For two events A and B associated with a sample space \(S\)

\(P(A|B)\)

Where,\(P(B)~≠~0\)

\(P(A∩B) ~= ~P(B)× P(A|B)\)

\(P(B|A)~ = ~\frac{P(B∩A)}{P(A)}\)

Where, \(P(A)~≠~0\)

\(P(B∩A) ~= ~P(A)× P(B|A)\)

Since, \(P(A∩B) ~=~ P(B∩A)\)

\(P(A∩B)~ = ~P(A)× P(B|A)\)

From (1) and (2), we get:

\(P(A∩B) ~=~ P(B)× P(A|B)~ = ~P(A)× P(B|A)\)

\(P(A)~≠~0,P(B)~≠~0\)

The above result is known as multiplication rule of probability.

For independent events \(A\)

\(P(A∩B)~ = ~P(B) ~×~ P(A)\)

Illustration 1: An urn contains 20 red and 10 blue balls. Two balls are drawn from a bag one after the other without replacement. What is the probability that both the balls drawn are red?

Solution: Let A and B denote the events that first and second ball drawn are red balls. We have to find \(P(A∩B)\)

\(P(A)~ = ~P\)

Now, only 19 red balls and 10 blue balls are left in the bag. Probability of drawing a red ball in second draw too is an example of conditional probability where drawing of second ball depends on the drawing of first ball.

Hence Conditional probability of \(B\)

\(P(B|A) ~=~\frac{19}{29}\)

By multiplication rule of probability,

\(P(A∩B) ~= ~P(A)~× ~P(B|A)\)

\(P(A∩B)~ =~ \frac{20}{30} ~× ~\frac{19}{29} ~=~ \frac{38}{87}\)

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