Four independent flips of a fair coin are made. Let $X$ denote the number of heads obtained. Plot the probability mass function of the random variable $X-2$.

Let $X$denote the number of heads obtained.

From the information, observe that a fair coin is flipped $4$times independently.

Consider $X$is the random variable that represents the number of heads obtained.

Consider $Y$is the random variable and it is defined as $Y=X-2$

Here, the range of $X$is $0$and $4$because there are $4$flips.

Now, the range of $Y$is,

$0\le X\le 4$

$0-2\le X-2\le 4-2$

$-2\le Y\le 2$

Therefore, the range of $Y$is $-2$and $+2$.

The random variable $X$follows binomial distribution with parameters $n=4$and $p=\frac{1}{2}$because the probability of getting any event is fixed for every flip and number flips is independent.

Calculate the probability distribution of $Y$.

$P(Y=-2)=P(X=0)$

$=\left(\begin{array}{l}n\\ 0\end{array}\right)p^0(1-p{)}^{n-0}$

$=\left(\begin{array}{l}4\\ 0\end{array}\right){\left(\frac{1}{2}\right)}^0{\left(1-\frac{1}{2}\right)}^{4-0}$

$=1\times 1\times {\left(\frac{1}{2}\right)}^4$

$=\frac{1}{16}$

$P(Y=-1)=P(X=1)$

$=\left(\begin{array}{l}n\\ 1\end{array}\right)p^1(1-p{)}^{n-1}$

$=\left(\begin{array}{l}4\\ 1\end{array}\right){\left(\frac{1}{2}\right)}^1{\left(1-\frac{1}{2}\right)}^{4-1}$

$=4\times \frac{1}{2}\times {\left(\frac{1}{2}\right)}^3$

$=\frac{4}{16}$

$P(Y=0)=P(X=2)$

$=\left(\begin{array}{l}n\\ 2\end{array}\right)p^2(1-p{)}^{n-2}$

$=\left(\begin{array}{l}4\\ 2\end{array}\right){\left(\frac{1}{2}\right)}^2{\left(1-\frac{1}{2}\right)}^{4-2}$

$=6\times {\left(\frac{1}{2}\right)}^2\times {\left(\frac{1}{2}\right)}^2$

$=\frac{6}{16}$

$P(Y=1)=P(X=3)$

$=\left(\begin{array}{l}n\\ 3\end{array}\right)p^3(1-p{)}^{n-3}$

$=\left(\begin{array}{l}4\\ 3\end{array}\right){\left(\frac{1}{2}\right)}^3{\left(1-\frac{1}{2}\right)}^{4-3}$

$=4\times {\left(\frac{1}{2}\right)}^3\times \left(\frac{1}{2}\right)$

$=\frac{4}{16}$

$P(Y=2)=P(X=4)$

$=\left(\begin{array}{l}n\\ 4\end{array}\right)p^4(1-p{)}^{n-4}$

$=\left(\begin{array}{l}4\\ 4\end{array}\right){\left(\frac{1}{2}\right)}^4{\left(1-\frac{1}{2}\right)}^{4-4}$

$=1\times {\left(\frac{1}{2}\right)}^4\times {\left(\frac{1}{2}\right)}^0$

$=\frac{1}{16}$

Therefore, the probability distribution is,

$\begin{array}{|cccccc|}\hline Y& -2& -1& 0& 1& 2\\ p\left(y\right)& \frac{1}{16}& \frac{4}{16}& \frac{6}{16}& \frac{4}{16}& \frac{1}{16}\\ \hline\end{array}$

The plot of the probability mass function is as follows: