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Question
A die is thrown three times. Events A and B are defined as below:
A : 5 on the first and 6 on the second throw.
B: 3 or 4 on the third throw.
Find the probability of B, given that A has already occurred.
Solution
A is an event of getting 5 on the first throw and 6 on the second throw.
Then
A={(5,6,1) (5,6,2) (5,6,3) (5,6,4) (5,6,5) (5,6,6)}
Also, B is an event of getting 3 or 4 on the third throw.
∴ A∩B={(5,6,3), (5,6,4)}
Required probability, `P(A|B)=(n(A∩B))/(n(A))=2/6=1/3`
Thus, the probability of B, given that A has already occurred is 1/3
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