# Conditional Probability

- If occurrence of one event (
*B*) is influenced by the occurrence of another event (*A*), then the two events*A*and*B*are known as dependent events*A*and*B*be two events. Then, conditional probability of â€˜*B*given*A*â€™ is the probability of happening of*B*when it is known that*A*has already happened. On the other hand, the probability of happening of*B*when nothing is known about the happening of*A*is called unconditional probability of*B*.*B*given*A*â€™ is denoted by*P*(*B*/*A*). The unconditional probability is*P*(*B*).*P*(*A*) > 0. Then, conditional probability of event â€˜*B*given*A*â€™ is defined as -
*P*(*A*) â‰ 0 and*P*(*A*) > 0*P*(*A*) = 0, the conditional probability*P*(*B*/*A*) is not defined. -
*P*(*B*) â‰ 0 and*P*(*B*) > 0

Example

Two fair dice are rolled. If the sum of the numbers obtained is 4, find the probability that the numbers obtained on both the dice are even.

Solution

Let

Let

Here, we have to find

When two dice are rolled together, there can be 6 Ã— 6 = 36 outcomes.

Event

Event (

âˆ´

We know that

Thus,

*A*be the event that the sum of the numbers is 4.Let

*B*be the event that the numbers on both the dice are even.Here, we have to find

*P*(*B*/*A*).When two dice are rolled together, there can be 6 Ã— 6 = 36 outcomes.

Event

*A*has 3 favourable outcomes, namely (1, 3), (2, 2) and (3, 1)Event (

*A*âˆ©*B*),*i.e.*, the event that the sum of the numbers is 4 and the numbers are even, has 1 favourable outcome, namely (2, 2).âˆ´

*P*[sum 4 and numbers even]We know that

Thus,

*P*[Numbers even given sum is 4]

**Note:** If *A* and *B* are two independent events, then

*P*(*A* âˆ© *B*) = *P*(*A*) Ã— *P*(*B*) or *P*(*B*/*A*) = *P*(*B*) or *P*(*A*/*B*) = *P*(*A*)

# Multiplication Rule or Compound Probability or Joint probability

The probability of occurrence of two events*A*and

*B*simultaneously is known as the

*compound probability*or

*joint probability*of

*A*and

*B*. Denoted by

*P*(

*A*âˆ©

*B*).

Let *A* and *B* be two events with respective probability *P*(*A*) and *P*(*B*). Let *P*(*B*/*A*) be the conditional probability of event *B*, given that event *A* has happened. Then, the probability of simultaneous occurrence of *A* and *B* is

*P*(*A* âˆ© *B*) = *P*(*B*) Â· *P*(*A*/*B*) or *P*(*A* âˆ© *B*) = *P*(*A*) Â· *P*(*B*/*A*) (when the events occur together)

Example

A box has 1 red and 3 white ball. Two balls are drawn one after the other from the box. Find the probability that the two balls drawn would be red if the first ball drawn is returned to the box before the second ball is drawn.

Solution

Let event

Event

Here, Also, since the first ball is returned before the second ball is drawn,

âˆ´

*A*: the first ball drawn is redEvent

*B*: the second ball drawn is redHere, Also, since the first ball is returned before the second ball is drawn,

âˆ´

*P*[Two balls are red]

**Note:** De Morganâ€™s law for probability:

*P*(*A*â€² âˆ© *B*â€²) = *P*(*A* âˆª *B*)â€² = 1 âˆ’ *P*(*A* âˆª *B*)

*P*(*A*â€² âˆª *B*â€²) = *P*(*A* âˆ© *B*)â€² = 1 âˆ’ *P*(*A* âˆ© *B*)

Probability that only event *A* occurs and event *B* does not occur is