Question

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 from it is found defective. The probability that it came from the box $A$ is

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

Answer 1

The correct option is D $\frac{42}{107}$

Let $A$ be the event that a screw drawn from the box and similar is for events $B$ and $C$.

Let $D$ be the event that a defective screw is drawn. Then

$P\left(D\right)=P(A\cap D)+P(B\cap D)+P(C\cap D)$

$=\frac{1}{3}.\frac{1}{5}+\frac{1}{3}.\frac{1}{6}+\frac{1}{3}.\frac{1}{7}$

$P\left(\frac{A}{D}\right)=\frac{P(A\cap D)}{P\left(D\right)}=\frac{\frac{1}{3}.\frac{1}{5}}{\frac{1}{15}+\frac{1}{18}+\frac{1}{21}}=\frac{42}{107}$