Question

suppose three test kits are tested at random. let D represent the defective test kit and let N represent the non defective test kit. if we let x be the random variable for the number of defective test kits, construct the probability distribution of the random variable X

Answer 1

$D$ and $N$ are equally likely to occur.

Then we have binomial distribution with parameters

$p=\dfrac{1}{2}$

$n=3$

Denote probability for $k$ defective test kits as $P(X=k)$.

Then

$P(X=k)=\dbinom{n}{k}p^k(1-p)^{n-k}$

Probabilities for each $k$ are

$P(X=0)=\dbinom{3}{0}\cdot\left(\dfrac{1}{2}\right)^0\cdot\left(\dfrac{1}{2}\right)^3=$ $\dfrac{3!}{0!3!}\cdot\dfrac{1}{8}=\dfrac{1}{8}$

$P(X=1)=\dbinom{3}{1}\cdot\left(\dfrac{1}{2}\right)^1\cdot\left(\dfrac{1}{2}\right)^2=$ $\dfrac{3!}{1!2!}\cdot\dfrac{1}{8}=\dfrac{3}{8}$

$P(X=2)=\dbinom{3}{2}\cdot\left(\dfrac{1}{2}\right)^2\cdot\left(\dfrac{1}{2}\right)^1=$ $\dfrac{3!}{2!1!}\cdot\dfrac{1}{8}=\dfrac{3}{8}$

$P(X=3)=\dbinom{3}{3}\cdot\left(\dfrac{1}{2}\right)^3\cdot\left(\dfrac{1}{2}\right)^0=$ $\dfrac{3!}{3!0!}\cdot\dfrac{1}{8}=\dfrac{1}{8}$

Check the sum of probabilities

$\displaystyle\sum\limits_{k=0}^3P(X=k)=\dfrac{1}{8}+\dfrac{3}{8}+\dfrac{3}{8}+\dfrac{1}{8}=1$

Construct the probability distribution

$\begin{array}{|c|c|c|c|c|}\hline \\X&0&1&2&3\\ \\ \hline \\P(X=k)&\dfrac{1}{8}&\dfrac{3}{8}&\dfrac{3}{8}&\dfrac{1}{8}\\ \\\hline\end{array}$