Question

Two different dice are thrown together, Find the probability that the numbers obtained have
(i) even sum, and
(ii) even product

Answer 1

When two different dice are thrown together, the sample space is given by

$\begin{array}{c}S=\\ \end{array}\left\{\begin{array}{c}(1,1)(1,2)(1,3)(1,4)(1,5)(1,6)\\ ::\\ ::\\ (6,1)(6,2)(6,3)(6,4)(6,5)(6,6)\end{array}\right\}$

$\therefore n\left(S\right)=36$

$A\equiv$ event that sum of nos is even.

$\begin{array}{c}A=\end{array}\left\{\begin{array}{c}(1,1)(1,3)(1,5)(2,2)(2,4)(2,6)\\ (3,1)(3,3)(3,5)(4,2)(4,4)(4,6)\\ (5,1)(5,3)(5,5)(6,2)(6,4)(6,6)\end{array}\right\}$

$\therefore n\left(A\right)=18$

$P\left(A\right)=\frac{n\left(A\right)}{n\left(S\right)}=\frac{18}{36}=\frac{1}{2}$

$B=$ event that the product is even.

$\begin{array}{c}B\equiv \\ \end{array}\left\{\begin{array}{c}(1,2)(1,4)(1,6)(2,1)(2,2)(2,3)\\ (2,4)(2,5)(2,6)(3,2)(3,4)(3,6)\\ (4,1)(4,2)(4,3)(4,4)(4,5)(4,6)(5,2)\\ (5,4)(5,6)(6,1)(6,2)(6,3)(6,4)(6,5)(6,6)\end{array}\right\}$

$\therefore n\left(B\right)=27$

$P\left(B\right)=\frac{n\left(B\right)}{n\left(S\right)}=\frac{27}{36}=\frac{3}{4}$