Question

A die is thrown two times and the sum of the scores appearing on the die is observed to be a multiple of $4$. Then the conditional probability that the score $4$ has appeared at least once is

• A. $\frac{1}{3}$
• B. $\frac{1}{4}$
• C. $\frac{1}{8}$
• D. $\frac{1}{9}$

Answer 1

The correct option is D

$\frac{1}{9}$

Explanation for the correct option:

Step1. Define events

Total outcomes $=\left\{\left(1,1\right),\left(1,2\right),\left(1,3\right),\left(1,4\right),\left(1,5\right),\left(1,6\right),......,\left(5,6\right),\left(6,1\right),\left(6,2\right),\left(6,3\right),\left(6,4\right),\left(6,5\right)\left(6,6\right)\right\}$

Total number of outcomes $=36$

Sum obtained is multiple of $4$ $=\mathrm{A}$

The score $4$ appeared at least once $=\mathrm{B}$

Step2. Calculate the probability of $\mathbf{A}\mathbf{}\mathit{a}\mathit{n}\mathit{d}\mathbf{}\left(A\cap B\right)$

$\begin{array}{rcl}\mathrm{A}& =& \left\{\left(1,3\right),\left(2,2\right),\left(3,1\right),\left(2,6\right),\left(3,5\right),\left(4,4\right),\left(5,3\right),\left(6,2\right),\left(6,6\right)\right\}\\ \mathrm{P}\left(\mathrm{A}\right)& =& \frac{9}{36}\left[\mathbf{\because }\mathbf{Probability}\mathbf{}\mathbf{=}\mathbf{}\frac{\mathbf{Number}\mathbf{}\mathbf{of}\mathbf{}\mathbf{favourable}\mathbf{}\mathbf{outcomes}}{\mathbf{Total}\mathbf{}\mathbf{number}\mathbf{}\mathbf{of}\mathbf{}\mathbf{outcomes}}\right]\\ \mathrm{B}& =& \left\{\left(1,4\right),\left(2,4\right),\left(3,4\right),\left(4,4\right),\left(5,4\right),\left(6,4\right),\left(4,1\right),\left(4,2\right),\left(4,3\right),\left(4,5\right),(4,6)\right\}\\ \mathrm{A}\cap \mathrm{B}& =& \left\{\left(4,4\right)\right\}\\ \mathrm{P}\left(\mathrm{A}\cap \mathrm{B}\right)& =& \frac{1}{36}\end{array}$

Step3. Calculate the $\mathrm{P}\left(\frac{B}{A}\right)$

$\mathrm{P}\left(\frac{B}{A}\right)=\frac{\mathrm{P}\left(\mathrm{A}\cap \mathrm{B}\right)}{\mathrm{P}\left(\mathrm{A}\right)}$

Substitute the required value in the formula

$\begin{array}{rcl}\mathrm{P}\left(\frac{B}{A}\right)& =& \frac{{\displaystyle \frac{1}{36}}}{{\displaystyle \frac{9}{36}}}\\ & =& \frac{1}{9}\end{array}$

Hence, Option(D) is the correct answer.