Q106 / 180 · COMEDK 2014 · Mathematics
A certain item is manufactured by 3 factory $F_{1}, F_{2}$ and $F_{3}$ with $30 \%$ of item made in $F_{1}, 20 \%$ in $F_{2}$ and $50 \%$ in $F_{3}$. It is found that $2 \%$ of the items produced by $F_{1} .3 \%$ of the items produced by $F_{2}$ and $4 \%$ of the items produced by $F_{3}$ are defective. Suppose that an items selected at random from the stock is found defective. What is the probability that the item came from $F_{1}$ ?
- A$\frac{1}{16}$
- B$\frac{1}{8}$
- C$\frac{1}{3}$
- Correct answerD$\frac{3}{16}$
Solution
Let events $T_{1}, T_{2}, T_{3}$ be the following
$T_{1}$ : The item is manufactured by factory $F_{1}$.
$T_{2}$ : The item is manufactured by factory $F_{2}$.
$T_{3}$ : : The item is manufactured by factory $F_{3}$.
Clearly, $T_{1}, T_{2}, T_{3}$ are mutually exclusive and exhaustive events.
$$
\begin{aligned}
&P\left(T_{1}\right)=30 \%=0.3, P\left(T_{2}\right)=20 \%=0.2 \\
&P\left(T_{3}\right)=50 \%=0.5
\end{aligned}
$$
Let $E$ be the event that item is defective.
$$
\text { Now, } \quad \begin{array}{ll}
& P\left(E \mid T_{1}\right)=2 \%=0.02 \\
& P\left(E \mid T_{2}\right)=3 \%=0.03 \\
& P\left(E \mid T_{3}\right)=4 \%=0.04
\end{array}
$$
Hence, by Bayes' theorem, we have
$$
\begin{aligned}
P\left(T_{1} \mid E\right) &=\frac{P\left(T_{1}\right) P\left(E \mid T_{1}\right)}{P\left(T_{1}\right) P\left(E \mid T_{1}\right)+P\left(T_{2}\right)\left(P\left(E \mid T_{2}\right)+P\left(T_{3}\right) P\left(E \mid T_{3}\right)\right.} \\
&=\frac{0.3 \times 0.02}{0.3 \times 0.02+0.2 \times 0.03+0.5 \times 0.04} \\
&=\frac{0.006}{0.006+0.006+0.20}=\frac{0.006}{0.032}=\frac{6}{32}=\frac{3}{16}
\end{aligned}
$$
Hence, the correct answer is option: 'D'.