A random variable X has the following probability distribution

Then the value of $k$ is

32

The correct answer is Option (1) → 8

Given probability distribution:

$X:$ 2  3  4  5

$P(X): \frac{5}{k}, \frac{7}{k}, \frac{9}{k}, \frac{11}{k}$

Since total probability = 1,

$\frac{5}{k} + \frac{7}{k} + \frac{9}{k} + \frac{11}{k} = 1$

$\frac{(5 + 7 + 9 + 11)}{k} = 1$

$\frac{32}{k} = 1$

$k = 32$

Therefore, the value of $k$ is 32.