Question

A pair of dice is thrown. What is the probability of getting an even number on the first die or a total of $8$?

Answer 1

The total number of possible outcomes is $36$.
i.e., $n\left(S\right)=36$
Let $A$: Event of getting an even number on the first die

For this condition, the first die can show $1$ out of $3$ even numbers and the second die can show any number among the $6$ possible outcomes.

$n\left(A\right)=3\times 6=18$

$B$=Event of getting the sum as $8$

Possible outcomes are=$\left\{\right(2,6),(6,2),(3,5),(5,3),(4,4\left)\right\}$

$n\left(B\right)=5$

And, $(A\cap B)=$$\left\{\right(2,6),(6,2),(4,4\left)\right\}$

$n(A\cap B)=3$

$\therefore$ Required probability

$=\frac{n\left(AorB\right)}{n\left(S\right)}$ $=\frac{n\left(A\right)+n\left(B\right)-n(A\cap B)}{n\left(S\right)}=\frac{20}{36}=\frac{5}{9}$