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If $P (X = x) = k {(\frac{3}{8})}^{X}, x = 1, 2, 3, \dots$ is the probability distribution functiion of a discrete random variable X, then k =

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$\frac{5}{8}$

$\frac{8}{3}$

$\frac{5}{3}$

$\frac{4}{3}$

answer is C.

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Detailed Solution

$P (X = x) = k {(\frac{3}{8})}^{X}, x = 1, 2, 3, \dots$

$\Rightarrow k {(\frac{3}{8})}^{1} + k {(\frac{3}{8})}^{2} + k {(\frac{3}{8})}^{3} + \cdot \cdot \cdot = 1$

$\Rightarrow 1 + {(\frac{3}{8})}^{1} + {(\frac{3}{8})}^{2} + {(\frac{3}{8})}^{3} + \cdot \cdot \cdot = \frac{1}{k} + 1$

$\Rightarrow \frac{1}{1 - \frac{3}{8}} = \frac{1}{k} + 1$

$k = \frac{5}{3}$

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