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; $P\left(B_3\right)=\frac{1}{3};P\left(\frac{S}{B_3}\right)=\frac{1}{7}$
$P\left(\frac{B_1}{S}\right)=\frac{\frac{1}{3}\times \frac{1}{5}}{\frac{1}{3}\times \frac{1}{5}+\frac{1}{3}\times \frac{1}{6}+\frac{1}{3}\times \frac{1}{7}}$
$=\frac{42}{107}$