The probability mass function of a random var | vedclass

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(B) We know that the sum of all probabilities in a probability mass function must be equal to $1$.
Thus,$\sum_{r = 0}^n P(X = r) = 1$.
$\frac{^n C_0

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$\frac{1}{2}$

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(B) We know that the sum of all probabilities in a probability mass function must be equal to $1$.
Thus,$\sum_{r = 0}^n P(X = r) = 1$.
$\frac{^n C_0 + ^n C_1 + ^n C_2 + \dots + ^n C_n}{32} = 1$.
Using the identity $\sum_{r = 0}^n n C_r = 2^n$,we get $\frac{2^n}{32} = 1$.
$2^n = 32$,which implies $2^n = 2^5$,so $n = 5$.
Now,we need to calculate $P[X \leq 2] = P(X = 0) + P(X = 1) + P(X = 2)$.
$P[X \leq 2] = \frac{^5 C_0}{32} + \frac{^5 C_1}{32} + \frac{^5 C_2}{32}$.
Substituting the values: $^5 C_0 = 1$,$^5 C_1 = 5$,and $^5 C_2 = \frac{5 	imes 4}{2} = 10$.
$P[X \leq 2] = \frac{1 + 5 + 10}{32} = \frac{16}{32} = \frac{1}{2}$.

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

The probability mass function of a random variable $X$ is given by $P[X = r] = \begin{cases} \frac{^n C_r}{32}, & r = 0, 1, 2, \dots, n \\ 0, & \text{otherwise} \end{cases}$. Then,$P[X \leq 2] = $

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