Question

A die is tossed thrice.

A success is getting $1$or $6$ on a toss.

The mean and the variance of the number of successes.

• A. $\mathrm{\mu }=1,{\mathrm{\sigma }}^2=\frac{2}{3}$
• B. $\mathrm{\mu }=\frac{2}{3},{\mathrm{\sigma }}^2=1$
• C. $\mathrm{\mu }=2,{\mathrm{\sigma }}^2=\frac{2}{3}$
• D. $\mathrm{None}\mathrm{of}\mathrm{these}$

Answer 1

The correct option is A

$\mathrm{\mu }=1,{\mathrm{\sigma }}^2=\frac{2}{3}$

The explanation of the correct option :

Step1. Define the possible events:

Total number of times die thrown $\left(n\right)=3$

Total number of outcomes$=6$

Getting $1\mathrm{or}6$ is a success $=\mathrm{p}$

Not getting $1\mathrm{or}6$ is a failure $=\mathrm{q}$

Step2. Calculate the probability of success and failure:

Probability of success $\text{'}\mathrm{p}\text{'}=\frac{2}{6}$

$=\frac{1}{3}$

Probability of failure $\text{'}\mathrm{q}\text{'}=\frac{4}{6}$

$=\frac{2}{3}$

Step3. Calculate mean and variance:

Mean$\mathbf{\text{'}}\mathbf{\mu }\mathbf{\text{'}}\mathbf{}\mathbf{=}\mathbf{}\mathit{n}\mathit{p}$

[Substitute the value of $\text{'}\mathrm{n}\text{'}\mathrm{and}\text{'}\mathrm{p}\text{'}$in the above formula]

$\therefore \mu =3\times \frac{1}{3}$

$\mathbf{=}\mathbf{1}$

And variance$\mathbf{\text{'}}{\mathbf{\sigma }}^{\mathbf{2}\mathbf{\text{'}}\mathbf{}}\mathbf{=}\mathbf{}\mathit{n}\mathit{p}\mathit{q}$

[Substitute the value of $\text{'}\mathrm{n}\text{'},\text{'}\mathrm{p}\text{'},\mathrm{and}\text{'}\mathrm{q}\text{'}$in the above formula]

$\therefore {\mathrm{\sigma }}^2=3\times \frac{1}{3}\times \frac{2}{3}$

$\mathbf{=}\mathbf{}\frac{\mathbf{2}}{\mathbf{3}}$

Hence, Option(A) is the correct answer.