 # Multiplication Rule of Probability

The multiplication rule of probability explains the condition between two events. 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 and B, P(B|A) = P(B). The equation (2) can be modified into,

P(A∩B) = P(B) × P(A)

## Multiplication Theorem in Probability

We have already learned the multiplication rules we follow in probability, such as;

P(A∩B) = P(A)×P(B|A) ; if P(A) ≠ 0

P(A∩B) = P(B)×P(A|B) ; if P(B) ≠ 0

Let us learn here the multiplication theorems for independent events A and B.

If A and B are two independent events for a random experiment, then the probability of simultaneous occurrence of two independent events will be equal to product of their probabilities. Hence,

P(A∩B) = P(A).P(B)

Now, from multiplication rule we know;

P(A∩B) = P(A)×P(B|A)

Since A and B are independent, therefore;

P(B|A) = P(B)

Therefore, again we get;

P(A∩B) = P(A).P(B)

Hence, proved.

### 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) = 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) = 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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