# From a lot of 30 bulbs which include 6 defectives, a sample of 4 bulbs is drawn at random with replacement..

Question:

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.

Solution:

It is given that out of 30 bulbs, 6 are defective.

⇒ Number of non-defective bulbs = 30 − 6 = 24

4 bulbs are drawn from the lot with replacement.

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

$\therefore P(X=0)=P(4$ non-defective and 0 defective $)={ }^{4} C_{0} \cdot \frac{4}{5} \cdot \frac{4}{5} \cdot \frac{4}{5} \frac{4}{5}=\frac{256}{625}$

$\mathrm{P}(\mathrm{X}=1)=\mathrm{P}(3$ non-defective and 1 defective $)={ }^{4} C_{1} \cdot\left(\frac{1}{5}\right) \cdot\left(\frac{4}{5}\right)^{3}=\frac{256}{625}$

$P(X=2)=P(2$ non-defective and 2 defective $)={ }^{4} C_{2} \cdot\left(\frac{1}{5}\right)^{2} \cdot\left(\frac{4}{5}\right)^{2}=\frac{96}{625}$

$P(X=3)=P(1$ non-defective and 3 defective $)={ }^{4} C_{3} \cdot\left(\frac{1}{5}\right)^{3} \cdot\left(\frac{4}{5}\right)=\frac{16}{625}$

$P(X=4)=P(0$ non-defective and 4 defective $)={ }^{4} C_{4} \cdot\left(\frac{1}{5}\right)^{4} \cdot\left(\frac{4}{5}\right)^{0}=\frac{1}{625}$

Therefore, the required probability distribution is as follows.