Question

Let A, B , and C be three independent events with $P\left(A\right)=\frac{1}{3},P\left(B\right)=\frac{1}{2},andP\left(C\right)=\frac{1}{4}.$ . 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) – 3 P(ABC)
Since A, B ,C are independent events,
Required probability = P(A)P(B) + P(B)P(C) +P(C)P(A) – 3 P(A)P(B)P(C)
$=\frac{1}{3}\times \frac{1}{2}+\frac{1}{2}\times \frac{1}{4}+\frac{1}{4}\times \frac{1}{3}-3\frac{1}{3}\times \frac{1}{2}\times \frac{1}{4}=\frac{1}{4.}$