A die is thrown two times and the sum of the scores appearing on the die is observed to be a multiple of 4 .…

A die is thrown two times.

Total number of possible outcomes$=36$.

Let $E$ is the event that sum of the score appearing on the die is multiple of $4$.

Possible outcomes in favor of $E=\left\{\left(1, 3\right), \left(3, 1\right), \left(2, 2\right), \left(2, 6\right), \left(6, 2\right), \left(3, 5\right), \left(5, 3\right), \left(4, 4\right), \left(6, 6\right)\right\}$.

Number of favorable outcomes$=9$.

$P\left(E\right)=\frac{1}{4}$.

Let $F$ is the event that at least one $4$ appear.

Possible outcomes in favor of $F=\left\{\left(1, 4\right), \left(4, 1\right), \left(2, 4\right), \left(4, 2\right), \left(3, 4\right), \left(4, 3\right), \left(4, 4\right), \left(4, 5\right), \left(5, 4\right), \left(4, 6\right), \left(6, 4\right)\right\}$

$E\cap F=\left\{\left(4, 4\right)\right\}$

$\Rightarrow P\left(E\cap F\right)=\frac{1}{36}$.

We have to find conditional probability of occurrence of at least one $4$ given that sum of scores is multiple of $4$.

$P\left(F/E\right)=\frac{P\left(E\cap F\right)}{P\left(E\right)}=\frac{{\displaystyle \frac{1}{36}}}{{\displaystyle \frac{1}{4}}}=\frac{1}{9}$.