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%.
Additional Resources
The following tutorials explain how to perform other calculations related to probabilities:
How to Find the Probability of A or B
How to Find the Probability of A and B
How to Find the Probability of A Given B
How to Find the Probability of “At Least One” Success