Question

Two fair dice are rolled simultaneously. If one of the dice shows four, then the probability of other die showing six, is equal to:

• A. $\frac{2}{11}$
• B. $\frac{1}{18}$
• C. $\frac{1}{6}$
• D. $\frac{1}{36}$

Answer 1

The correct option is A $\frac{2}{11}$

Let A be the event of getting four on at least one of the dice and B be the event of showing 6 on other dice.
Then we have to find $P\left(\frac{B}{A}\right)$
$P\left(\frac{B}{A}\right)=\frac{P(B\cap A)}{P\left(A\right)}=\frac{P(B\cap A)}{P\left(A\right)}A=(1,4),(4,1),(4,2),(2,4),(3,4),(4,3),(4,4),(4,5),(5,4),(6,4),(4,6)$
and $A\cap B=\left\{\right(4,6),(6,4\left)\right\}$.
$\therefore P\left(\frac{B}{A}\right)=\frac{2}{11}$