# How To Find The Probability Of Neither A Nor B

Probability is a branch of mathemetics that** **deals with the occurrence of a random event. The Probability of neither event A nor B occurring is the Probability that A does not happen and B does not happen. To find the solution for this Probability, it is important to revisit some basic laws of Probability.

The Probability of event A occurring is denoted as P(A), and the Probability of B occurring is denoted as P(B). The Probability of any set (all possible outcomes) is always equal to 1: P(S)=1. For any event, A, P(A) ≤1. It follows that the Probability of A not occurring, also known as the complement of A, is denoted P(¯A) = 1-P(A).

If A and B are mutually exclusive, the two events do not have any elements in common. Therefore, the Probability of event A and event B happening is zero since the events cannot occur simultaneously. The intersection of the two events is an empty set, meaning P(A∩B) = 0. An example of mutually exclusive events is heads and tails when tossing a single coin; both cannot happen simultaneously.

For mutually exclusive events A and B, the Probability of either A or B occurring is the Probability that A occurs, or B occurs; P (A or B) =P(A)+P(B). Using the rule P(¯A) = 1-P(A) to find the Probability of neither A nor B occurring for the exclusive events, the equation becomes;

P [not A or not B] = P (not A) + P (not B); where P (not A) = P(¯A) hence;

P (¯A or ¯B) = 1- P (A or B) =1-{P(A)+P(B)}

It makes sense since the Probability of neither A nor B is simply the Probability of A not happening and B not happening; hence the sum of their compliments.

## Finding the Probability of Neither A nor B When Not Mutually Exclusive

When events are not mutually exclusive, they can occur simultaneously because the outcome of one event does not affect the possibility of the other event occurring. Let events A and B be two independent events; the Probability of both A and B occurring is greater than zero. The intersection of the two events is not an empty set as seen above in mutually exclusive events; P(A∩B)>0. An example is the occurrence of even and prime numbers on a dice.

For two independent events, A and B, the Probability of either A or B occurring is;

P (A or B) =P(A)+P(B)- P (A and B)

This general formula is modified for mutually exclusive events to exclude the last part since P (A and B) is always zero as they cannot co-occur.

As mentioned above, finding the Probability of neither A nor B occurring requires using the rule P(¯A) = 1-P(A). The equation becomes:

P [not A or not B] = P (not A) + P (not B) + P (not A and not B); where P (not A) = P (¯A) hence;

P (¯A or ¯B) = 1- P (A or B) = P(¯A) + P(¯B)- P (¯A and ¯B)

The simplified form of the probability formula becomes:

=1-{P(A)+P(B)-P (A and B)}

It makes sense since it is the Probability of A not happening, B not happening, and A and B not happening. Therefore, neither A nor B happens.

## The Difference Between the Probability of Either A or B, and Neither A Nor B

The Probability that either event A or B occurs is the Probability of one event or the other occurring. On the other hand, the Probability of neither A nor B occurring is the chance that none of the two events occurs.

### Sample Questions and Solutions on the Probability of Neither A nor B

*Question 1: What is the Probability of either A or B occurring, and neither A nor B will happen if A and B are mutually exclusive, P(A) = 0.30 and P(B) = 0.20?*

*Question 1: What is the Probability of either A or B occurring, and neither A nor B will happen if A and B are mutually exclusive, P(A) = 0.30 and P(B) = 0.20?*

- Solution:
- For mutually exclusive;
- P (Either A or B): P (A or B) =P(A)+P(B) = 0.30+0.20= 0.50
- P (Neither A nor B): P (¯A or ¯B) = 1- P (A or B) = 1-0.50=0.50
- ANS: 0.50

*Question 1: Out of 300 students in a school, 95 play Cricket only, 120 play football only, 80 play volleyball only, and 5 play no games. If one student is chosen at random, find the Probability that (i) he plays volleyball, (ii) he plays either cricket or volleyball (iii) he plays neither football nor volleyball*.

*Question 1: Out of 300 students in a school, 95 play Cricket only, 120 play football only, 80 play volleyball only, and 5 play no games. If one student is chosen at random, find the Probability that (i) he plays volleyball, (ii) he plays either cricket or volleyball (iii) he plays neither football nor volleyball*.

Solution:

- P (A=Cricket only) =95/300
- P (B=Football Only) =120/300
- P (C=Volleyball Only) = 80/300
- P (No games) =5/300

*i) Probability that he plays volleyball*

- P(C) =80/300 =0.27

*(ii) he plays either cricket or volleyball*

- = P(A)+P(c)= 95/300+80/300=0.58

*(iii) he plays neither football nor volleyball*

- P (Neither A nor B) =1- P (A or B) =1-0.58=0.42

*Question 3: The probability of happening and event A is 0.5 and that of B is 0.3. If A and Bare are mutually exclusive events, then the probability of happening neither A nor B is;*

*Question 3: The probability of happening and event A is 0.5 and that of B is 0.3. If A and Bare are mutually exclusive events, then the probability of happening neither A nor B is;*

*Suppose, the probability of an event A occurring is PA and that of Bis PB**PA=0.5 and PB=0.3**Now we need to find the probability of neither A nor B, which is PA’∩B’**PA’∩B’=PA∪B¯**⇒PA’∩B’=1-PA∪B**⇒PA’∩B’=1-[PA+PB]**Now the events are mutually exclusive, so PA∩B=0**PA’∩B’=1-0.5+0.3**⇒PA’∩B’=1-0.8**⇒PA’∩B’=0.2*- ANS: 0.2

*Question 4: Given P(A) = 0.4, P(B) = 0.35, and P(A intersect B) = 0.14, what is the probability that neither A nor B takes place?*

*Question 4: Given P(A) = 0.4, P(B) = 0.35, and P(A intersect B) = 0.14, what is the probability that neither A nor B takes place?*

*Explanation:*

*As P(A)=0.4,P(B)=0.35, and P(A∩B)=0.14,**P(A∪B)=P(A)+P(B)−P(A∩B)**= 0.4+0.35−0.14=0.61**As P(A∪B)=0.61,**the probability that neither A nor B takes place is 1−0.61=0.39*- ANS: 0.39

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