Question

Two dice are thrown once. Find the probability of getting an even number on the $1st$ die or a total of $8$.

Answer 1

Step 1: Find the number of favourable outcomes:

Sample space for throwing two dice will be,

$$\begin{gathered} n\left(S\right)={\text{(number of possible outcomes of throwing a dice)}}^{\text{number of dices thrown}} \\ n\left(S\right)={\left(6\right)}^2 \\ n\left(S\right)=36 \end{gathered}$$

Let A be the event of getting an even number on the first die and B be the event of getting a total of $8$.

The favourable outcomes of A will be,

$=(2,1),(2,2),(2,3),(2,4),(2,5),(2,6),(4,1),(4,2),(4,3),(4,4),(4,5),(4,6),(6,1),(6,2),(6,3),(6,4),(6,5),(6,6)$

Therefore, the number of favourable outcomes will be $18$.

The favourable outcomes of B will be,

$=(2,6),(3,5),(4,4),(5,3),(6,2)$

Therefore, the number of favourable outcomes will be $5$.

Step 2: Find the number of favourable outcomes which are common to both A and B:

$A\cap B=(2,6),(4,4),(6,2)$

Therefore, the number of common favourable outcomes are $n(A\cap B)=3$.

Step 3: Use the formula for probability:

$\text{Probability =}\frac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}}$

The probability of each outcome will be,

$$\begin{gathered} P\left(A\right)=\frac{18}{36} \\ P\left(B\right)=\frac{5}{36} \\ P(A\cap B)=\frac{3}{36} \end{gathered}$$

Thus, the probability will be,

$$\begin{gathered} P\left(AUB\right)=P\left(A\right)+P\left(B\right)–P(A\cap B) \\ =\frac{18}{36}+\frac{5}{36}-\frac{3}{36} \\ =\frac{18+5-3}{36} \\ =\frac{20}{36} \\ =\frac{5}{9} \end{gathered}$$

Final answer:

Hence, the probability of getting an even number on the $1st$ die or a total of $8$ is $\frac{5}{9}$.