Question

From a lot of 30 bulbs which include 6 defective, a sample of 4 bulbs is drawn at random with replacement. Find the probability distribution of the number of defective bulbs.

Answer 1

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

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

4bulbs are drawn from the lot with replacement.

Let p= P (obtaining a defective bulb when a bulb is drawn) = $\frac{6}{30}=\frac{1}{5}$

q = P(obtaining a non-defective bulb when a bulb is drawn)

P(X=0)= P (no defective bulb in the sample)

${=}^4{C}_0{p}^0{q}^4={\left(\frac{4}{5}\right)}^4=\frac{256}{625}$ (Using Binomial distribution )

P(X=1)=P (one defective bulb in the sample)

${=}^4{C}_1{p}^1{q}^3=4\left(\frac{1}{5}\right)\left(\frac{4}{5}\right)=\frac{256}{625}$ (Using Binomial distribution )

P(X=2)=P (two defective and two non-defective bulbs are drawn)

${=}^4{C}_2{p}^2{q}^2=6{\left(\frac{1}{5}\right)}^2{\left(\frac{4}{5}\right)}^2=\frac{96}{625}$ (Using Binomial distribution )

P(X=3)=P (three defective and one non-defective bulbs are drawn)

=${=}^4{C}_3{p}^3{q}^1=4{\left(\frac{1}{5}\right)}^3\left(\frac{4}{5}\right)=\frac{16}{625}$ (Using Binomial distribution)

P(X=3)=P (four defective bulbs are drawn) = ${}^4{C}_4{p}^4{q}^0$

$1{\left(\frac{1}{5}\right)}^4{\left(\frac{4}{5}\right)}^0=\frac{1}{625}$

Therefore, the required prpbability distribution is as follows.

$\begin{array}{cccccc}X& 0& 1& 2& 3& 4\\ P\left(X\right)& \frac{256}{625}& \frac{256}{625}& \frac{96}{625}& \frac{16}{625}& \frac{1}{625}\end{array}$