Question

Four fair dice are thrown independently $27$ times.

Then the expected number of times, at least two dice show up as a three or a five, is

Answer 1

Explanation for the correct option:

Step 1. Probability of getting $3$ or $5$:

Probability of getting $3$or $5$ $\left(p\right)=\frac{2}{6}$

$=\frac{1}{3}$

Probability of not showing $3$or $5$ $\left(q\right)=1-\frac{1}{3}$

$=\frac{2}{3}$

Step 2. Find the expected number of times :

Use the binomial distribution formula:

$\mathit{P}\mathbf{\left(}\mathit{x}\mathbf{\right)}\mathbf{=}{}^{\mathbf{n}}\mathit{C}{}_{\mathbf{x}}\mathit{p}^{\mathbf{x}}{\mathit{q}}^{\mathbf{n}\mathbf{-}\mathbf{x}}$

Where,

$n=$Total number of event

$x=$Total number of successful events

$p=$Probability of success

$q=$Probability of failure

The experiment is performed with four dices independently.

Step 3. Find the probability of showing $3$or $5$ at least twice in one throw of each dice is

$\Rightarrow {}^{4}C_4{p}^4+{}^{4}C_{3}q{p}^3+{}^{4}C_2{q}^2{p}^2=1\times p^4+4\times q{p}^3+6\times q^2{p}^2$

Putting the value of $p$ and $q$

$$\begin{gathered} {\left(\frac{1}{3}\right)}^4+4\times \frac{2}{3}\times {\left(\frac{1}{3}\right)}^3+6\times {\left(\frac{2}{3}\right)}^2\times {\left(\frac{1}{3}\right)}^2 \\ =\frac{1}{81}+\frac{8}{81}+\frac{24}{81} \\ =\frac{33}{81} \end{gathered}$$

Such experiment was performed $27$ times

So expected outcome

$\mu =\frac{33}{81}\times 27=11$

Hence, the expected number of times, at least two dice show up as a three or a five is $\mathbf{11}$.