Theorem: If A and B are two dependent events then the probability of occurrence of A given that B has already occurred and is denoted by P(A/B) is given by
Similarly, the probability of occurrence of B given that A has already occurred is given by
Proof: Let S be the sample space.
Interchange A and B in equation (i), we get
Example: Find the probability of drawing a heart on each of two consecutive draws from well shuffled-packs of cards if the card is not replaced after the draw.
Solution: Let event A is a heart on the first draw, and event B is a heart on the second draw.
When we get a heart on the first draw, the second draw has 51 outcomes and 12 are favorable.
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