Use Bayes' rule to find the indicated probability. A person must select one of three boxes, each filled with clocks. The probability of box \( A \) being selected is 0.31 , of box \( B \) being selected is 0.13 , and of box \( C \) being selected is 0.56 . The probability of finding a red clock in box \( A \) is 0.2 , in box \( B \) is 0.4 , and in box \( C \) is 0.9 . A box is selected. Given that the box contains a red clock, what is the probability that box A was chosen? 0.062 0.1 0.31 0.333 0.133

We start by using Bayes' rule, which states that \[ P(A|\text{red})=\frac{P(\text{red}|A)P(A)}{P(\text{red})}, \] where \( P(\text{red}|A)=0.2 \) and \( P(A)=0.31 \). First, compute \( P(\text{red}) \) using the law of total probability: \[ P(\text{red}) = P(\text{red}|A)P(A) + P(\text{red}|B)P(B) + P(\text{red}|C)P(C). \] Given: \[ P(\text{red}|A)=0.2,\quad P(A)=0.31,\\ P(\text{red}|B)=0.4,\quad P(B)=0.13,\\ P(\text{red}|C)=0.9,\quad P(C)=0.56. \] Now substitute these values: \[ P(\text{red}) = (0.2)(0.31) + (0.4)(0.13) + (0.9)(0.56). \] Compute each term: \[ 0.2 \times 0.31 = 0.062,\\ 0.4 \times 0.13 = 0.052,\\ 0.9 \times 0.56 = 0.504. \] Sum the terms to get: \[ P(\text{red}) = 0.062 + 0.052 + 0.504 = 0.618. \] Now, apply Bayes' rule: \[ P(A|\text{red})=\frac{(0.2)(0.31)}{0.618} = \frac{0.062}{0.618}. \] Divide to obtain: \[ P(A|\text{red})\approx 0.1. \] Thus, the probability that box \( A \) was chosen given that a red clock is found is approximately \( 0.1 \).