Question

From a lot of 30 bulbs which include 6 defectives, a sample of 2 bulbs are drawn at random with replacement. Find the probability distribution of the number of defective bulbs.

Answer 1

The question says that out of $30$ bulbs, $6$ are defective.

So, Number of non-defective bulbs =$30-6=24$.

$2$ bulbs are drawn from the lot at random with replacement.

Let $X$ be the random variable that denotes the number of defective bulbs in the selected bulbs.

Hence,

$P(X=0)=$ P ($2$ non- defective and $0$ defective)

$=C(2,0)$.$\frac{24}{30}.$$\frac{24}{30}$=$\frac{16}{25}$

$P(X=1)=$ P ($1$ non- defective and $1$ defective)

$=C(2,1)$.$\frac{24}{30}.$$\frac{6}{30}$ =$\frac{8}{25}$

$P(X=2)=$ P ($0$ non- defective and $2$ defective)

=C(2,2).$\frac{6}{30}.$$\frac{6}{30}$=$\frac{1}{25}$

Therefore, the required probability distribution is as follows.

X	0	1	2
P(X)	$\frac{16}{25}$	$\frac{8}{25}$	$\frac{1}{25}$