Eduprobe
Question:
From a lot of 15 bulbs which include 5 defectives, a sample of 4 bulbs is drawn one by one with replacement. Find the probability distribution of the number of defective bulbs. Hence find the mean of the distribution.
Solution:
Let X = number of defective bulbs out of 4 drawn = 0, 1, 2, 3, 4
Probability of defective bulb = 5/15 = 1/3
Probability of non-defective bulb = 10/15 = 2/3
P(X=0) = 4C0 (2/3)^4 = 16/81
P(X=1) = 4C1 (1/3) (2/3)^3 = 32/81
Similarly, find the probability for x=2, 3, 4.
Probability distribution is shown in the table:
| X | P(X) |
|---|---|
| 0 | 16/81 |
| 1 | 32/81 |
| 2 | 24/81 |
| 3 | 8/81 |
| 4 | 1/81 |
| Mean = ΣxP(x) = (0 * 16/81) + (1 * 32/81) + (2 * 24/81) + (3 * 8/81) + (4 * 1/81) = 80/81 = 4/3 |