The chances of defective screws in three boxes A,B,C are 15,16,17 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

# 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

1. A

$16/29$

2. B

$1/15$

3. C

$27/59$

4. D

$42/107$

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### Solution:

Let  and ${E}_{3}$ denote the events of selecting box  respectively and $A$ be the event that a screw selected at random is defective. Then,

By Baye's rule, we have

Required probability $=P\left({E}_{1}/A\right)$

$=\frac{P\left({E}_{1}\right)P\left(A/{E}_{1}\right)}{P\left({E}_{1}\right)P\left(A/{E}_{1}\right)+P\left({E}_{2}\right)P\left(A/{E}_{2}\right)+P\left({E}_{3}\right)P\left(A/{E}_{3}\right)}$

$=\frac{\frac{1}{3}×\frac{1}{5}}{\frac{1}{3}×\frac{1}{5}+\frac{1}{3}×\frac{1}{6}+\frac{1}{3}×\frac{1}{7}}=\frac{42}{107}$  Register to Get Free Mock Test and Study Material

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