The p.m.f. of a random variable X is given by | vedclass

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(D) The given probability mass function is $P(X=x) = \frac{\binom{5}{x}}{2^5}$ for $x \in \{0, 1, 2, 3, 4, 5\}$.$P(X=0) = \frac{\binom{5}{0}}{32} = \f

Vedclass · Accepted Answer

$P(X \leq 2) > P(X \geq 3)$

Vedclass · Answer

(D) The given probability mass function is $P(X=x) = \frac{\binom{5}{x}}{2^5}$ for $x \in \{0, 1, 2, 3, 4, 5\}$.$P(X=0) = \frac{\binom{5}{0}}{32} = \frac{1}{32}$ and $P(X=5) = \frac{\binom{5}{5}}{32} = \frac{1}{32}$. Thus,$P(X=0) = P(X=5)$ is correct.$P(X \leq 1) = P(X=0) + P(X=1) = \frac{1}{32} + \frac{5}{32} = \frac{6}{32}$.$P(X \geq 4) = P(X=4) + P(X=5) = \frac{5}{32} + \frac{1}{32} = \frac{6}{32}$. Thus,$P(X \leq 1) = P(X \geq 4)$ is correct.$P(X \leq 2) = P(X=0) + P(X=1) + P(X=2) = \frac{1}{32} + \frac{5}{32} + \frac{10}{32} = \frac{16}{32} = \frac{1}{2}$.$P(X \geq 3) = P(X=3) + P(X=4) + P(X=5) = \frac{10}{32} + \frac{5}{32} + \frac{1}{32} = \frac{16}{32} = \frac{1}{2}$.Since $P(X \leq 2) = \frac{1}{2}$ and $P(X \geq 3) = \frac{1}{2}$,the statement $P(X \leq 2) > P(X \geq 3)$ is incorrect.

$X = x_i$	$1$	$2$	$3$	$5$
$P(X = x_i)$	$2k^2$	$k$	$k$	$k^2$

$X$	$0$	$2$	$4$	$6$	$8$
$P(X)$	$a$	$2a$	$a+b$	$2b$	$3b$

The p.m.f. of a random variable $X$ is given by $P(X=x) = \begin{cases} \frac{\binom{5}{x}}{2^5}, & x = 0, 1, 2, 3, 4, 5 \\ 0, & \text{otherwise} \end{cases}$. Then which of the following is not correct?

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