# A and B are two independent events. The probability that both A and B occur is 1/6 and the probability that neither of them occurs  is 1/3. Find the probability of the occurrence of A.

Jitender Pal
10 years ago
Hello Student,
Given that A and B are independent events
∴ P (A ∩ B) = P (A). P (B) . . . . . . . . . . . . . . . . . . . . . . . . . (1)
Also given that P (A ∩ B) = 1/6 . . . . . . . . . . . . . . . . . . . . . . . . . (2)
And P ($\underset{A}{\rightarrow} \ \cap \underset{B}{\rightarrow}$) = 1/3 . . . . . . . . . . . . . . . . . . . . . . . . . . (3)
Also P ($\underset{A}{\rightarrow} \ \cap \underset{B}{\rightarrow}$) = 1 – P ($\underset{A}{\rightarrow} \ \cup \underset{B}{\rightarrow}$)
⇒ P ($\underset{A}{\rightarrow} \ \cap \underset{B}{\rightarrow}$) = 1 – P (A) - P (B) P ($\underset{A}{\rightarrow} \ \cap \underset{B}{\rightarrow}$)
⇒ 1/3 = 1 - P (A) - P (B) + 1/6
⇒ P (A) + P (B) . . . . . . . . . . . . . . . . . . . . . . . . . . (4)
From (1) and (2) we get
P (A). P (B) = 1/6 . . . . . . . . . . . . . . . . . . . . . . . . . . . (5)
Let P (A) = x and P (B) = y then eq’ s (4) and (5) become
x + y = 5/6, xy = 1/6
⇒ x - y = ± √(x + y)2 – 4xy
= ± √25/36 – 4/6 = ± 1/6
∴ We get x = 1/2 and y = 1/3
Or x = 1/3 and y = 1/2
Thus P (A) = 1/2 and P (B) = 1/3
Or P (A) = 1/3 and P (B) = 1/2 .

Thanks
Jitender Pal