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 and C?

• A. $\frac{25}{69},\frac{28}{69},\frac{16}{69}.$
• B. $\frac{25}{69},\frac{27}{69},\frac{17}{69}.$
• C. $\frac{28}{69},\frac{25}{69},\frac{16}{69}.$
• D. $\frac{27}{69},\frac{26}{69},\frac{16}{69}.$

Answer 1

Answer: (A)

$\frac{25}{69},\frac{28}{69},\frac{16}{69}.$

Here $P\left(A\right)=\frac{25}{100},P\left(B\right)=\frac{35}{100},P\left(C\right)=\frac{40}{100}$

$P\left(\frac{D}{A}\right)=\frac{5}{100},P\left(\frac{D}{B}\right)=\frac{4}{100},P\left(\frac{D}{C}\right)=\frac{2}{100}$

where $D$ denotes defective bolts
Now $P\left(D\right)=P\left(A\right).P\left(\frac{D}{A}\right)+P\left(B\right).P\left(\frac{D}{B}\right)+P\left(C\right).P\left(\frac{D}{C}\right)$

$=\frac{25}{100}.\frac{5}{100}+\frac{35}{100}.\frac{4}{100}+\frac{40}{100}.\frac{2}{100}=0.0345$

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

$=\frac{\frac{25}{100}.\frac{5}{100}}{0.0345}=\frac{25}{69}$

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

$=\frac{\frac{35}{100}.\frac{4}{100}}{0.0345}=\frac{28}{69}$
$P\left(\frac{C}{D}\right)=\frac{P\left(C\right).P\left(\frac{D}{C}\right)}{P\left(A\right).P\left(\frac{D}{A}\right)+P\left(B\right).P\left(\frac{D}{B}\right)+P\left(C\right).P\left(\frac{D}{C}\right)}$

$=\frac{\frac{40}{100}.\frac{2}{100}}{0.0345}=\frac{16}{69}$