A company manufactured fans with three units A, B and C. Unit A, B and C produces 2% , 10% and 14% defective fans respectively. Units A, B and C produces 40%, 40% and 20% fans of total products respectively. A fan is randomly chosen that is found to be defective then the probability that the chosen fan is produced from unit C, is
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → \( \displaystyle\frac{7}{19} \)
The fan is known to be defective, and we want the probability that it came from C, so Bayes’ theorem applies. Let D be the event that the fan is defective.
\( \displaystyle P(D)=0.4\times 0.02+0.4\times 0.10+0.2\times 0.14 \)
\( \displaystyle =0.008+0.040+0.028=0.076 \)
\( \displaystyle P(C\mid D)=\frac{P(C)\,P(D\mid C)}{P(D)} \)
\( \displaystyle =\frac{0.028}{0.076} \)
\( \displaystyle =\frac{28}{76}=\frac{7}{19} \)
Shortcut: The defective fans from A, B and C are in the ratio 40 × 2 : 40 × 10 : 20 × 14 = 80 : 400 : 280 = 2 : 10 : 7. So the probability that a defective fan came from C is \( \frac{7}{2+10+7}=\frac{7}{19} \).