Question

A random variable $x$ has following probability distribution

Then the value of K is

• A. $\frac{1}{8}$
• B. $\frac{3}{2}$
• C. $\frac{1}{2}$
• D. $\frac{1}{4}$

Answer 1

Answer: (A)

$\frac{1}{8}$

Answer 2

The sum of probabilities must equal 1:

$K + \frac{1}{6} + \frac{3}{8} + 2K + \frac{1}{12} = 1$.

Combine the terms:

$3K + \left(\frac{1}{6} + \frac{3}{8} + \frac{1}{12}\right) = 1$.

Find a common denominator (24):

$\frac{1}{6} = \frac{4}{24}, \quad \frac{3}{8} = \frac{9}{24}, \quad \frac{1}{12} = \frac{2}{24}$.

Thus,

$3K + \frac{4+9+2}{24} = 3K + \frac{15}{24} = 1 \implies 3K + \frac{5}{8} = 1$.

Solve for $K$:

$3K = 1 - \frac{5}{8} = \frac{3}{8} \implies K = \frac{1}{8}$.