Theorem of total probability will present independent event, suppose event A and event B. There are two specific situations for this theory to prove:

  • If two events A and B are mutually exclusive and
  • If the two events A and B are not mutually exclusive.

We’ll explain them one by one:

Firstly, if those two events are mutually exclusive then the occurrence of either A or B is

P (A + B) = P (A) + P (B)

Let us considern for all possible outcomes in an experiment which are mutually exclusive. If n1is the outcome for event A, and n2for event B, then:

P(A) = n1/n,                     P(B) = n2/n

For this first experiment we have taken events A and B are mutually exclusive, so we’ll definitely point out that the two outcomes n1 and n2 are completely different from each other. When calculating total outcomes for events either A or B is n1+n2.

P(A + B) = n1 + n2/ n = n1/n + n2/n = P(A) + P(B)

Secondly, if the two events A and B are not mutually exclusive, then probability of occurrence of events A and B is

P (A +B) = P (A) + P (B) – P (AB)

Note: Here the event A + B means the occurrence of one of the following mutually exclusive events: AB, AB and AB. Hence,

P (A +B) = P (AB + AB + AB) = P (AB) + P (AB) + P (AB)

Then it’s:

 P (A) = P (AB) + P (AB)

=>               P (AB) = P (A) – P (AB)

And              P (B) = P (AB) + P (AB)

=>               P (AB) = P (B) – P (AB)

So we get:

P(A + B)= P(AB) +[P(A)- P(AB)]+[P(B)- P(AB)]

= P (A) + P (B) – P (AB)

If for example, events A, B and C are not mutually exclusive, then we’ll use:

P (A + B+ C) = P (A) + P (B) + P(C) – P (AB)-P (AC) – P (BC) + P (ABC)

To go further into the theorem of total probability we have to memorize two other equations:

  • Boole’s inequality: Here we’ll consider P (A +B) ≤ P (A) + P (B). One thing must be noted here. Since we are placing P (AB) = 0, we are trying to provide information on even A and B that are mutually exclusive.
  • Bonferroni’s inequality: Here we’ll consider P (AB) ≥ P (A) + P (B) – I

 

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