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From a lot of 30 bulbs which include 6 defectives, a sample of 4 bulbs is drawn at random with replacement. Find the probability distribution of the number of defective bulbs.
Given: A lot of 30 bulbs which include 6 defectives.
Then number of non-defective bulbs = 30 – 6 = 24
As 4 bulbs are drawn at random with replacement.
Let X denotes the number of defective bulbs from the selected bulbs.
Clearly, X can take the value of 0, 1, 2, 3 or 4.
P(X = 0) = P(4 are non defective and 0 defective)
P(X = 1) = P(3 are non defective and 1 defective)
P(X = 2) = P(2 are non defective and 2 defective)
P(X = 3) = P(1 are non defective and 3 defective)
P(X = 4) = P(0 are non defective and 4 defective)
Hence, the required probability distribution is,
X | 0 | 1 | 2 | 3 | 4 |
P(X) |
A random variable X has the following probability distribution:
X | 0 | 1 | 2 | 3 | 4 | 5 | 6 | 7 |
P(X) | 0 | k | 2k | 2k | 3k | K2 | 2 K2 | 7 K2 + k |
Determine
(i) k (ii) P(X < 3)
(iii) P(X > 6) (iv) P(0 < X < 3)