Question

Let ‘A’, ‘B’ and ‘C’ be three independent events with $P\left(A\right)=\frac{1}{3},P\left(B\right)=\frac{1}{2}$ and $P\left(C\right)=\frac{1}{4}$. The probability of exactly 2 of these events occurring, is equal to:

• A. $\frac{1}{4}$
• B. $\frac{7}{24}$
• C. $\frac{3}{4}$
• D. $\frac{17}{24}$

Answer 1

The correct option is A $\frac{1}{4}$

P (exactly two of A,B and C)
= P (AB) + P (BC) + P (CA) - 3P (ABC)
Since A, B, C are independent events, required probability
= P(A) P(B) + P (B) P (C) + P(C) P (A)
-3P (A) P(B) P(C)
$=\frac{1}{3}\times \frac{1}{2}+\frac{1}{2}\times \frac{1}{4}+\frac{1}{4}\times \frac{1}{3}-3\times \frac{1}{3}\times \frac{1}{2}\times \frac{1}{4}=\frac{6}{24}=\frac{1}{4}$