Question

From a lot of 15 bulbs which include 5 defective, 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. [CBSE 2014]

Answer 1

Let getting a defective bulb in a trial be a success.

We have,

$$\begin{gathered} p=\mathrm{probability}\mathrm{of}\mathrm{getting}\mathrm{a}\mathrm{defective}\mathrm{bulb}=\frac{5}{15}=\frac{1}{3}\mathrm{and} \\ q=\mathrm{probability}\mathrm{of}\mathrm{getting}\mathrm{non}-\mathrm{defective}\mathrm{bulb}=1-p=1-\frac{1}{3}=\frac{2}{3} \\ \mathrm{Let}X\mathrm{denote}\mathrm{the}\mathrm{number}\mathrm{of}\mathrm{success}\mathrm{in}\mathrm{a}\mathrm{sample}\mathrm{of}4\mathrm{trials}.\mathrm{Then}, \\ X\mathrm{follows}\mathrm{binomial}\mathrm{distribution}\mathrm{with}\mathrm{parameters}n=4\mathrm{and}p=\frac{1}{3} \\ \therefore \mathrm{P}\left(X=r\right)={}^4\mathrm{C}_r{p}^r{q}^{\left(4-r\right)}={}^4\mathrm{C}_r{\left(\frac{1}{3}\right)}^r{\left(\frac{2}{3}\right)}^{\left(4-r\right)}=\frac{{}^4\mathrm{C}_r{2}^{\left(4-r\right)}}{3^4},\mathrm{where}r=0,1,2,3,4 \\ \mathrm{i}.\mathrm{e}. \\ \mathrm{P}\left(X=0\right)=\frac{{}^4\mathrm{C}_0{2}^4}{3^4}=\frac{16}{81}, \\ \mathrm{P}\left(X=1\right)=\frac{{}^4\mathrm{C}_1{2}^3}{3^4}=\frac{32}{81}, \\ \mathrm{P}\left(X=2\right)=\frac{{}^4\mathrm{C}_2{2}^2}{3^4}=\frac{24}{81}, \\ \mathrm{P}\left(X=3\right)=\frac{{}^4\mathrm{C}_3{2}^1}{3^4}=\frac{8}{81}, \\ \mathrm{P}\left(X=4\right)=\frac{{}^4\mathrm{C}_4{2}^0}{3^4}=\frac{1}{81} \end{gathered}$$

So, the probability distribution of X is given as follows:

X:	0	1	2	3	4
P(X):	$\frac{16}{81}$	$\frac{32}{81}$	$\frac{24}{81}$	$\frac{8}{81}$	$\frac{1}{81}$

Now,

$$\begin{gathered} \mathrm{Mean},\mathrm{E}\left(X\right)=0\times \frac{16}{81}+1\times \frac{32}{81}+2\times \frac{24}{81}+3\times \frac{8}{81}+4\times \frac{1}{81} \\ =\frac{32+48+24+4}{81} \\ =\frac{108}{81} \\ =\frac{4}{3} \\ \mathrm{Note}:\mathrm{We}\mathrm{can}\mathrm{also}\mathrm{calculate}\mathrm{the}\mathrm{mean}\mathrm{of}\mathrm{the}\mathrm{binomial}\mathrm{distribution}\mathrm{by} \\ \mathrm{Mean},\mathrm{E}\left(X\right)=np=4\times \frac{1}{3}=\frac{4}{3} \end{gathered}$$