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

Answer 1

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.`