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

Hint: First we will analyze the question and see that the event of picking is a Bernoulli trial and that it has a binomial distribution. After that we will write it in a binomial distribution formula that is $P\left( X=x \right){{=}^{n}}{{C}_{x}}{{q}^{n-x}}{{p}^{x}}$, we will take $p=\dfrac{6}{30}$ and then using $p+q=1$, we will find $q$ also. We will substitute $x$ from 0 to 4 and then find out the probability distribution table.

Complete step by step answer:
Let X be the number of defective bulbs.
Now picking bulb is a Bernoulli trial that means that the repeated trials of the experiment will give exactly two possible two outcomes that is either Success or failure and it has a binomial distribution.
Therefore, X has binomial distribution. That is the probability distribution of the number of defective balls:
$P\left( X=x \right){{=}^{n}}{{C}_{x}}{{q}^{n-x}}{{p}^{x}}\text{ }.........\text{ Equation 1}\text{.}$
Where: $n$ = number of times we pick a bulb = $4$
$p$ = Probability of getting defective bulb = $\dfrac{\text{number of defectives}}{\text{Total number of bulb}}=\dfrac{6}{30}=\dfrac{1}{5}$
$q=1-p=\left( 1-\dfrac{1}{5} \right)=\dfrac{4}{5}$
Now putting these values in equation 1 we will get :
$P\left( X=x \right){{=}^{4}}{{C}_{x}}{{\left( \dfrac{4}{5} \right)}^{4-x}}{{\left( \dfrac{1}{5} \right)}^{x}}$
Now, finding the probability distribution for $X=0$
$P\left( X=0 \right){{=}^{4}}{{C}_{0}}{{\left( \dfrac{4}{5} \right)}^{4-0}}{{\left( \dfrac{1}{5} \right)}^{0}}=\left( \dfrac{4!}{0!\left( 4-0 \right)!} \right)\times {{\left( \dfrac{4}{5} \right)}^{4-0}}\times 1=\dfrac{256}{625}$
Similarly, we will now substitute $X=1$ :
$P\left( X=1 \right){{=}^{4}}{{C}_{1}}{{\left( \dfrac{4}{5} \right)}^{4-1}}{{\left( \dfrac{1}{5} \right)}^{1}}=\left( \dfrac{4!}{1!\left( 4-1 \right)!} \right)\times {{\left( \dfrac{4}{5} \right)}^{3}}\times \left( \dfrac{1}{5} \right)=\dfrac{256}{625}$
Further, we will again find the probability distribution for $X=2$:
$P\left( X=2 \right){{=}^{4}}{{C}_{2}}{{\left( \dfrac{4}{5} \right)}^{4-2}}{{\left( \dfrac{1}{5} \right)}^{2}}=\left( \dfrac{4!}{2!\left( 4-2 \right)!} \right)\times {{\left( \dfrac{4}{5} \right)}^{2}}\times {{\left( \dfrac{1}{5} \right)}^{1}}=\dfrac{96}{625}$
Now, we will find out for $X=3$
$P\left( X=3 \right){{=}^{4}}{{C}_{3}}{{\left( \dfrac{4}{5} \right)}^{4-3}}{{\left( \dfrac{1}{5} \right)}^{3}}=\left( \dfrac{4!}{3!\left( 4-3 \right)!} \right)\times \left( \dfrac{4}{5} \right)\times {{\left( \dfrac{1}{5} \right)}^{3}}=\dfrac{16}{625}$
And finally we will take $X=4$ :
$P\left( X=4 \right){{=}^{4}}{{C}_{4}}{{\left( \dfrac{4}{5} \right)}^{4-4}}{{\left( \dfrac{1}{5} \right)}^{4}}=\left( \dfrac{4!}{4!\left( 4-4 \right)!} \right)\times {{\left( \dfrac{4}{5} \right)}^{0}}\times {{\left( \dfrac{1}{5} \right)}^{4}}=\dfrac{1}{625}$

So, the probability distribution of the number of defective bulbs is:

Note: Though there is no complexity in the calculation, students can make mistakes while finding the factorials of different numbers. Note that the value of $0!=1$ . Also, in word problems of probability, it is better to fully describe the process so that the examiner can understand every step that you are writing.