State whether the following table represents a probability distribution of a random variable $X$. Give reasons for your answer.
$X$ $0$ $1$ $2$ $P(X)$ $0.4$ $0.4$ $0.2$
| $X$ | $0$ | $1$ | $2$ |
| $P(X)$ | $0.4$ | $0.4$ | $0.2$ |
(A) For a table to represent a probability distribution of a random variable $X$,it must satisfy two conditions:
$1$. Each probability $P(X_i)$ must be non-negative,i.e.,$P(X_i) \ge 0$ for all $i$.
$2$. The sum of all probabilities must be equal to $1$,i.e.,$\sum P(X_i) = 1$.
Checking the given table:
$1$. All probabilities $(0.4, 0.4, 0.2)$ are $\ge 0$.
$2$. The sum of probabilities is $0.4 + 0.4 + 0.2 = 1.0$.
Since both conditions are satisfied,the given table is a valid probability distribution of the random variable $X$.
$1$. Each probability $P(X_i)$ must be non-negative,i.e.,$P(X_i) \ge 0$ for all $i$.
$2$. The sum of all probabilities must be equal to $1$,i.e.,$\sum P(X_i) = 1$.
Checking the given table:
$1$. All probabilities $(0.4, 0.4, 0.2)$ are $\ge 0$.
$2$. The sum of probabilities is $0.4 + 0.4 + 0.2 = 1.0$.
Since both conditions are satisfied,the given table is a valid probability distribution of the random variable $X$.
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$X$ $4k$ $\frac{30}{7}k$ $\frac{32}{7}k$ $\frac{34}{7}k$ $\frac{36}{7}k$ $\frac{38}{7}k$ $\frac{40}{7}k$ $6k$ $P(X)$ $\frac{2}{15}$ $\frac{1}{15}$ $\frac{2}{15}$ $\frac{1}{5}$ $\frac{1}{15}$ $\frac{2}{15}$ $\frac{1}{5}$ $\frac{1}{15}$
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