Question

A fair die is tossed repeatedly until a six is obtained. Let $X$ denote the number of tosses required. The probability that $X\ge 3$ equals

• A. $\frac{125}{216}$
• B. $\frac{25}{36}$
• C. $\frac{5}{36}$
• D. $\frac{25}{216}$

Answer 1

The correct option is B

$\frac{25}{36}$

Explanation for correct option:

Step 1: Define possible outcomes:

Possible outcomes of throwing a die $=\{1,2,3,4,5,6\}$

A fair die is tossed repeatedly until a six is obtained

Sample space $=\left\{6,\left(1,6\right),\left(2,6\right),\left(3,6\right),\left(4,6\right),\left(5,6\right),\left(1,1,6\right),\left(1,2,6\right)...........\infty \right\}$

Step 2: Calculate the probability that $X\ge 3$:

As we know,

$$\begin{gathered} \mathrm{P}\left(1\right)+\mathrm{P}\left(2\right)+\mathrm{P}\left(3\right)+\mathrm{P}\left(4\right)+.....+\mathrm{P}\left(\infty \right)=1 \\ \Rightarrow \mathrm{P}\left(\mathrm{X}\ge 3\right)=1-\left[\mathrm{P}\left(1\right)+\mathrm{P}\left(2\right)\right] \\ \Rightarrow \mathrm{P}\left(\mathrm{X}\ge 3\right)=1-\left[\frac{1}{6}+\left(\frac{1}{6}\right)\left(\frac{5}{6}\right)\right] \\ \Rightarrow \mathrm{P}\left(\mathrm{X}\ge 3\right)=1-\left[\frac{11}{36}\right] \\ \Rightarrow \mathrm{P}\left(\mathrm{X}\ge 3\right)=\mathbf{}\frac{25}{36} \end{gathered}$$

Hence, Option(B) is the correct answer.