Question

In a bolt factory, machines $A,B$ and $C$ manufacture $25$%, $35$%, $40$% respectively. Of the total of their output $5$%,$4$% and $2$% are defective. A bolt is drawn and is found to be defective. What are the probabilities that it was manufactured by the machines $A,B,C$.

• A. $\frac{25}{69},\frac{28}{69},\frac{16}{69}$
• B. $\frac{28}{69},\frac{25}{69},\frac{16}{69}$
• C. $\frac{25}{69},\frac{16}{69},\frac{28}{69}$
• D. $\frac{16}{69},\frac{28}{69},\frac{25}{69}$

Answer 1

The correct option is A $\frac{25}{69},\frac{28}{69},\frac{16}{69}$

$P\left(A\right)=0.25$
$P\left(B\right)=0.35$
$P\left(C\right)=0.40$
Hence
Applying baye's theorem, we get to find the possibility that the bolt was manufactured by A is
$=\frac{0.25\times 0.05}{\left(0.25\right(0.05\left)\right)+\left(0.35\right(0.04\left)\right)+\left(0.4\right(0.02\left)\right)}$
$=\frac{0.0125}{0.0345}$
$=\frac{25}{69}$
Similarly
Applying baye's theorem, we get to find the possibility that the bolt was manufactured by B is
$=\frac{0.35\times 0.04}{\left(0.25\right(0.05\left)\right)+\left(0.35\right(0.04\left)\right)+\left(0.4\right(0.02\left)\right)}$
$=\frac{0.014}{0.0345}$
$=\frac{28}{69}$
Applying baye's theorem, we get to find the possibility that the bolt was manufactured by C is
$=\frac{0.4\times 0.02}{\left(0.25\right(0.05\left)\right)+\left(0.35\right(0.04\left)\right)+\left(0.4\right(0.02\left)\right)}$
$=\frac{0.008}{0.0345}$
$=\frac{16}{69}$