Addition Theorem of Probability

Addition Theorem of Probability

Theorem 8.4 : (Addition Theorem of Probability for Two Events)

If A and B are any two events in a random experiment, then

P(A∪ B) = P(A) + P(B) – P(A∩B)

Proof:

For any two events A and B, the shaded region in fig. 8.9 represents the event A ∪ B.

A∪ B =A∪(B-(A∩B))

The events A and B-(A ∩B) are mutually exclusive.

Using Axiom 3∪

P(A∪B) = P(A ∪[B-(A ∩B)])

= P(A) + P[B-(A∩B)]

Since(A∩B)⊂B,,

B = (A+B)∪(B –(A+B))

The events on the right hand side are disjoints. Hence by axiom 3

P(B) = P ( A ∩B) + P( B–(A∩B))

i.e. P[B–(A∩B)] = P(B) – P(A∩B) …………..(8.2)

Substituting (8.2) in (8.1)∪ it follows that

P(A∪B) = P(A) + P(B) – P(A∩B).

Corollary: If A∪ B and C are any three events, then

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

Example 8.12

In the Annual sports meet, among the 260 students in XI standard in the school, 90 participated in Kabadi, 120 participated in Hockey, and 50 participated in Kabadi and Hockey. A Student is selected at random. Find the probability that the student participated in (i) Either Kabadi or Hockey, (ii) Neither of the two tournaments, (iii) Hockey only, (iv) Kabadi only, (v) Exactly one of the tournaments.

Solution:

n(s)=260

 Let A : the event that the student participated in Kabadi

 B : the event that the student participated in Hockey.

 n(A) = 90; n(B) = 120; n(A∪B) = 50

(i) The probability that the student participated in either Kabadi or Hockey is

 P (A∪B) = P(A) + P(B) - P (A∩B)

(ii) The probability that the student participated in neither of the two tournaments in

(iii) The probability that the student participated in Hockey only is

(iv) The probability that the student participated in Kabadi only

(v) The probability that the student participated in exactly one of the tournaments is