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(B) Given the probability function $P(X=r) = r/k$ for $r = 1, 2, 3, 4, 5$. Since the sum of all probabilities must be $1$, we have: $\sum_{r=1}^{5} P(

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$P(X=4 	ext{ or } X=k/5)$

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(B) Given the probability function $P(X=r) = r/k$ for $r = 1, 2, 3, 4, 5$. Since the sum of all probabilities must be $1$, we have: $\sum_{r=1}^{5} P(X=r) = 1 \Rightarrow \frac{1}{k} + \frac{2}{k} + \frac{3}{k} + \frac{4}{k} + \frac{5}{k} = 1$ $\frac{1+2+3+4+5}{k} = 1 \Rightarrow \frac{15}{k} = 1 \Rightarrow k = 15$. We need to find $P(X=2 	ext{ or } X=k/3)$. Since $k=15$, $k/3 = 15/3 = 5$. So, $P(X=2 	ext{ or } X=5) = P(X=2) + P(X=5) = \frac{2}{15} + \frac{5}{15} = \frac{7}{15}$. Now, let us check the options: Option $B$: $P(X=4 	ext{ or } X=k/5) = P(X=4 	ext{ or } X=15/5) = P(X=4 	ext{ or } X=3) = \frac{4}{15} + \frac{3}{15} = \frac{7}{15}$. Thus, $P(X=2 	ext{ or } X=k/3) = P(X=4 	ext{ or } X=k/5)$.

$x$	$0$	$1$	$2$	$3$	$4$	$5$
$P(x)$	$k$	$0.3$	$0.15$	$0.15$	$0.1$	$2k$

$X$	$4k$	$\frac{30}{7}k$	$\frac{32}{7}k$	$\frac{34}{7}k$	$\frac{36}{7}k$	$\frac{38}{7}k$	$\frac{40}{7}k$	$6k$
$P(X)$	$\frac{2}{15}$	$\frac{1}{15}$	$\frac{2}{15}$	$\frac{1}{5}$	$\frac{1}{15}$	$\frac{2}{15}$	$\frac{1}{5}$	$\frac{1}{15}$

If the probability function of a discrete random variable $X$ is $P(X=r) = r/k$ for $r = 1, 2, 3, 4, 5$, then $P(X=2 \text{ or } X=k/3)$ is equal to:

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