Question

A die is thrown twice and the sum of the numbers appearing is observed to be 8. What is the conditional probability that the number 5 has appeared at least once

Answer 1

It is given that a die is thrown twice.

So the total outcome $=6^2=36.$

Consider $P(A)$ as the probability of getting the number $5$ at least once. Consider $P(B)$ as the probability of getting the sum $=8.$

Sample space of B $\{(2,6), (3,5), (4,4), (5,3),(2,6)\}.$

$$P(B)=\dfrac{5}{36}$$

Consider $P(A\cap B)$ as the probability of getting the number $5$ at least once and the sum equal to $8.$

Sample space of $(A\cap B)=\{(3,5), (5,3)\}.$

$$P(A\cap B)=\dfrac{2}{36}=\dfrac{1}{18}$$

The conditional probability that the number 5 has appeared at least once given that the sum is $8$ is

$$P(A|B)=\dfrac{P(A\cap B)}{P(B)}=\dfrac{\dfrac{1}{18}}{\dfrac{5}{36}}=\dfrac{2}{5}$$