The probability distribution of a random variable x is given below.

Then the value of k is

$\frac{84}{103}$

The correct answer is Option (2) → $\frac{84}{103}$

Given:

$P(x) = \frac{k}{3}, \frac{k}{2}, \frac{k}{4}, \frac{k}{7}$

Total probability = 1

$\Rightarrow \frac{k}{3} + \frac{k}{2} + \frac{k}{4} + \frac{k}{7} = 1$

Take LCM = 84

$\Rightarrow k\left(\frac{28 + 42 + 21 + 12}{84}\right) = 1$

$\Rightarrow k\left(\frac{103}{84}\right) = 1$

$\Rightarrow k = \frac{84}{103}$

Required value of $k = \frac{84}{103}$