The probability distribution of a discrete random variable $X$ is $\begin{aligned} \begin{array}{||c|c|c|c|c|c||} \hline X & 0 & 1 & 2 & 3 & 4 \ \hline P(X=x) & 2k & k & 2k & 4k & k \ \hline \end{array} \end{aligned}$ If $a=P(x

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The probability distribution of a discrete random variable $X$ is $\begin{aligned} \begin{array}{||c|c|c|c|c|c||} \hline X & 0 & 1 & 2 & 3 & 4 \ \hline P(X=x) & 2k & k & 2k & 4k & k \ \hline \end{array} \end{aligned}$ If $a=P(x < 3)$ and $b=P(2 < x < 4)$, then

Aicharya · Accepted Answer

A discrete random variable $X$ has the probability distribution:

$X$	0	1	2	3	4
$P(X=x)$	$2k$	$k$	$2k$	$4k$	$k$

Since the sum of all probabilities must equal $1$, we have $2k + k + 2k + 4k + k = 10k = 1$, so $k = \frac{1}{10}$.

The value $a$ is defined as $P(X < 3)$, which is $P(X=0) + P(X=1) + P(X=2) = 2k + k + 2k = 5k = 5 \cdot \frac{1}{10} = \frac{1}{2}$.

The value $b$ is $P(2 < X < 4)$, which corresponds to $P(X=3) = 4k = 4 \cdot \frac{1}{10} = \frac{2}{5}$.

Comparing $a = \frac{1}{2}$ and $b = \frac{2}{5}$, we see that $\frac{1}{2} = 0.5 > 0.4 = \frac{2}{5}$, so $a > b$.

Final answer: $\boxed{B}$

The probability distribution of a discrete random variable $X$ is $\begin{aligned}…

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