Two die having 6 faces each when tossed simultaneously will have a total outcome of 62 = 36
Let P(A) be the probability of getting a sum greater than 8.
Let P(B) be the probability of getting 4 on the first die.
The sample space of B = {(4,1),(4,2),(4,3),(4,4),(4,5),(4,6)}
\(\therefore P(B)=\frac{6}{36}=\frac{1}{6}\)
Let P(A ∩ B) be the probability of getting 4 on the first die and the sum greater than or equal to 8
The sample space of (A ∩ B) = {(4,4),(4,5),(4,6)}
\(\therefore P(A \cap B)=\frac{6}{36}=\frac{1}{12}\)
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 sum of the numbers is greater than or equal to 8 given that 4 was thrown first:
