Question

In a bolt factory, three machines $A,B$ and $C$ produce $25\%,35\%$ and $40\%$ of total output respectively. It was found that $5\%,4\%$ and $2\%$ are defective bolts in the production by machines $A,B,C$ respectively. If a bolt is chosen at random from the total output and is found to be defective, then the chance that the bolt comes from the machine:

• A. $A\text{is}\frac{125}{449}$
• B. $B\text{is}\frac{125}{449}$
• C. $C\text{is}\frac{128}{449}$
• D. $A\text{is}\frac{128}{449}$

Answer 1

Answer: (A), (C)

$C\text{is}\frac{128}{449}$

Let $E_1,E_2,E_3$ denotes that the bolt is produced by machines $A,B,C$ respectively.
Let '$A$' be the event that the bolt is defective.
$P\left(E_1\right)=\frac{25}{100},P\left(E_2\right)=\frac{35}{100},P\left(E_3\right)=\frac{40}{100}$
$P\left(\frac{A}{E_1}\right)=\frac{5}{100}\cdot \frac{25}{100}\text{,}P\left(\frac{A}{E_2}\right)=\frac{4}{100}\cdot \frac{35}{100}\text{and}P\left(\frac{A}{E_3}\right)=\frac{2}{100}\cdot \frac{40}{100}$
Using Baye's theorem, we have:
$P\left(E_1/A\right)=\frac{P\left(E_1\right).P\left(A/E_1\right)}{\sum P\left(E_i\right).P\left(A/E_i\right)}=\left[\frac{\frac{25}{100}\times \frac{5}{100}\times \frac{25}{100}}{\frac{25}{100}\times \frac{5}{100}\times \frac{25}{100}+\frac{35}{100}\times \frac{4}{100}\times \frac{35}{100}+\frac{40}{100}\times \frac{2}{100}\times \frac{40}{100}}\right]=\frac{25\times 125}{25\times (125+196+128)}=\frac{125}{449}$
Similarly, we have: $P\left(E_2/A\right)=\frac{196}{449}\text{and}P\left(E_3/A\right)=\frac{128}{449}$