# Multiplication Law of Probability

Multiplication law in probability applies to combination of events. When the events have to occur together then we make use of the multiplication law of probability. Now two cases arise: whether the events are independent or dependent.

The events are said of be independent only when the occurrence of one event does not change the probability of occurrence of any other event with it. The given events are said to be dependent when the occurrence of one event changes the probability of occurrence of any other event.

## Rules

The multiplication rule states that:
“The probability of occurrence of given two events or in other words the probability of intersection of two given events is equal to the product obtained by finding the product of the probability of occurrence of both events.”
This implies that if A and B are two events given then:

P (A and B) = P (A $\cap$ B) = P (A) * P (B)

This rule is applicable in all the cases, that is, when events are independent or dependent. In case when we have dependent events we have to be very careful in determining the probability of the second event after the occurrence of first event.

In such case the multiplication rule is modified as:

P (A and B) = P (A $\cap$ B) = P (A) * P (B|A)

Here, P (B|A) is the probability of occurrence of the second event B when the first event A has already occurred.

Let us see an example to understand this. We have two events of drawing two candies from a box one by one but with replacement. It is clear that here the given events are completely independent and thus we can multiply the probabilities of the given two events for finding out the probability of combined events.

Now, suppose the candies are taken from the box without putting the first one back. It is clear that the events are dependent and thus we need to find the conditional probability for finding the probability of occurrence of combined event.

## Problems

Let us see some examples on the multiplication law of probability.
Example 1:

A bag contains 3 pink candies and 7 green candies. Two candies are taken out from the bag with replacement. Find the probability that both candies are pink.

Solution:

Let A = event that first candy is pink and B = event that second candy is pink.

$\rightarrow$ P (A) = $\frac{3}{10}$ …(i)

Since the candies are taken out with replacement, this implies that the given events A and B are independent.

$\rightarrow$ P (B|A) = P (B) = $\frac{3}{10}$ …(ii)

Hence by the multiplication law we get,

P (A $\cap$ B) = P (A) * P (B|A)

$\rightarrow$ P (A $\cap$ B) = $\frac{3}{10}$ * $\frac{3}{10}$ [using (i) and (ii)]

= $\frac{9}{100}$ = 0.09

Example 2:

A bag has 4 white cards and 5 blue cards. We draw two cards from the bag one by one without replacement. Find the probability of getting both cards white.

Solution:

Let A = event that first card is white and B = event that second card is white.

From question, P (A) = $\frac{4}{9}$.

Now P (B) = P (B|A) because the events given are dependent on each other.

$\rightarrow$ P (B) = $\frac{3}{8}$.

So, P (A $\cap$ B) = $\frac{4}{9}$ * $\frac{3}{8}$ = $\frac{1}{6}$.