If the p.m.f. of a random variable X is given | vedclass

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(C) The given probability mass function is $P(X=x) = \frac{\binom{5}{x}}{2^5}$ for $x \in \{0, 1, 2, 3, 4, 5\}$.Calculating the probabilities:$P(X=0)

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$P(X \leq 3) \leq P(X \geq 3)$

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(C) The given probability mass function is $P(X=x) = \frac{\binom{5}{x}}{2^5}$ for $x \in \{0, 1, 2, 3, 4, 5\}$.Calculating the probabilities:$P(X=0) = \frac{\binom{5}{0}}{32} = \frac{1}{32}$,$P(X=1) = \frac{\binom{5}{1}}{32} = \frac{5}{32}$,$P(X=2) = \frac{\binom{5}{2}}{32} = \frac{10}{32}$,$P(X=3) = \frac{\binom{5}{3}}{32} = \frac{10}{32}$,$P(X=4) = \frac{\binom{5}{4}}{32} = \frac{5}{32}$,$P(X=5) = \frac{\binom{5}{5}}{32} = \frac{1}{32}$.Now evaluate the options:$A$. $P(X \leq 1) = P(0) + P(1) = \frac{1+5}{32} = \frac{6}{32}$ and $P(X \geq 4) = P(4) + P(5) = \frac{5+1}{32} = \frac{6}{32}$. Thus,$P(X \leq 1) = P(X \geq 4)$ is true.$B$. $P(X \leq 2) = P(0) + P(1) + P(2) = \frac{1+5+10}{32} = \frac{16}{32} = 0.5$. $P(X \geq 4) = \frac{6}{32} = 0.1875$. Since $0.5 \geq 0.1875$,this is true.$C$. $P(X \leq 3) = P(0) + P(1) + P(2) + P(3) = \frac{1+5+10+10}{32} = \frac{26}{32} = 0.8125$. $P(X \geq 3) = P(3) + P(4) + P(5) = \frac{10+5+1}{32} = \frac{16}{32} = 0.5$. Since $0.8125 \leq 0.5$ is false,option $C$ is not true.$D$. $P(X \leq 2) = \frac{16}{32} = 0.5$ and $P(X \geq 3) = \frac{16}{32} = 0.5$. Thus,$P(X \leq 2) = P(X \geq 3)$ is true.

$X$	$1$	$2$	$3$	$4$	$5$
$P(X=x)$	$K$	$2K$	$3K$	$2K$	$K$

$X=x_i$	$-2$	$-1$	$0$	$1$	$2$
$P(X=x_i)$	$0.2$	$0.3$	$0.15$	$0.25$	$0.1$

$X = x_i$	$0$	$1$	$2$	$3$	$4$
$P(X = x_i)$	$0.4$	$0.3$	$0.1$	$0.1$	$0.1$

If the p.m.f. of a random variable $X$ is given by $P(X=x) = \frac{\binom{5}{x}}{2^{5}}$ for $x = 0, 1, 2, \ldots, 5$ and $0$ otherwise,then which of the following is not true?

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