The chances of defective screws in three boxes A, B, C are $\frac{1}{5}, \frac{1}{6}, \frac{1}{7}$ respectively. A box is selected at random and a screw drawn from it at random is found to be defective. Then, the probability that it came from box A, is

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$\frac{1}{15}$

$\frac{27}{59}$

$\frac{16}{29}$

$\frac{42}{107}$

answer is D.

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Detailed Solution

Let $E_{1}, E_{2} a n d E_{3}$ denote the events of selecting boxes A, B and C respectively and A be the event that a screw selected at random is defective. Then,

$P (E_{1}) = \frac{1}{3}, P (E_{2}) = \frac{1}{3}, P (E_{3}) = \frac{1}{3}, P (A / E_{1}) = \frac{1}{5}$

$P (A / E_{2}) = \frac{1}{6}, P (A / E_{3}) = \frac{1}{7}$

Now, by Bayes' theorem, the required probability.

$P (E_{1} / A) = \frac{P (E_{1}) P (A / E_{1})}{P (E_{1}) P (A / E_{1}) + P (E_{2}) P (A / E_{2}) + P (E_{3}) P (A / E_{3})}$

$\Rightarrow P (E_{1} / A) = \frac{\frac{1}{3} \cdot \frac{1}{5}}{\frac{1}{3} \cdot \frac{1}{5} + \frac{1}{3} \cdot \frac{1}{6} + \frac{1}{3} \cdot \frac{1}{7}} = \frac{\frac{1}{5}}{\frac{1}{5} + \frac{1}{6} + \frac{1}{7}} = \frac{42}{107}$