# How to Find the Probability of Neither A Nor B

Given two events, A and B, to “find the probability of neither A nor B” means to find the probability that neither event A nor event B occurs.

We use the following formula to calculate this probability:

P(Neither A Nor B) = 1 – ( P(A) + P(B) – P(A∩B) )

where:

• P(A): The probability that event A occurs.
• P(B): The probability that event B occurs.
• P(A∩B): The probability that event A and event B both occur.

The following examples show how to use this formula in practice.

## Example 1: Probability of Neither A Nor B (Basketball Players)

Suppose the probability that a given college basketball player gets drafted into the NBA is 0.03.

Also suppose the probability that a given college basketball player has a 4.0 GPA is 0.25.

Also suppose the probability that a given college basketball player has a 4.0 GPA and gets drafted into the NBA is 0.005.

If we randomly select some college basketball player, what is the probability that they neither get drafted nor have a 4.0 GPA?

Solution:

• P(drafted) = 0.03
• P(4.0 GPA) = 0.25
• P(drafted ∩ 4.0 GPA) = 0.005

Thus, we can calculate:

• P(Neither drafted Nor 4.0 GPA) = 1 – ( P(drafted) + P(4.0 GPA) – P(drafted ∩ 4.0 GPA) )
• P(Neither drafted Nor 4.0 GPA) = 1 -(.03 + .25 – .005)
• P(Neither drafted Nor 4.0 GPA) = 0.715

If we randomly select some college basketball player, the probability that they neither get drafted nor have a 4.0 GPA is 0.715 or 71.5%.

## Example 2: Probability of Neither A Nor B (Exam Scores)

Suppose the probability that a given student receives a perfect score on a final exam is 0.13.

Also suppose the probability that a given student used a new studying method is 0.35.

Also suppose the probability that a given student received a perfect score and used a new studying method is 0.04.

If we randomly select some student, what is the probability that they neither received a perfect score nor used a new studying method?

Solution:

• P(perfect score) = 0.13
• P(new method) = 0.35
• P(perfect score ∩ new method) = 0.04

Thus, we can calculate:

• P(Neither perfect score Nor new method) = 1 – ( P(perfect score) + P(new method) – P(perfect score ∩ new method) )
• P(Neither perfect score Nor new method) = 1 – (0.13 + 0.35 – 0.04)
• P(Neither perfect score Nor new method) = 0.56

If we randomly select some student, the probability that they neither received a perfect score nor used a new studying method is 0.56 or 56%.