Conditional Probabilities defective items

conditional probabilityprobabilitystatistics

A company sends 30% of its product to Client A and 70% to Client B. Client A reports that 5% of the products it received are defective, whereas Client B reports that 4% of products received are defective. The defective products are returned back to the company.

What is the probability of a returned defective product coming from client B?

probability of an item being sent to customer A = $P(C_a) = 0.30$ & $P(D|C_a) = 0.05$

probability of an item being sent to customer B = $P(C_b) = 0.70$ & $P(D|C_b) = 0.04$

$P$(Returned) = $P$(Defective)= $P(D)$ = $0.3 \times 0.05 + 0.7\times 0.04 = 0.043$

My thinking is that I need to find $P$( $C_b$ | Defective ) $= P( C_b | D )$

$P( C_b | D ) = \frac{P(D \,\cap \,C_b)}{P(D)} = \frac{P(D \,\cap \,C_b)}{0.043}\;$ so I need to find $P(D \,\cap \,C_b)$

$P(D|C_b) = \frac{P(D \,\cap \,C_b)}{P(C_b)}\;$ so $0.04 = \frac{P(D \,\cap \,C_b)}{0.70}\;$ which implies $P(D \,\cap \,C_b) = 0.028$

$P( C_b | D ) = \frac{0.028}{0.043} = \frac{28}{43}$

however I feel my answer may be wrong. Am i on the right track?

Best Answer

That's right. You could just say (.04)(.70) / [(.04)(.70) + (.05)(.30)] = .028 / .043 = .651