# Conditional Probabilities defective items

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?

• Looks correct to me. Nov 8, 2020 at 15:11
• Nov 8, 2020 at 20:40

## 1 Answer

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