For two events A and B associated with a sample space \(S\), the set \(A∩B\) denotes the events in which both event \(A\) and event \(B\) have occurred. Hence, \((A∩B)\) denotes the simultaneous occurrence of the events \(A\) and \(B\). The event A∩B can be written as \(AB\). The probability of event \(AB\) is obtained by using the properties of conditional probability. We know that the conditional probability of event \(A\) given that \(B\) has occurred is denoted by \(P(A|B)\) and is given by:

\(P(A|B)\) = \(\frac{P(A∩B)}{P(B)}\)

Where,\(P(B)~≠~0\).

\(P(A∩B) ~= ~P(B)× P(A|B)\) ……………………………………..(1)

\(P(B|A)~ = ~\frac{P(B∩A)}{P(A)}\)

Where, \(P(A)~≠~0\).

\(P(B∩A) ~= ~P(A)× P(B|A)\)

Since, \(P(A∩B) ~=~ P(B∩A)\)

\(P(A∩B)~ = ~P(A)× P(B|A)\) ………………………………………(2)

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

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

\(P(A)~≠~0,P(B)~≠~0\).

The above result is known as multiplication rule of probability.

For independent events \(A\) and \(B\), \(P(B|A)~ = ~P(B)\). The equation (2) can be modified into,

\(P(A∩B)~ = ~P(B) ~×~ P(A)\)

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 \(P(A∩B)\) or \(P(AB)\).

\(P(A)~ = ~P\)(red balls in first draw) = \(\frac{20}{30}\)

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 \(B\) on \(A\) will be,

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

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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