Target Exam

CUET

Subject

-- Applied Mathematics - Section B2

Chapter

Probability Distributions

Question:

A discrete random variable X has the following probability distribution:

X 1 2 3 4 5 6
P(X) $\frac{2}{k}$ $\frac{4}{k}$ $\frac{1}{k}$ $\frac{2}{k}$ $\frac{3}{k}$ $\frac{5}{k}$

The value of k is :

Options:

$\frac{2}{17}$

17

5

$\frac{1}{17}$

Correct Answer:

$\frac{1}{17}$

Explanation:

The correct answer is Option (4) → $\frac{1}{17}$

Sum of all probabilities must be 1.

$2k+4k+1k+2k+3k+5k=1$

$17k=1$

$k=\frac{1}{17}$