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Define the random variable:
Let X be the number of defective bulbs in the sample of 2.
X can take values 0, 1, or 2.
Calculate the probabilities:
Total number of bulbs = 30
Number of defective bulbs = 6
Number of non-defective bulbs = 30 - 6 = 24
Since the bulbs are drawn with replacement, the probabilities remain constant for each draw.
P(X = 0): Probability of getting 0 defective bulbs (both non-defective)
P(X = 0) = (24/30) * (24/30) = (4/5) * (4/5) = 16/25
P(X = 1): Probability of getting 1 defective bulb
This can happen in two ways: Defective then Non-defective OR Non-defective then Defective
P(X = 1) = (6/30) * (24/30) + (24/30) * (6/30) = 2 * (1/5) * (4/5) = 8/25
P(X = 2): Probability of getting 2 defective bulbs
P(X = 2) = (6/30) * (6/30) = (1/5) * (1/5) = 1/25
Probability Distribution:
The probability distribution of X is:
| X | P(X) |
|---|---|
| 0 | 16/25 |
| 1 | 8/25 |
| 2 | 1/25 |
Calculate the mean:
Mean (μ) = Σ [X * P(X)]
μ = (0 * 16/25) + (1 * 8/25) + (2 * 1/25) = 0 + 8/25 + 2/25 = 10/25 = 2/5 = 0.4
Correct Answer: 0.4