The probability mass function of a random var | vedclass

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(B) Given the probability mass function $P(X=x) = \frac{{}^{5}C_{x}}{2^{5}}$ for $x \in \{0, 1, 2, 3, 4, 5\}$.We need to calculate $P(X \leq 2) = P(X=

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$P(X \geq 3)$

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(B) Given the probability mass function $P(X=x) = \frac{{}^{5}C_{x}}{2^{5}}$ for $x \in \{0, 1, 2, 3, 4, 5\}$.We need to calculate $P(X \leq 2) = P(X=0) + P(X=1) + P(X=2)$.$P(X=0) = \frac{{}^{5}C_{0}}{2^{5}} = \frac{1}{32}$$P(X=1) = \frac{{}^{5}C_{1}}{2^{5}} = \frac{5}{32}$$P(X=2) = \frac{{}^{5}C_{2}}{2^{5}} = \frac{10}{32}$Summing these values: $P(X \leq 2) = \frac{1+5+10}{32} = \frac{16}{32} = \frac{1}{2}$.Now,check the options:$P(X \geq 3) = P(X=3) + P(X=4) + P(X=5) = \frac{{}^{5}C_{3} + {}^{5}C_{4} + {}^{5}C_{5}}{2^{5}} = \frac{10+5+1}{32} = \frac{16}{32} = \frac{1}{2}$.Since $P(X \leq 2) = \frac{1}{2}$ and $P(X \geq 3) = \frac{1}{2}$,we have $P(X \leq 2) = P(X \geq 3)$.

$X$	$0$	$1$	$2$	$3$	$4$	$5$	$6$	$7$
$P(X)$	$0$	$k$	$2k$	$3k$	$3k^2$	$k^2$	$2k^2$	$7k^2+k$

The probability mass function of a random variable $X$ is given by $P(X=x) = \frac{{}^{5}C_{x}}{2^{5}}$ for $x = 0, 1, 2, 3, 4, 5$ and $0$ otherwise. Then,$P(X \leq 2)$ is equal to:

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