Question

Two fair dice are rolled simultaneously. One of the dice shows four. The probability of other dice 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 the other dice
We have to find $P\left(B/A\right)=\frac{P(A\bigcap B)}{P\left(A\right)}=\frac{n(A\bigcap B)}{n\left(A\right)}$
$A=\left\{\right(1,4),(4,1\left)\right(2,4\left)\right(4,2\left)\right(3,4\left)\right(4,3\left)\right(4,4\left)\right(4,5\left)\right(5,4\left)\right(6,4\left)\right(4,6\left)\right\}andB=\left\{\right(4,6\left)\right(6,4\left)\right\}$
$\therefore P\left(\frac{B}{A}\right)=\frac{2}{11}.$