A die has 6 faces and its sample space S = {1,2,3,4,5,6}.
The total number of outcomes = 6.
Let P(A) be the probability of getting an even number.
The sample space of A = {2,4,6}
\(\therefore P(A)=\frac{3}{6}=\frac{1}{2}\)
Let P(B) be the probability of getting a number whose value is greater than 2.
The sample space of B = {3,4,5,6}
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 of getting a number greater than 2 given that the outcome is even is given by:
