# 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 number of defective bulbs. Hence find the mean of the distribution. - Mathematics and Statistics

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 number of defective bulbs. Hence find the mean of the distribution.

#### Solution

Total number of bulbs = 15
Number of defective bulbs, i.e., n(D) = 5

∴ P(D) = 5/15=1/3

Number of non-defective bulbs, n(ND) = 10

∴ P(ND) = 10/15=2/3

Let X be a random variable that shows the number of defective bulbs in a draw of 4 bulbs.
Clearly, X can take the values 0, 1, 2, 3 and 4.

P(X = 0) = P (no defective bulbs)

      = (2/3)^4=16/81

P(X = 1) = P (1 defective and 3 non-defective bulbs)

= 4 xx1/3xx(2/3)^3=32/81

P(X = 2) = P (2 defective and 2 non-defective bulbs)

= 6xx(1/3)^2xx(2/3)^2=24/81

P(X = 3) = P (3 defective and 1 non-defective bulb)

  = 4xx (1/3)^3xx 2/3=8/81

P(X = 4) = P (All bulbs are defective)

 = (1/3)^4=1/81

Now, probability distribution is given by

 X 0 1 2 3 4 pi 16/81 32/81 24/81 8/81 1/81

∴ Mean = ∑_i x_i p_i  = 0xx16/81 + 1xx32/81 + 2xx24/81 + 3xx8/81 + 4xx1/81

= 0 + 32/81 + 48/81 + 24/81 + 4/81

=108/81=4/3

Hence, the mean of the distribution is 4/3.

Concept: Random Variables and Its Probability Distributions
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