For any two event A, B the Probability of A union B equals to probability of A added to probability of B minus probability of A intersection B.

**P (A∪B) = P(A) + P(B) – P(A∩B)**

#### Proof :-

Suppose, the simple events in A∩B are γ_{1}, γ_{2}, γ_{3},…………,γ_{k}

simple events in A are

α_{1}, α_{2}, α_{3},……….,α_{m}, γ_{1}, γ_{2}, γ_{3},…………,γ_{k}

simple events in B are β_{1}, β_{2}, β_{3},……….,β_{n}, γ_{1}, γ_{2}, γ_{3},…………,γ_{k}

Then simple events in A∪B = α_{1}, α_{2}, α_{3},……….,α_{m}, β_{1}, β_{2}, β_{3},……….,β_{n}, γ_{1}, γ_{2}, γ_{3},…………,γ_{k}

Number of simple events in A∩B = k

Number of simple events in A = m+k

Number of simple events in B = n+k

Number of simple events in A∪B = m+n+k

If we suppose number of simple events in S = N, then

- P(A∩B) = k/N
- P(A∪B) = (m+n+k)/N
- P(A) = (m+k)/N
- P(B) = (n+k)/N

But according to the theorem we have

→ P(A) + P(B) – P(A∩B) = P (A∪B)

considering L.H.S(left hand side of equation above)

P(A) + P(B) – P(A∩B)

= (m+k)/N + (n+k)/N – k/N

= (m+n+k)/N

= P (A∪B)

Note that, if A and B are exclusive events then P (A∪B) = P(A) + P(B)

**Remark:- ** For any three events A,B,C

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

And

- If A,B,C are mutually exclusive events, then P(A∪B∪C) = P(A) + P(B) + P(C)
- If A,B,C are mutually exclusive and exhaustive events, then P(A∪B∪C) = P(A) + P(B) + P(C) = 1

In general, if we consider A_{1}, A_{2}, A_{3},………,A_{n} are any n elements then

**Note :-**

- If A
_{1}, A_{2}, A_{3},……..,A_{n}are any n*mutually exclusive events*, then - If A
_{1}, A_{2}, A_{3},……..,A_{n}are mutually exclusive and exhaustive events then

Since A and complimentary element of A are exclusive and exhaustive for any event A we get,

If A,B are two events such that A is subset of B then P(B-A) = P(B) – P(A)

Note that, for any two events A,B P(B-A) = P(B) – P(A∩B)