Question

A random variable $x$ has following probability distribution

• 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

Given the probabilities for $x$:

$$K,\, \frac{1}{6},\, \frac{3}{8},\, 2K,\, \frac{1}{12}$$

Since these must sum to 1:

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

Combine $K$ terms:

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

Find a common denominator for the fractions (LCM of 6, 8, 12 is 24):

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

Thus:

$$\frac{4}{24} + \frac{9}{24} + \frac{2}{24} = \frac{15}{24} = \frac{5}{8}$$

So, we have:

$$3K + \frac{5}{8} = 1 \quad \Longrightarrow \quad 3K = 1 - \frac{5}{8} = \frac{3}{8}$$

Thus:

$$K = \frac{3}{8} \div 3 = \frac{1}{8}$$