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

A
$16 / 29$
B
$1 / 15$
C
$27 / 59$
D
$42 / 107$

Solution:

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

$\begin{array}{l} P (E_{1}) = P (E_{2}) = P (E_{3}) = 1 / 3 \\ P (A / E_{1}) = \frac{1}{5}, P (A / E_{2}) = \frac{1}{6}, P (A / E_{3}) = \frac{1}{7} \end{array}$

By Baye's rule, we have

Required probability $= P (E_{1} / A)$

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

$= \frac{\frac{1}{3} \times \frac{1}{5}}{\frac{1}{3} \times \frac{1}{5} + \frac{1}{3} \times \frac{1}{6} + \frac{1}{3} \times \frac{1}{7}} = \frac{42}{107}$

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

Solution:

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