Question

State whether the given table is not the probability distributions of a random variable. Give reasons for your answer.
$$\begin{array}{cccc}X& 0& 1& 2\\ P\left(X\right)& 0.4& 0.4& 0.2\end{array}$$

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

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

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

Answer 1

We know that conditions for each probability distribution, distribution is 1.

Sum of the probabilities = 0.4 + 0.4 +0.2 = 1
therefore , the given table is a probability distribution of a random variable.

It can be seen that for X = 3, P(X) = - 0.1
It is known that probability of any observation is not negative therefore, the given table is not a probability distribution of random variable

Here, all the probabilities are positive but th esum of probabilities
0.6 + 0.1 +0.2 = 0.8 <1
Thus, the given distribution is not a probability distribution,

Here, all th probabilities are positive but the sum of probabilities
= 0.3 + 0.2 +0.4 +0.1+0.05 = 1.05> 1
Thus, the given distribution is not a probability distribution,