Question

In a bolt factory, three machines A, B, and C manufacture 25%, 35% and 40% of the total production respectively. Of their respective outputs, 5%, 4% and 2% are defective. A bolt is drawn at random from the total production and it is found to be defective. Find the probability that it was manufactured by machine C.

Answer 1

	Machine A	Machine B	Machine C
% of total production	25%	35%	40%
% defective of their output	5%	4%	2%

∴ P(A) = ${\displaystyle \frac{25}{100}=\frac{1}{4}}$,

P(B) = ${\displaystyle \frac{35}{100}=\frac{7}{20}}$

and P(C) = ${\displaystyle \frac{40}{100}=\frac{2}{5}}$

P(D/A) = ${\displaystyle \frac{5}{100}\times \frac{1}{4}=\frac{1}{80}}$

P(D/B) = ${\displaystyle \frac{4}{100}\times \frac{7}{20}=\frac{7}{500}}$

P(D/C) = ${\displaystyle \frac{2}{100}\times \frac{2}{5}=\frac{1}{125}}$

P(C/D) = ${\displaystyle \frac{\text{P}\left(\text{C}\right)·\text{P}\left(\frac{\text{D}}{\text{C}}\right)}{(\text{P}\left(\text{A}\right)·\text{P(}D\text{/}A\text{) +}P\text{(}B\text{) ·}P\text{(}D\text{/}B)+\text{P}\left(\text{C}\right)·\text{P}\left(\frac{\text{D}}{\text{C}}\right)}}$

= ${\displaystyle \frac{\frac{2}{5}\times \frac{1}{125}}{\frac{1}{4}\times \frac{1}{80}+\frac{7}{20}\times \frac{7}{500}+\frac{2}{5}\times \frac{1}{125}}}$

=  ${\displaystyle \frac{\frac{2}{625}}{\frac{250+392+256}{80000}}}$

= ${\displaystyle \frac{2\times 80000}{625\times 898}=\frac{128}{449}=0.2851}$