A die thrown twice will have a total outcome of 62 = 36.
Let P(A) be the probability of getting the number 5 at least once.
Let P(B) be the probability of getting sum = 8.
The sample space of B = {(2,6),(3,5),(4,4),(5,3),(6,2)}
\(\therefore P(B)=\frac{5}{36}\)
Let P(A ∩ B) be the probability of getting the number 5 at least once and the sum equal to 8
The sample space of (A ∩ B) = {(3,5)(5,3)}
\(\therefore P(A \cap B)=\frac{2}{36}=\frac{1}{18}\)
Tip – By conditional probability, P(A/B) = \(\frac{P(A \cap B)}{P(B)}\) where P(A/B) is the probability of occurrence of the event
A given that B has already occurred.
The probability that the number 5 have appeared at least once given that the sum = 8:
