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.

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#### 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

Concept: Conditional Probability

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