Question

In a bolt factory, three machines A,B, and C manufacture 25,35 and 40 percent of the total bolts respectively. Out of the total bolts manufactured by the machines 5,4 and 2 percent are defective from machine A, B & C respectively. A bolt is drawn at random and is found to be defective. Find the probability that it was manufactured by (i) Machine A or C (ii) Machine B.

Answer 1

Consider the problem

Let

$A:$ bolt manufactured from machine $A$

$B:$bolt manufactured from machine $B$

$C:$bolt manufactured from machine $C$

$D:$ bolt is defective

WE need find the probability that the bolt is manufactured by machine $AorC$ and machine $B$, if it is defective.

That is $P\left(\frac{B}{D}\right)$

So, $P\left(\frac{B}{D}\right)=\frac{P\left(B\right).P\left(\frac{D}{B}\right)}{P\left(A\right).P\left(\frac{D}{A}\right)+P\left(B\right).P\left(\frac{D}{B}\right)+P\left(C\right).P\left(\frac{D}{C}\right)}$

$P\left(A\right)=$ probability that the bolt is made by machine $A$

$\begin{array}{c}=25\%=\frac{25}{100}\\ =0.25\end{array}$

$P\left(B\right)=$ probability that the bolt is made by machine $B$

$\begin{array}{c}=35\%=\frac{35}{100}\\ =0.35\end{array}$

$P\left(C\right)=$ probability that the bolt is made by machine $C$

$\begin{array}{c}=40\%=\frac{40}{100}\\ =0.40\end{array}$

$P\left(AorC\right)=$ probability that the bolt is made by machine $AorC$

$\begin{array}{c}=65\%=\frac{65}{100}\\ =0.65\end{array}$

And,

$P\left(\frac{D}{A}\right)=$ probability of a defective bolt from machine $A$

$\begin{array}{c}=5\%=\frac{5}{100}\\ =0.05\end{array}$

$P\left(\frac{D}{B}\right)=$ probability of a defective bolt from machine $B$

$\begin{array}{c}=4\%=\frac{4}{100}\\ =0.04\end{array}$

$P\left(\frac{D}{C}\right)=$ probability of a defective bolt from machine $C$

$\begin{array}{c}=2\%=\frac{2}{100}\\ =0.02\end{array}$

$P\left(\frac{D}{AorC}\right)=$ probability of a defective bolt from machine $AorC$

$=0.07$

Now apply Bayes' Theorem

$\begin{array}{c}P\left(\frac{B}{D}\right)=\frac{0.35\times 0.04}{0.25\times 0.05+0.35\times 0.04+0.04\times 0.02}\\ =\frac{0.014}{0.0125+0.014+0.008}\\ =\frac{0.014}{0.0345}\\ =\frac{140}{345}\\ =\frac{28}{69}\end{array}$

Therefore, required probability by the machine $B$ is $\frac{28}{69}$

And Machine $AorC$ is

Apply Bayes' Theorem

$\begin{array}{c}P\left(\frac{AorC}{D}\right)=\frac{0.65\times 0.07}{0.65\times 0.07+0.35\times 0.04}\\ =\frac{13}{17}\end{array}$