Question

A fair die is tossed repeatedly until a six is obtained. Let $X$ denote the number of tosses required. The conditional probability that $X\ge 6$ given $X>3$ equals

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

Answer 1

The correct option is D

$\frac{25}{36}$

Explanation for correct option:

Step 1: Define possible outcomes:

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

The number of tosses requires to get six on the die $=X$

Probability of not obtaining six on a fair die $=\frac{5}{6}$

Step 2: Find the conditional probability that $X\ge 6$ is given $X>3$:

As per conditional probability,

$P\left(\frac{X\ge 6}{X>3}\right)=\frac{P\left[\left(X\ge 6\right)\cap \left(X>3\right)\right]}{P\left(X>3\right)}....\left(1\right)\mathbf{[}\mathbf{\because }\mathbf{P}\mathbf{\left(}\frac{\mathbf{A}}{\mathbf{B}}\mathbf{\right)}\mathbf{}\mathbf{=}\frac{\mathbf{P}\mathbf{(}\mathbf{A}\mathbf{\cap }\mathbf{B}\mathbf{)}}{\mathbf{P}\mathbf{\left(}\mathbf{B}\mathbf{\right)}}\mathbf{]}$

We know,

$$\begin{gathered} \Rightarrow P\left(X\ge 6\right)=P\left(6\right)+P\left(7\right)+P\left(8\right)+......+\infty \\ \Rightarrow P\left(X>3\right)=P\left(4\right)+P\left(5\right)+P\left(6\right)+P\left(7\right)+......+\infty \end{gathered}$$

$$\begin{gathered} \therefore P\left[\left(X\ge 6\right)\cap \left(X>3\right)\right]=P\left(6\right)+P\left(7\right)+.....+\infty \\ =P\left(X\ge 6\right) \end{gathered}$$

Step 3: Calculate the value of all possible conditions:

$P\left(X\ge 6\right)=\frac{5^5}{6^6}+\frac{5^6}{6^7}+\frac{5^7}{6^8}+......+\infty$

$\begin{array}{rcl}& =& \frac{5^5}{6^6}\left(\frac{1}{1-{\displaystyle \frac{5}{6}}}\right)\end{array}$ $\begin{array}{rcl}& & \left[\because S_{\infty }=\frac{a}{1-r}\right]\end{array}$ where $a$is the first term and $r$ is the common difference

$\begin{array}{rcl}& =& \frac{5^5}{6^6}\left(\frac{6}{1}\right)\\ & =& \frac{5^5}{6^5}......\left(2\right)\end{array}$

$P\left(X>3\right)=\frac{5^3}{6^4}+\frac{5^4}{6^5}+\frac{5^5}{6^6}+......+\infty$

$$\begin{gathered} =\frac{5^3}{6^4}\left(\frac{1}{1-{\displaystyle \frac{5}{6}}}\right) \\ =\frac{5^3}{6^4}\left(\frac{6}{1}\right) \\ =\frac{5^3}{6^3}......\left(3\right) \end{gathered}$$

Substitute the value of equations $\left(3\right)\mathrm{and}\left(2\right)\mathrm{in}\left(1\right)$ we have,

$\mathrm{P}\left(\frac{\mathrm{X}\ge 6}{\mathrm{X}>3}\right)=\frac{{\displaystyle \frac{5^5}{6^5}}}{{\displaystyle \frac{5^3}{6^3}}}$

$$\begin{gathered} =\frac{5^2}{6^2} \\ =\frac{\mathbf{25}}{\mathbf{36}} \end{gathered}$$

Hence, Option(D) is the correct answer.