Answer:
1) The probability of at least 1 defective is approximately 45.621%
2) The probability that there will be exactly 3 shipments each containing at least 1 defective device among the 20 devices that are tested from the shipment is approximately 16.0212%
Explanation:
The given parameters are;
The defective rate of the device = 3%
Therefore, the probability that a selected device will be defective, p = 3/100
The probability of at least one defective item in 20 items inspected is given by binomial theorem as follows;
The probability that a device is mot defective, q = 1 - p = 1 - 3/100 = 97/100 = 0.97
The probability of 0 defective in 20 = ₂₀C₀(0.03)⁰·(0.97)²⁰ ≈ 0.543794342927
The probability of at least 1 = 1 - The probability of 0 defective in 20
∴ The probability of at least 1 = 1 - 0.543794342927 = 0.45621
The probability of at least 1 defective ≈ 0.45621 = 45.621%
2) The probability of at least 1 defective in a shipment, p ≈ 0.45621
Therefore, the probability of not exactly 1 defective = q = 1 - p
∴ q ≈ 1 - 0.45621 = 0.54379
The probability of exactly 3 shipment with at least 1 defective, P(Exactly 3 with at least 1) is given as follows;
P(Exactly 3 with at least 1) = ₁₀C₃(0.45621)³(0.54379)⁷ ≈ 0.160212
Therefore, the probability that there will be exactly 3 shipments each containing at least 1 defective device among the 20 devices that are tested from the shipment is 16.0212%
