Question

A fair die is rolled. The probability that the first time $1$ occurs at the even throw is

• A. $\frac{1}{6}$
• B. $\frac{5}{11}$
• C. $\frac{6}{11}$
• D. $\frac{5}{36}$

Answer 1

The correct option is B

$\frac{5}{11}$

Explanation for correct option:

Step1. Define events:

Event of getting $1$, when a fair die is rolled$=E$

Event of not getting $1$, when a fair die is rolled $=\overline{E}$

Possible outcomes$=\left\{1,2,3,4,5,6\right\}$

Step2. Calculate the probability:

$$\begin{gathered} P\left(E\right)=\frac{1}{6} \\ P\left(\overline{E}\right)=\frac{5}{6} \end{gathered}$$

The probability that the first time $1$ occurs at the even throw

Even throws are two times, four times…….$\infty$$$\begin{gathered} P\left(\overline{E}\right)P\left(E\right)+P\left(\overline{E}\right)P\left(\overline{E}\right)P\left(\overline{E}\right)P\left(E\right)+P\left(\overline{E}\right)P\left(\overline{E}\right)P\left(\overline{E}\right)P\left(\overline{E}\right)P\left(\overline{E}\right)P\left(E\right)+......+\infty \\ =\left(\frac{5}{6}\right)\left(\frac{1}{6}\right)+\left(\frac{5}{6}\right)\left(\frac{5}{6}\right)\left(\frac{5}{6}\right)\left(\frac{1}{6}\right)+..........+\infty \\ =\left(\frac{5}{6}\right)\left(\frac{1}{6}\right)\left[1+\frac{25}{36}+{\left(\frac{25}{36}\right)}^2+....+\infty \right] \\ =\frac{{\displaystyle \frac{5}{36}}}{\left(1-{\displaystyle \frac{25}{36}}\right)}\left[\mathbf{\because }\mathit{I}\mathit{n}\mathbf{}\mathit{g}\mathit{e}\mathit{o}\mathit{m}\mathit{e}\mathit{t}\mathit{r}\mathit{i}\mathit{c}\mathbf{}\mathit{p}\mathit{r}\mathit{o}\mathit{g}\mathit{r}\mathit{e}\mathit{s}\mathit{s}\mathit{i}\mathit{o}\mathit{n}\mathbf{,}\mathbf{}{\mathit{S}}_{\mathbf{\infty }}\mathbf{=}\frac{\mathbf{a}}{\mathbf{1}\mathbf{-}\mathbf{r}}\mathbf{}\mathit{w}\mathit{h}\mathit{e}\mathit{r}\mathit{e}\mathbf{}\mathbf{\text{'}}\mathit{a}\mathbf{\text{'}}\mathbf{=}\mathit{f}\mathit{i}\mathit{r}\mathit{s}\mathit{t}\mathbf{}\mathit{t}\mathit{e}\mathit{r}\mathit{m}\mathbf{}\mathbf{,}\mathbf{}\mathit{r}\mathbf{=}\mathbf{}\mathit{c}\mathit{o}\mathit{m}\mathit{m}\mathit{o}\mathit{n}\mathbf{}\mathit{r}\mathit{a}\mathit{t}\mathit{i}\mathit{o}\right] \end{gathered}$$$=\frac{\mathbf{5}}{\mathbf{11}}$

Hence, Option(B) is the correct answer.