If $X$ is a random variable which can assume values 0, 1, 2, 3 or 4 such that $P(X = 1) = P(X = 2)$ and $3P(X = 3) = 4P(X = 4) = P(X = 0) = \frac{1}{8}$ then $P(X > 0)$ is:

$\frac{7}{8}$

The correct answer is Option (4) → $\frac{7}{8}$

Given:

• $P(X = 0) = \frac{1}{8}$
• $P(X = 1) = P(X = 2)$
• $3P(X = 3) = 4P(X = 4) = \frac{1}{8}$

From the last condition:

$P(X = 3) = \frac{1}{24},\quad P(X = 4) = \frac{1}{32}$

Let $P(X = 1) = P(X = 2) = x$

Total probability:

$\frac{1}{8} + x + x + \frac{1}{24} + \frac{1}{32} = 1$

Then:

$\frac{12}{96} + 2x + \frac{4}{96} + \frac{3}{96} = 1$

$2x = 1 - \frac{19}{96} = \frac{77}{96}$

$x = \frac{77}{192}$

Now compute $P(X > 0)$:

$P(X > 0) = P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4)$

$= 2x + \frac{1}{24} + \frac{1}{32}$

$= \frac{154}{192} + \frac{4}{96} + \frac{3}{96}$

$= \frac{154}{192} + \frac{7}{96} = \frac{77}{96} + \frac{7}{96} = \frac{84}{96} = \frac{7}{8}$