Question

An item is tested on a device for its defectiveness. The probability that such an item is defective is 0.3. The device gives an accurate result in 8 out of 10 such tests. If the device reports that an item tested is not defective, then the probability that it is actually defective is

• A. $\frac{2}{15}$
• B. $\frac{3}{29}$
• C. $\frac{3}{31}$
• D. $\frac{4}{51}$

Hint:

Apply Bayes’ theorem, $P(A \mid B) = \frac{P(B \mid A) P(A)}{P(B)}$, for conditional probabilities involving test accuracy. Compute $P(B)$ using the law of total probability.

Answer 1

The Correct Option is C

Let $D$ be the event that the item is defective, and $N$ be the event that the device reports not defective. Given: - $P(D) = 0.3$, so $P(D') = 0.7$. - Device accuracy is 0.8, so: - $P(N \mid D') = 0.8$ (correctly reports not defective when not defective). - $P(N' \mid D) = 0.8$ (correctly reports defective when defective), so $P(N \mid D) = 1 - 0.8 = 0.2$. We need $P(D \mid N)$. Using Bayes’ theorem: \[ P(D \mid N) = \frac{P(N \mid D) P(D)}{P(N)} \] \[ P(N) = P(N \mid D) P(D) + P(N \mid D') P(D') = (0.2 \cdot 0.3) + (0.8 \cdot 0.7) = 0.06 + 0.56 = 0.62 \] \[ P(D \mid N) = \frac{0.2 \cdot 0.3}{0.62} = \frac{0.06}{0.62} = \frac{6}{62} = \frac{3}{31} \] Option (3) is correct. Options (1), (2), and (4) do not match the computed probability.