Question

Given independent events A, B, and C with rtain intersection probabilities, find $P(A)$, $P(B)$, and $P(C)$.

• A. $\frac{1}{2}, \frac{1}{4}, \frac{1}{5}$
• B. $\frac{1}{2}, \frac{1}{2}, \frac{1}{3}$
• C. $\frac{1}{2}, \frac{1}{3}, \frac{1}{4}$
• D. $\frac{1}{3}, \frac{1}{4}, \frac{1}{5}$

Hint:

Independent Events. Multiply individual probabilities of complements for compound independent events.

Answer 1

The Correct Option is C

Given: \[ P(A \cap B^c \cap C^c) = \frac{1}{4},\quad P(A^c \cap B \cap C^c) = \frac{1}{8},\quad P(A^c \cap B^c \cap C^c) = \frac{1}{4} \] Using independence: \begin{align*} P(A \cap B^c \cap C^c) &= x(1-y)(1-z) = \frac{1}{4}
P(A^c \cap B \cap C^c) &= (1-x)y(1-z) = \frac{1}{8}
P(A^c \cap B^c \cap C^c) &= (1-x)(1-y)(1-z) = \frac{1}{4} \end{align*} Solving these gives: \[ P(A) = \frac{1}{2}, \quad P(B) = \frac{1}{3}, \quad P(C) = \frac{1}{4} \]