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.

Answer 1

Given: $6$ bulbs are defective out of $30.$
Then,
Number of non-defective bulbs
$=30-6=24$
A sample of $4$ bulbs is drawn at random with replacement.
Let
$X$: be a number of defective bulbs.
Hence,value of $X$ are $0,1,2,3\&4$
Here,
picking bulbs is a Bernoulli trial

So,$X$ has binomial distribution

$P(X=x){=}^n{C}_x{a}^{n-x}{b}^x...\left(1\right)$

$n=$ number of times we pick a bulb $=4$

$a=$ probability of getting defective bulb

$a=\frac{6}{30}=\frac{1}{5}$

$b=$ probability of getting non defective ball

$b=1-a=1-\frac{1}{5}=\frac{4}{5}$

Substituting values of $a\&b\text{in}\left(1\right).$
Hence,

$P(X=x)={}^4{C}_x{\left(\frac{1}{5}\right)}^x{\left(\frac{4}{5}\right)}^{4-x}$
Now,
$P(X=0)={}^4{C}_0{\left(\frac{1}{5}\right)}^0{\left(\frac{4}{5}\right)}^{4-0}$

$\Rightarrow P(X=0){=}^4{C}_0{\left(\frac{1}{5}\right)}^0{\left(\frac{4}{5}\right)}^4=\frac{256}{625}$

$P(X=1){=}^4{C}_1{\left(\frac{1}{5}\right)}^1{\left(\frac{4}{5}\right)}^{4-1}$

$P(X=1){=}^4{C}_1{\left(\frac{1}{5}\right)}^1{\left(\frac{4}{5}\right)}^3$

$\Rightarrow P(X=1)=4\times \frac{64}{625}=\frac{256}{625}$

$P(X=2){=}^4{C}_2{\left(\frac{1}{5}\right)}^2{\left(\frac{4}{5}\right)}^{4-2}$

$P(X=2){=}^4{C}_2{\left(\frac{1}{5}\right)}^2{\left(\frac{4}{5}\right)}^2$

$\Rightarrow P(X=2)=6\times \frac{16}{625}=\frac{96}{625}$

$P(X=3)={}^4{C}_3{\left(\frac{1}{5}\right)}^3{\left(\frac{4}{5}\right)}^{4-3}$

$P(X=3){=}^4{C}_3{\left(\frac{1}{5}\right)}^3{\left(\frac{4}{5}\right)}^1$

$\Rightarrow P(X=3)=4\times \frac{4}{625}=\frac{16}{625}$

5) $P(X=4){=}^4{C}_4{\left(\frac{1}{5}\right)}^4{\left(\frac{4}{5}\right)}^{4-4}$

$\Rightarrow P(X=4){=}^4{C}_4{\left(\frac{1}{5}\right)}^4{\left(\frac{4}{5}\right)}^0=\frac{1}{625}$

Therefore, probability distribution is

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