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Question 1186302: A manufacturing company regularly conducts quality control
checks at specified periods on the products it manufactures.
Historically, the failure rate for LED light bulbs that the company
manufactures is 5%. Suppose a random sample of 10 LED light
bulbs is selected. What is the probability that
a. none of the LED light bulbs are defective?
b. exactly one of the LED light bulbs is defective?
c. two or fewer of the LED light bulbs are defective?
d. three or more of the LED light bulbs are defective?
Answer by CPhill(1959)   (Show Source):
You can put this solution on YOUR website! This is a binomial probability problem. We have a fixed number of trials (n = 10 light bulbs), each trial is independent, and there are only two outcomes (defective or not defective). The probability of a defective bulb (p) is 0.05, and the probability of a non-defective bulb (q) is 1 - p = 0.95.
The binomial probability formula is:
P(x) = (nCx) * p^x * q^(n-x)
Where:
* P(x) is the probability of exactly x successes (defective bulbs in this case)
* n is the number of trials (10)
* x is the number of successes (defective bulbs)
* p is the probability of success (0.05)
* q is the probability of failure (0.95)
* nCx is the binomial coefficient, calculated as n! / (x! * (n-x)!)
Here are the solutions:
**a. None of the LED light bulbs are defective (x = 0):**
P(0) = (10C0) * (0.05)^0 * (0.95)^10
P(0) = 1 * 1 * 0.5987
P(0) ≈ 0.5987
**b. Exactly one of the LED light bulbs is defective (x = 1):**
P(1) = (10C1) * (0.05)^1 * (0.95)^9
P(1) = 10 * 0.05 * 0.6302
P(1) ≈ 0.3151
**c. Two or fewer of the LED light bulbs are defective (x = 0, 1, or 2):**
We need to calculate P(0), P(1), and P(2) and then add them together. We already have P(0) and P(1) from parts a and b.
P(2) = (10C2) * (0.05)^2 * (0.95)^8
P(2) = 45 * 0.0025 * 0.6634
P(2) ≈ 0.0746
P(x ≤ 2) = P(0) + P(1) + P(2)
P(x ≤ 2) ≈ 0.5987 + 0.3151 + 0.0746
P(x ≤ 2) ≈ 0.9884
**d. Three or more of the LED light bulbs are defective (x = 3, 4, ..., 10):**
It's easier to calculate the complement of this event (0, 1, or 2 defective bulbs) and subtract it from 1, since the total probability is equal to 1.
P(x ≥ 3) = 1 - P(x ≤ 2)
P(x ≥ 3) = 1 - 0.9884
P(x ≥ 3) ≈ 0.0116
**In summary:**
* a. P(0) ≈ 0.5987
* b. P(1) ≈ 0.3151
* c. P(x ≤ 2) ≈ 0.9884
* d. P(x ≥ 3) ≈ 0.0116
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