Two dice are rolled, If both dice have six faces numbered $1,2,3,5,7,11$, then the probability that the sum…
Two dice are rolled, If both dice have six faces numbered $1,2,3,5,7,11$, then the probability that the sum of the numbers on upper most face is prime is
$\frac{1}{4}$
$\frac{3}{4}$
$\frac{1}{9}$
$\frac{2}{7}$
Solution
Two dice are rolled.
$\therefore \quad \mathrm{n}(\mathrm{S})=36$
A : Event that the sum of the numbers on upper most face is prime.
$\begin{aligned} \therefore \quad & A=\{(1,1),(1,2),(2,1),(2,3),(2,5),(2,11), \\ & (3,2),(5,2),(11,2)\}\end{aligned}$
$\therefore \quad \mathrm{P}(\mathrm{A})=\frac{9}{36}=\frac{1}{4}$