A pair of fair and ordinary dice is rolled simultaneously. It is found that they show different outcomes. The probability that the sum of the out comes will be either 6 or 10, is equal to

$\frac{1}{5}$

$E_1$ : Event that dice show different outcomes.

$E_2$ : Outcome is either 6 or 10

$P\left(E_1\right)=\frac{5}{6}, P\left(E_1 \cap E_2\right)=\frac{4+2}{36}=\frac{1}{6}$

As 6 can occur in four ways namely

(2, 4), (4, 2) (1, 5), (5, 1)

and ten can occur in 2 ways namely (6, 4) and (4, 6)

Thus, required probability,

$P\left(E_1 / E_2\right)=\frac{P\left(E_1 \cap E_2\right)}{P\left(E_1\right)}=\frac{1 / 6}{5 / 6}=\frac{1}{5}$