# The chance of throwing a total of 3 or 5 or 11 wih two dice is

$\begin {array} {1 1} (A)\;\large\frac{5}{36} & \quad (B)\;\large\frac{1}{9} \\ (C)\;\large\frac{2}{9} & \quad (D)\;\large\frac{19}{36} \end {array}$

3 can be thrown as (1,2)(2,1)
5 can be thrown as (1,4)(4,1)(2,3)(3,2)
11 can be thrown as (5,6)(6,11)
$\therefore$ Required probability = $\large\frac{8}{36}$
$= \large\frac{2}{9}$
Ans (C)