Question

In a bolt factory, machine A, B and C manufacture respectively 25%, 35% and 40% of the total bolts. If their respective output 5%, 4% and 2% are defective. A bolt is drawn at random from the product and is found to be defective, the probability that is was manufactured by machine B is

• A. 0.41
• B. 0.21
• C. 0.61
• D. 0.71

Answer 1

The correct option is A 0.41

A: Bolt is manufactured by machine A.
B: Bolt is manufactured by machine B.
C: Bolt is manufactured by machine C.
P(A) = 0.25, P(B) = 0.35, P(C) = 0.40
The probability of drawing a defective bolt manufactured by machine A is $P\left(\frac{D}{A}\right)=0.05$
Similarly, $P\left(\frac{D}{B}\right)=0.04$ and $P\left(\frac{D}{C}\right)=0.02$
By Bay's theorem
$P\left(\frac{B}{D}\right)=\frac{P\left(B\right)P\left(\frac{D}{B}\right)}{P\left(A\right)P\left(\frac{D}{A}\right)+P\left(B\right)P\left(\frac{D}{B}\right)+P\left(C\right)P\left(\frac{D}{C}\right)}$
$=\frac{0.35\times 0.04}{0.25\times 0.05+0.35\times 0.04+0.40\times 0.02}$
$=0.40579\approx 0.41$