \\ x_{1} p_{1} p_{1} x_{1} p_{1} x_{1}^{2} \\ 0 \frac{8}{27} 00 \\ 1 \frac{12}{27} \frac{12}{27} \frac{12}{27} \\ 2 \frac{6}{27} \frac{12}{27} \frac{24}{27} \\ 3 \frac{1}{27} \frac{3}{27} \frac{9}{27} \end{array}$
$1 \frac{45}{27}$
Mean $\mu=\sum p_{1} x_{1}=1$
$\begin{array}{l} \text { Variance }=\sigma^{2}=\Sigma \mathrm{p}_{1} \mathrm{x}_{1}{ }^{2}-\mu \\ \Rightarrow 5 / 3-1 / 1 \end{array}$