Question

Three machines A, B, and C-are used to produce a large quantity of identical parts at a factory. Machine A produces 60% of the parts, while Machines B and C produce 30% and 10% of the parts, respectively. Historical records indicate that 10% of the parts pro- duced by Machine A are defective, compared with 30% for Machine B and 40% for Machine C. Suppose we randomly select a part produced by one of these three machines at random. ​

What's the probability that the part is defective?

a) 60%
b) 10%
c) 30%
d) 40%

Answer 1

The overall probability of selecting a defective part is calculated using the law of total probability and is 19%. None of the options provided in the question are correct.

Step-by-step explanation:

To calculate the overall probability that a part is defective from machines A, B, and C, we apply the law of total probability. Each machine has a different probability of producing a defective part, and they contribute to the total production proportionally. The overall probability is found by multiplying the likelihood of a part coming from each machine by the probability it is defective and summing these products.

The probabilities are:

• P(A defect) = P(A) × P(defect | A) = 0.6 × 0.1 = 0.06
• P(B defect) = P(B) × P(defect | B) = 0.3 × 0.3 = 0.09
• P(C defect) = P(C) × P(defect | C) = 0.1 × 0.4 = 0.04

Adding these together gives the total probability of a defective part:

P(defective) = P(A defect) + P(B defect) + P(C defect) = 0.06 + 0.09 + 0.04 = 0.19 or 19%

Therefore, none of the options a) 60%, b) 10%, c) 30%, or d) 40% is correct. The correct probability of selecting a defective part is 19%.