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

Let $X=$ number of defective bulbs out of 4 drawn $=0,1,2,3,4$

Probability of defective bulb = $\frac{5}{15}=\frac{1}{3}$

Probability of non-defective bulb = $1-\frac{1}{3}=\frac{2}{3}$

$P(X=0){=}^4{C}_0.(\frac{2}{3}{)}^4=\frac{16}{81}$

$P(X=1){=}^4{C}_1.\frac{1}{3}.(\frac{2}{3}{)}^3=\frac{32}{81}$

Similarly, find the probability for $x=2,3,4.$

Probability distribution is shown in the table:

Mean $=$ $\sum xP\left(x\right)=\frac{108}{81}\Rightarrow \frac{4}{3}.$