For Mutually Exclusive Events

The additive theorem of probability states if A and B are two mutually exclusive events then the probability of either A or B is given by

P(A or B)=P(A)+P(B)P(A∪B)=P(A)+P(B)P(A or B)=P(A)+P(B)P(A∪B)=P(A)+P(B)

The theorem can he extended to three mutually exclusive events also as

P(A∪B∪C)=P(A)+P(B)+P(C)P(A∪B∪C)=P(A)+P(B)+P(C)

Example

Problem Statement:

A card is drawn from a pack of 52, what is the probability that it is a king or a queen?

Solution:

Let Event (A) = Draw of a card of king

Event (B) Draw of a card of queen

P (card draw is king or queen) = P (card is king) + P (card is queen)

P(A∪B)=P(A)+P(B)=452+452=213+213=413P(A∪B)=P(A)+P(B)=452+452=213+213=413

For Non-Mutually Exclusive Events

In case there is a possibility of both events to occur then the additive theorem is written as:

P(A or B)=P(A)+P(B)−P(A and B)P(A∪B)=P(A)+P(B)−P(AB)P(A or B)=P(A)+P(B)−P(A and B)P(A∪B)=P(A)+P(B)−P(AB)

Example

Problem Statement:

A shooter is known to hit a target 3 out of 7 shots; whet another shooter is known to hit the target 2 out of 5 shots. Find the probability of the target being hit at all when both of them try.

Solution:

Probability of first shooter hitting the target P (A) = 3737

Probability of second shooter hitting the target P (B) = 2525

Event A and B are not mutually exclusive as both the shooters may hit target. Hence the additive rule applicable is

P(A∪B)=P(A)+P(B)−P(A∩B)=37+25−(37×25)=2935−635=2335