Two dice are thrown simultaneously. The probability of getting a multiple of 2 on one dice and multiple of 3…
Two dice are thrown simultaneously. The probability of getting a multiple of 2 on one dice and multiple of 3 on the other dice is
$\frac{5}{12}$
$\frac{11}{36}$
$\frac{13}{36}$
$\frac{5}{36}$
Solution
Two dice are thrown simultaneously.
$n(S)=6 \times 6=36$
Let $E$ be the event of getting multiple of 2 on one die and multiple of 3 on another.
$\begin{aligned} \Rightarrow \quad E_1= & \{(2,3),(3,2),(4,3),(3,4),(6,3),(3,6), \\ & (2,6),(6,2),(4,6),(6,4),(6,6)\}\end{aligned}$
$\begin{aligned} & \Rightarrow \quad n(E)=11 \\ & \therefore \quad P(E)=\frac{n(E)}{n(S)}=\frac{11}{36}\end{aligned}$