If A and B are two events associated with a random experiment such that P(A ∪ B) = 0.8, P(A ∩ B) = 0.3 and \(P(\bar A)\)= 0.5 , find P(B).

**(i) Given :** P(A) = 0.25, P(A or B) = 0.5 and P(B) = 0.4

**To find :** P(A and B)

**Formula used : **P(A or B) = P(A) + P(B) – P(A and B)

Substituting in the above formula we get,

0.5 = 0.25 + 0.4 – P(A and B)

0.5 = 0.65 – P(A and B)

P(A and B) = 0.65 – 0.5

P(A and B) = 0.15

P(A and B) = 0.15

**(ii) Given :** P(A) = 0.25, P(A and B) = 0.15 ( from part (i))

**To find :** P(A and \(\overline{B}\) )

**Formula used :** P(A and \(\overline{B}\) ) = P(A) – P(A and B)

Substituting in the above formula we get,

P(A and \(\overline{B}\) ) = 0.25 – 0.15

P(A and \(\overline{B}\) ) = 0.10

P(A and \(\overline{B}\)) = 0.10

Given A and B are two events

And, P(A’) = 0.5 P(A ∩ B) = 0.3 P(A ∪ B) = 0.8

∵ P(A’) = 1 – P(A) ⇒ P(A) = 1 – 0.5 = 0.5

We need to find P(B).

By definition of P(A or B) under axiomatic approach(also called addition theorem) we know that:

P(A ∪ B) = P(A) + P(B) – P(A ∩ B)

∴ P(B) = P(A ∪ B) + P(A ∩ B) – P(A)

⇒ P(B) = 0.8 + 0.3 – 0.5 = 1.1 – 0.5 = 0.6

∴ P(B) = 0.6