# Multiplication Rule of Probability

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.

## Multiplication Rule of Probability Statement and proof

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,

$P(A∩B)~ = ~P(B) ~×~ P(A)$

### Multiplication Rule of Probability Example

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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