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Odds in Favor and Odds against an Event

The value of ...

Question

The value of k, if the probability distribution of a $X = x : 1, 2, 3$ random variable X is $p (X = x) : \frac{1}{k}, \frac{2}{k}, \frac{3}{k}$ is

\frac{1}{6}

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6

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\frac{1}{3}

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\frac{1}{\sqrt{6}}

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Solution

The correct option is A $6$
concept : { sum of probability distribution against random variable always is equal to 1 }
$\sum_{i = 1}^{n} P (x_{i}) = 1$

$\frac{1}{k} + \frac{2}{k} + \frac{3}{k} = 1$

$\frac{6}{k} = 1$

$k = 6$

value of k is 6.

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