Question

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

• A. $\frac{16}{29}$
• B. $\frac{1}{15}$
• C. $\frac{27}{59}$
• D. $\frac{42}{107}$

Answer 1

Answer: (D)

Let $S=$ screws drawn is found to be defective
$B_{1}=A$ box is selected; $P\left(B_{1}\right)=\frac{1}{3};P\left(\frac{S}{B_{1}}\right)=\frac{1}{5}$
$B_{2}=B$ box is selected; $P\left(B_{2}\right)=\frac{1}{3};P\left(\frac{S}{B_{2}}\right)=\frac{1}{6}$
$B_{3}=C$ box is selected; $P\left(B_{3}\right)=\frac{1}{3};P\left(\frac{S}{B_{3}}\right)=\frac{1}{7}$
$P\left(\frac{B_{1}}{S}\right)=\frac{\frac{1}{3} \times \frac{1}{5}}{\frac{1}{3} \times \frac{1}{5} + \frac{1}{3} \times \frac{1}{6} + \frac{1}{3} \times \frac{1}{7}}$
$=\frac{42}{107}$