State whether the given table is not the probability distributions of a random variable. Give reasons for your answer.
$\begin{matrix} Y & - 1 & 0 & 1 P (Y) & 0.6 & 0.1 & 0.2 \end{matrix}$

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Solution

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

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Similar questions

State whether the given table is not the probability distributions of a random variable. Give reasons for your answer.
$\begin{matrix} x & 0 & 1 & 2 & 3 & 4 P (X) & 0.1 & 0.5 & 0.2 & - 0.1 & 0.3 \end{matrix}$

State whether the given table is not the probability distributions of a random variable. Give reasons for your answer.
$\begin{matrix} Z & 3 & 2 & 1 & 0 & - 1 P (Z) & 0.3 & 0.2 & 0.4 & 0.1 & 0.05 \end{matrix}$

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

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

$\begin{matrix} Y & - 1 & 0 & 1 P (Y) & 0.6 & 0.1 & 0.2 \end{matrix}$

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

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

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

(i)

X	0	1	2
P (X)	0.4	0.4	0.2

(ii)

X	0	1	2	3	4
P (X)	0.1	0.5	0.2	− 0.1	0.3

(iii)

Y	−1	0	1
P (Y)	0.6	0.1	0.2

(iv)

Z	3	2	1	0	−1
P (Z)	0.3	0.2	0.4	0.1	0.05

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State whether the given table is not the probability distributions of a random variable. Give reasons for your answer. Y−101P(Y)0.60.10.2

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

State whether the given table is not the probability distributions of a random variable. Give reasons for your answer.
$\begin{matrix} Y & - 1 & 0 & 1 P (Y) & 0.6 & 0.1 & 0.2 \end{matrix}$

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