Question

Given P(A) = 0.4 and P(A ∪ B) = 0.7. Find P(B) if 

(i) A and B are mutually exclusive 

(ii) A and B are independent events 

(iii) P(A/B) = 0.4 

(iv) P(B/A) = 0.5

Answer 1

P(A) = 0.4, P(A ∪ B) = 0.7 

(i) When A and B are mutually exclusive 

P(A ∪ B) = P(A) P(B) 

(i.e.,) 0.7 = 0.4 + P(B)

0.7 – 0.4 = P(B) 

(i.e.,) P(B) = 0.3

(ii) Given A and B are independent 

⇒ P(A ∩ B) = P(A). P(B) 

Now, P(A ∪ B) = P(A) + P(B) – P (A ∩ B) 

(i.e.,) 0.7 = 0.4 + P(B) – (0.4) (P(B)) 

(i.e.,) 0.7 – 0.4 = P(B) (1 – 0.4) 

0.3 = P (B) 0.6 

⇒ P(B) = 0.3/0.6 = 3/6 = 0.5

(iii) P(A/B) = 0.4 

(i.e.,) P(A ∩ B)/P(B) = 0.4 

⇒ P(A ∩ B) = 0.4 [P(B)] … (i) 

But We know P(A ∪ B) = P(A) + P(B) – P(A ∩ B) 

P(A ∩ B) = P(A) + P(B) – P(A ∪ B) 

⇒ P(A ∩ B) = 0.4 + P(B) – 0.7 

= P(B) – 0.3 … (ii) 

From (i) and (ii) (equating R.H.S) 

We get 

0.4 [P(B)] = P(B) – 0.3

0.3 = P(B) (1 – 0.4) 

0.6 (P(B)) = 0.3 ⇒ P(B) = 0.3/0.6 = 3/6 = 0.5

(iv) P(B/A) = 0.5 

(i.e.,) P(A ∩ B)/P(A) = 0.5 

(i.e.,) P(A ∩ B) = 0.5 × P(A) 

= 0.5 × 0.4 = 0.2 

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

⇒ 0.7 = 0.4 + P(B) – 0.2 

⇒ 0.7 = P(B) + 0.2 

⇒ P(B) = 0.7 – 0.2 = 0.5