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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

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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,$P(X=2 	ext{ or } X=k/3) = P(X=4 	ext{ or } X=k/5)$.

$X$	$1$	$2$	$3$	$4$	$5$	$6$	$7$	$8$
$P(X)$	$0.15$	$0.23$	$0.12$	$0.10$	$0.20$	$0.08$	$0.07$	$0.05$

$X = x$	$0$	$1$	$2$	$3$	$4$
$P(X = x)$	$k$	$2k$	$4k$	$2k$	$k$

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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The probability distribution of a discrete random variable $X$ is given by the following table:
$X = x$ $0$ $1$ $2$ $3$ $4$
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