Q52 / 120 · VITEEE 2010 · Mathematics
The chances of defective screws in three boxes
$\mathrm{A}, \mathrm{B}, \mathrm{C}$ are $\frac{1}{5}, \frac{1}{6}, \frac{1}{7}$ respectively. A box isselected at random and ascrew drawn from it at random is found to be defective. Then, the probability that it came from box $\mathrm{A}$, is
- A$\frac{16}{29}$
- B$\frac{1}{15}$
- C$\frac{27}{59}$
- Correct answerD$\frac{42}{107}$
Solution
Let $E_{1}, E_{2}$ and $E_{3}$ denote the events of selecting boxes $A, B, C$ respectively and $A$ be the event that a screw selected at random is defective. Then,
$$
\begin{array}{l}
P\left(E_{1}\right)=\frac{1}{3}, P\left(E_{2}\right)=\frac{1}{3}, P\left(E_{3}\right)=\frac{1}{3} \\
P\left(\frac{A}{E_{1}}\right)=\frac{1}{5}, P\left(\frac{A}{E_{2}}\right)=\frac{1}{6}, P\left(\frac{A}{E_{3}}\right)=\frac{1}{7}
\end{array}
$$
By Baye's rule, the required probability
$$
\begin{aligned}
P\left(\frac{E_{1}}{A}\right)=& \frac{P\left(E_{1}\right) P\left(\frac{A}{E_{1}}\right)}{P\left(E_{1}\right) P\left(\frac{A}{E_{1}}\right)+P\left(E_{2}\right) P\left(\frac{A}{E_{2}}\right)+P\left(E_{3}\right) P\left(\frac{A}{E_{3}}\right)} \\
\Rightarrow P\left(\frac{E_{1}}{A}\right) &=\frac{\frac{1}{3} \cdot \frac{1}{5}}{\frac{1}{3} \cdot \frac{1}{5}+\frac{1}{3} \cdot \frac{1}{6}+\frac{1}{3} \cdot \frac{1}{7}} \\
&=\frac{\frac{1}{5}}{\frac{1}{5}+\frac{1}{6}+\frac{1}{7}}=\frac{42}{107}
\end{aligned}
$$
Hence, the correct answer is option: 'D'.