**Multiplication rule of probability**

Let E and F be two events associated with a sample space S. Clearly, the set E ∩ F denotes the event that both E and F have occurred. In other words, E ∩ F denotes the simultaneous occurrence of the events E and F. The event E ∩ F is also written as EF. Very often we need to find the probability of the event EF. For example, in the experiment of drawing two cards one after the other, we may be interested in finding the probability of the event ‘a king and a queen’. The probability of event EF is obtained by using the conditional probability as obtained below :

We know that the conditional probability of event E given that F has occurred is denoted by P(E|F) and is given by

From this result, we can write

(1)

Also, we know that

or

Thus

(2)

Combining (1) and (2), we find that

The above result is known as the **multiplication rule of probability**.

**Example** An urn contains 10 black and 5 white balls. Two balls are drawn from the urn one after the other without replacement. What is the probability that both drawn balls are black?

**Solution** Let E and F denote respectively the events that first and second ball drawn are black. We have to find P(E ∩ F) or P (EF).

Now

Also given that the first ball drawn is black, i.e., event E has occurred, now there are 9 black balls and five white balls left in the urn. Therefore, the probability that the second ball drawn is black, given that the ball in the first draw is black, is nothing but the conditional probability of F given that E has occurred.

i.e.

By multiplication rule of probability, we have

**Multiplication rule of probability for more than two events** If E, F and G are three events of sample space, we have

Similarly, the multiplication rule of probability can be extended for four or more events.

**Example** Three cards are drawn successively, without replacement from a pack of 52 well shuffled cards. What is the probability that first two cards are kings and the third card drawn is an ace?

**Solution** Let K denote the event that the card drawn is king and A be the event that the card drawn is an ace. Clearly, we have to find P (KKA)

Now

Also, P (K|K) is the probability of second king with the condition that one king has already been drawn. Now there are three kings in (52 − 1) = 51 cards.

Lastly, P(A|KK) is the probability of third drawn card to be an ace, with the condition that two kings have already been drawn. Now there are four aces in left 50 cards.

By multiplication law of probability, we have