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,

$$P(A∩B) = P(B) × P(A)$$

Example: 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,

$$P(A∩B) = P(A) × P(B|A)$$

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