# 7. Conditional Probability

If *E*_{1} and *E*_{2} are two events, the probability that *E*_{2} occurs given that *E*_{1} has occurred is denoted by *P*(*E*_{2}|*E*_{1}).

*P*(*E*_{2}|*E*_{1}) is called the **conditional probability** of *E*_{2} given that *E*_{1} has occurred.

## Calculating Conditional Probability

Let *E*_{1} and *E*_{2} be any two events defined in a sample space *S* such that *P*(*E*_{1}) > 0.

The **conditional probability** of *E*_{2}, assuming *E*_{1} has already occurred, is given by

`P(E_2|E_1)=(P(E_2\ "and"\ E_1))/(P(E_1))`

### Example 1

Let *A* denote the event 'student is female' and let *B* denote the event 'student is French'. In a class of `100` students suppose `60` are French, and suppose that `10` of the French students are females. Find the probability that if I pick a French student, it will be a girl, that is, find *P*(*A*|*B*).

Answer

Since `10` out of `100` students are both French and female, then

`P(A\ "and"\ B) = 10/100`

Also, `60` out of the `100` students are French, so

`P(B) = 60/100`

So the required probability is:

`P(A|B)=frac{P(A\ "and"\ B)}{P(B)}=frac{10/100}{60/100}=1/6`

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### Example 2

What is the probability that the total of two dice will be greater than `8`, given that the first die is a `6`?

Answer

Let *E*_{1} = first die is `6`;

Let *E*_{2} = total of two dice is ` > 8`

Then "*E*_{1} and *E*_{2}" will be given by `(6, 3),\ (6, 4),\ (6, 5),\ (6, 6)`.

There are `36` possible outcomes when we throw 2 dice.

So

`P(E_2\ "and"\ E_1)=4/36=1/9`

Therefore

`P(E_2|E_1)` `= frac{P(E_2\ "and"\ E_1)}{P(E_1)}` `=frac{1/9}{1/6}=6/9` `=2/3`

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