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A large shipment of disposable flashlights contains 1% that are defective. Use the Poisson approximation of the binomial distribution to find the probability that among 200 flashlights randomly selected from the shipment:

a) Exactly 3 will be defective.

b) At most 2 will be defective.

c) At least 3 will be defective.

User Prasanna
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Final answer:

To find the probabilities, we use the Poisson approximation of the binomial distribution. For exactly 3 defective flashlights, the probability is approximately 0.1804. For at most 2 defective flashlights, the probability is approximately 0.6767. And for at least 3 defective flashlights, the probability is approximately 0.3233.

option c is the correct

Step-by-step explanation:

In this problem, we can use the Poisson approximation of the binomial distribution to find the probabilities.

a) To find the probability that exactly 3 flashlights will be defective, we use the formula P(X = x) = (e^(-λ) * λ^x) / x!, where λ = np. In this case, n = 200 and p = 0.01. So, λ = 200 * 0.01 = 2. The probability is P(X = 3) = (e^(-2) * 2^3) / 3! ≈ 0.1804.

b) To find the probability that at most 2 flashlights will be defective, we calculate P(X ≤ 2) = P(X = 0) + P(X = 1) + P(X = 2). Using the same formula, we find P(X = 0) ≈ 0.1353, P(X = 1) ≈ 0.2707, and P(X = 2) ≈ 0.2707. Adding these probabilities, we get P(X ≤ 2) ≈ 0.6767.

c) To find the probability that at least 3 flashlights will be defective, we calculate 1 - P(X ≤ 2). In this case, 1 - 0.6767 ≈ 0.3233. Therefore, the probability that at least 3 flashlights will be defective is approximately 0.3233.

User Petros Kyriakou
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8.3k points
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