Question

A dealer gets refrigerators from 3 different manufacturing companies \(C_1, C_2, C_3\). 25\% of his stock is from \(C_1\), 35\% from \(C_2\), and 40\% from \(C_3\). The percentages of receiving defective refrigerators from \(C_1, C_2, C_3\) are 3\%, 2\%, and 1\%, respectively. If a refrigerator sold at random is found to be defective, then the probability that it is from \(C_2\) is:

• A. \( \frac{29}{37} \)
• B. \( \frac{8}{37} \)
• C. \( \frac{14}{37} \)
• D. \( \frac{15}{37} \)

Hint:

Use Bayes’ theorem for conditional probability calculations, especially in defect analysis problems.

Answer 1

The Correct Option is C

Step 1: Define events.
Let \(D\) be the event that a refrigerator is defective. Using the total probability theorem: \[ P(D) = P(D | C_1) P(C_1) + P(D | C_2) P(C_2) + P(D | C_3) P(C_3) \] Substituting given values: \[ P(D) = (0.03 \times 0.25) + (0.02 \times 0.35) + (0.01 \times 0.40) \] \[ = 0.0075 + 0.007 + 0.004 = 0.0185 \] Step 2: Compute conditional probability.
Using Bayes' theorem: \[ P(C_2 | D) = \frac{P(D | C_2) P(C_2)}{P(D)} \] \[ = \frac{(0.02 \times 0.35)}{0.0185} = \frac{0.007}{0.0185} = \frac{14}{37} \] Thus, the required probability is: \[ \mathbf{\frac{14}{37}} \]