Question

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

Answer 1

Hint-Since there are 2 dices so total possible outcomes are 36 then count the probability of getting even number on the first die and the probability of getting a sum of 8, add both the probabilities then subtract it from the probability of getting an even number on first die and a sum of 8.

Complete step-by-step answer:
Given that 2 dices are thrown so the total number of possible outcomes n(S)=$6 \times 6$ $ = 36$.Now let us consider that A=getting an even number on first die and B= getting sum of 8
n(A)$ = 18$ ,B=(2,6),(6,2),(3,5),(5,3),(4,4)$ \Rightarrow $ n(B)=$5$ then,
P(A)=n(A)/n(S) =$\dfrac{{18}}{{36}}$ and P(B)=n(B)/n(S)=$\dfrac{5}{{36}}$
Getting an even number on first die and total of 8 is (2,6),(6,2)(4,4)
Now the event of getting both A and B n(A∩B)=3,then
P(A∩B)=n(A∩B)/n(S)=$\dfrac{3}{{36}}$
Therefore required possibility =P(A)+P(B)-P(A∩B)=$\dfrac{{18}}{{36}} + \dfrac{5}{{36}} - \dfrac{3}{{36}} = \dfrac{{20}}{{36}} = \dfrac{5}{9}$.

Note: You can also solve this question by directly using the formula$ [{n(A)+n(B)-n(A∩B)}/n(S)]$ and you will get the required answer.