The p.m.f of a random variable X is P(X=x)=fr | vedclass

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(D) Given the p.m.f $P(X=x) = \frac{1}{32} \binom{5}{x}$ for $x \in \{0, 1, 2, 3, 4, 5\}$.$P(X \leq 2) = P(X=0) + P(X=1) + P(X=2)$$= \frac{1}{32} \lef

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

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(D) Given the p.m.f $P(X=x) = \frac{1}{32} \binom{5}{x}$ for $x \in \{0, 1, 2, 3, 4, 5\}$.$P(X \leq 2) = P(X=0) + P(X=1) + P(X=2)$$= \frac{1}{32} \left[ \binom{5}{0} + \binom{5}{1} + \binom{5}{2} \right] = \frac{1}{32} (1 + 5 + 10) = \frac{16}{32} = \frac{1}{2}$.$P(X \geq 3) = P(X=3) + P(X=4) + P(X=5)$$= \frac{1}{32} \left[ \binom{5}{3} + \binom{5}{4} + \binom{5}{5} \right] = \frac{1}{32} (10 + 5 + 1) = \frac{16}{32} = \frac{1}{2}$.Since $P(X \leq 2) = \frac{1}{2}$ and $P(X \geq 3) = \frac{1}{2}$,it follows that $P(X \leq 2) = P(X \geq 3)$.

$X=x$	$0$	$1$	$2$	$3$
$P(X=x)$	$\frac{1}{3}$	$\frac{1}{2}$	$0$	$\frac{1}{6}$

$X=x$	$-3$	$0$	$1$	$\alpha$
$P(X=x)$	$\frac{1}{4}$	$K$	$\frac{1}{4}$	$\frac{1}{3}$

The p.m.f of a random variable $X$ is $P(X=x)=\frac{1}{2^5}\binom{5}{x}$,where $x=0, 1, 2, 3, 4, 5$ and $P(X=x)=0$ otherwise. Then:

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