Question

From a lot of $10$ bulbs, which includes $3$ defectives bulbs, a sample of $2$ bulbs is drawn at random. Find the probability distribution of the number of defective bulbs.

Answer 1

There are 3 defective bulbs and 7 non-defective

bulbs.

Let x denote the random vanable of the

no.of defective bulb.

Then x can take values 0,1,2 since bulbs

are replaced.

$p=p\left(D\right)\frac{3}{10\$}q=p\left(D\right)=1-\frac{3}{10}=\frac{7}{10}$

$p(x=0)=$ $\frac{7c_2\times 3c_0}{10c_2}$ =$\frac{7\times 6}{10\times 4}$=$\frac{7}{15}$

p(x=1)=$\frac{7_{1}3c_2}{{10}_2}$ = $\frac{1\times 3\times 2}{10\times 9}$ $=\frac{7}{15}$

$p(x=2)=\frac{7c_0\times 3c_2}{{10}_{12}}$ $=\frac{1\times 3\times 2}{10\times 9}$= $\frac{1}{15}$

$\therefore$ Requride probability distlibution is .

x	0	1	2
p(x)	7/15	7/15	1/15