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Addition and Multiplication Theorems

Addition theorem on probability:

If A and B are any two events then the probability of happening of at least one of the events is defined as P(AUB) = P(A) + P(B)- P(A∩B).


Since events are nothing but sets,

From set theory, we have

n(AUB) = n(A) + n(B)- n(A∩B).

Dividing the above equation by n(S), (where S is the sample space)

n(AUB)/ n(S) = n(A)/ n(S) + n(B)/ n(S)- n(A∩B)/ n(S)

Then by the definition of probability,

P(AUB) = P(A) + P(B)- P(A∩B).


If the probability of solving a problem by two students George and James are 1/2 and 1/3 respectively then what is the probability of the problem to be solved.


Let A and B be the probabilities of solving the problem by George and James respectively.

Then P(A)=1/2 and P(B)=1/3.

The problem will be solved if it is solved at least by one of them also.

So, we need to find P(AUB).

By addition theorem on probability, we have

P(AUB) = P(A) + P(B)- P(A∩B).

P(AUB) = 1/2 +.1/3 – 1/2 * 1/3 = 1/2 +1/3-1/6 = (3+2-1)/6 = 4/6 = 2/3


If A and B are any two mutually exclusive events then P(A∩B)=0.

Then P(AUB) = P(A)+P(B).


Multiplication theorem on probability

If A and B are any two events  of a sample space such that P(A) ≠0 and P(B)≠0, then

P(A∩B) = P(A) * P(B|A) = P(B) *P(A|B).

Example:  If P(A) =  1/5  P(B|A) =  1/3  then what is P(A∩B)?

Solution: P(A∩B) = P(A) * P(B|A) = 1/5 * 1/3 = 1/15


Two events A and B are said to be independent if there is no change in the happening of an event with the happening of the other event.

i.e. Two events A and B are said to be independent if

P(A|B) = P(A) where P(B)≠0.

P(B|A) = P(B) where P(A)≠0.

i.e. Two events A and B are said to be independent if

P(A∩B) = P(A) * P(B).


While laying the pack of cards, let A be the event of drawing a diamond and B be the event of drawing an ace.

Then P(A) =  13/52 = 1/4 and P(B) =  4/52=1/13

Now, A∩B = drawing a king card from hearts.

Then P(A∩B) =  1/52

Now, P(A/B) = P(A∩B)/P(B) = (1/52)/(1/13) = 1/4 = P(A).

So, A and B are independent.

[Here, P(A∩B) = =    = P(A) * P(B)]


(1)    If 3 events A,B and C are independent the

P(A∩B∩C) = P(A)*P(B)*P(C).

(2)    If A and B are any two events, then P(AUB) = 1-P(A’)P(B’).


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