Question:

Machines A, B, C make 20%, 30%, 50% of bolts; defect rates 3%, 4%, 2%. Given a bolt is defective, probability it's from C is

$\dfrac{5}{14}$

$\dfrac{9}{28}$

$\dfrac{3}{7}$

$\dfrac{2}{7}$

Step-by-Step Solution

Key Concept: $P(D)=0.2\times0.03+0.3\times0.04+0.5\times0.02=0.006+0.012+0.01=0.028$. $P(C|D)=\frac{0.5\times0.02}{0.028}$.

$P(C|D)=\frac{5}{14}$.

Correct Answer: 1