When probability of an event A is given and we have to find the probability of other event B based on that event A, then the probability obtained is called conditional probability of B given A. It is denoted by P(B/A).

P(B/A) is read as probability of B given A.Example: -

Let us consider an experiment of rolling a die. Let A denote the event of getting 1 or 3.The outcomes of tossing a die is {1, 2, 3, 4, 5, 6}. Let B denote the event of getting an odd number. Suppose that the event B has already occurred.

If it is known that we got an odd number, the outcomes favorable are {1, 3, 5}

So the probability of getting 1 or 3 is 2/3. This is under the condition we got an odd number.

Therefore P(A/B) = 2/3 is the conditional probability of event A given event B has already occurred.

## Conditional Probability Formula

**P(A/B) = P(A∩B)/P(B)**

## Multiplication rule of Probability

### Multiplication rule for any two events

If A and B are any two events, then the combined probability of both A and B, denoted by

**P(A∩B) is given by**

**P(A∩B)**

**=**

**P(A/B)x**

**P(B)**

It can be also written as

**P(A∩B)**

**=**

**P(B/A)x**

**P(A)**

### Multiplication rule for two independent events

When A and B are two independent events( that is events that does not dependent on each other), then the combined probability of A and B is given by

**P(A∩B)**

**=**

**P(A)x**

**P(B)**

Let us consider a few examples.

Example 1: -

If P(A) = 1/12, P(B) = 1/3 and P(A∩B) = 1/52 find

(i) P(A/B)

(ii) P(B/A)

Solution: -

Give that P(A) = 1/12, P(B) = 1/3 and P(A∩B) = 1/52

We know that for any two events

P(A/B) = P(A∩B)/P(B) and

P(A/B) = P(A∩B)/P(B) and

**P(B/A) = P(A∩B)/P(A)**

(i) P(A/B) = (1/52)/(1/3)

=

**3/52**

(ii) P(B/A) = (1/52)/(1/12)

= 12/52

=

**3/13**

Example 2: -

If P(A) = 0.5, P(B) = 0.2 and P(A∩B) = 0.1 find

(i) P(A/B)

(ii) P(B/A)

Solution: -

Give that P(A) = 0.5, P(B) = 0.2 and P(A∩B) = 0.1

We know that for any two events

P(A/B) = P(A∩B)/P(B) and

P(A/B) = P(A∩B)/P(B) and

**P(B/A) = P(A∩B)/P(A)**

(i) P(A/B) = 0.1 / 0.5

= 1/5 =

**0.2**

(ii) P(B/A) = 0.1 / 0.2

= 1/2

=

**0.5**

Example 3: -

A box 'A' contains 2 white and 4 black marbles. Another box 'B' contains 5 white and 7 black marbles. A marble is transferred from box 'A' to box 'B' and a ball is drawn at random from box 'B'. What is the probability that it will be black?

Solution: -

P( drawing a white marble from the box A) = 2/6

P( drawing a black marble from the box A) = 4/6

P(drawing a black marble from the box B if the transferred marble is white)

= 7/13

P(drawing a black marble from the box B if the transferred marble is black)

= 8/13

P( drawing a black marble from box B)

= P(drawing a black marble from the box B if the transferred marble is white or

drawing a black marble from the box B if the transferred marble is black)

= 2/6 * 7/13 + 4/6 * 8/13

=14/78 + 32/78

= 46/78

=

Try Yourself: -

A bag contains 5 white and 3 black balls. Another contains 3 white and 5 black balls. One ball is drawn from each bag and the color is noted. Find the probability that

(i) both are of same color

(ii) one is white and other is black

P(drawing a black marble from the box B if the transferred marble is white)

= 7/13

P(drawing a black marble from the box B if the transferred marble is black)

= 8/13

P( drawing a black marble from box B)

= P(drawing a black marble from the box B if the transferred marble is white or

drawing a black marble from the box B if the transferred marble is black)

= 2/6 * 7/13 + 4/6 * 8/13

=14/78 + 32/78

= 46/78

=

**23/39****P(drawing a black marble from box B after transferring a marble from box A)= 23/39**Try Yourself: -

A bag contains 5 white and 3 black balls. Another contains 3 white and 5 black balls. One ball is drawn from each bag and the color is noted. Find the probability that

(i) both are of same color

(ii) one is white and other is black