Question

State which of the following are not the probability distributions of a random variable. Given reasons for your answer.

$\left(i\right)\begin{array}{cccc}X& & & \\ P\left(X\right)& & & \end{array}$

$\left(ii\right)\begin{array}{cccccc}X& & & & & \\ P\left(X\right)& & & & & \end{array}$

$\left(iii\right)\begin{array}{cccc}Y& & & \\ P\left(Y\right)& & & \end{array}$

$\left(iv\right)\begin{array}{cccccc}Z& & & & & \\ P\left(Z\right)& & & & & \end{array}$

Answer 1

$\left(i\right)$

Given:
$\begin{array}{cccc}X& 0& 1& 2\\ P\left(X\right)& 0.4& 0.4& 0.2\end{array}$

We know that sum of all probabilites of random variables are $1$.
Here,
Sum of the possibilites =$0.4+0.4+0.2=1$
Hence,
The given probability distributions table is of a random varaible.

$\left(ii\right)$

Given:

$\begin{array}{cccccc}X& 0& 1& 2& 3& 4\\ P\left(X\right)& 0.1& 0.5& 0.2& -0.1& 0.3\end{array}$

Always, $P\left(X\right)>0$
So, the given table is not the probability distributions of a random variable

$\left(iii\right)$

Given:

$\begin{array}{cccc}Y& -1& 0& 1\\ P\left(Y\right)& 0.6& 0.1& 0.2\end{array}$

We know that sum of all probability distributions is $1$.

Sum of probabilites

$=0.6+0.1+0.2=0.9\ne 1$

Hence, $Y$ is not a probability distribution.

$\left(iv\right)$

Given:

$\begin{array}{cccccc}Z& 3& 2& 1& 0& -1\\ P\left(Z\right)& 0.3& 0.2& 0.4& 0.1& 0.05\end{array}$

We know that sum of all probability distribution is $1$

Sum of probabilites

$=0.3+0.2+0.4+0.1+0.05=1.05\ne 1$
Hence, $Z$ is not a probability distribution.