 # Conditional Probability How to find the conditional probability: formula, examples, and their solutions.

## Formula P(B | A) means
the probability of (A and B)

So P(B | A) = P(A and B)/P(A).

This is why it is called the [conditional probability].

[B | A] is read as
[B bar A] or [B given A].

Probability of (A and B, Intersection)

## Example 1 Numbers from 1 to 10 are given.

So there are 10 numbers
that can be picked.

So n(S) = 10.

It says
the picked number is already an odd number.

So set the event A as
picking an [odd] number.

Those numbers are
{1, 3, 5, 7, 9}.

And it says
find the probability
that the picked (odd) number is a prime number.

So, if the event B is picking a [prime] number,

then picking an [odd] and a [prime] number is
[A and B].

So find [A and B]
by finding the prime numbers [B]
from {1, 3, 5, 7, 9} [A].

So [A and B] is {3, 5, 7}.

Probability of (A and B, Intersection)

A: {1, 3, 5, 7, 9}
So n(A) = 5.

A and B: {3, 5, 7}
So n(A and B) = 3.

n(S) = 10
n(A) = 5
n(A and B) = 3

So P(A) = 5/10.

And P(A and B) = 3/10.

Probability

P(A) = 5/10
P(A and B) = 3/10

So P(B | A) = [3/10] / [5/10].

Multiply 10
to both of the numerator and the denominator.

So P(B | A) = 3/5.

## Example 2 So set the event A as
Sam [oversleeping].

Then P(A) = 0.04.

Set the event B as
Sam [being late for school].

Then Sam [oversleeping] and [being late for school] is
[A and B].

So P(A and B) = 0.03.

Probability of (A and B, Intersection)

P(A) = 0.04
P(A and B) = 0.03

So P(B | A) = 0.03/0.04.

So P(B | A) = 3/4.

This means
if Sam overslept,
then the probability of him
being late for school is 3/4 (= 75%).