Practicing Success

Target Exam

CUET

Subject

-- Mathematics - Section A

Chapter

Probability

Question:

A random variable X has the following distribution :

X 0 1 2 3 4 5 6 7
P(X) 0 $3k$ $k$ $2k$ $3k^2$ $k^2$ $6k^2$ $3k$

The value of k is :

Options:

$\frac{1}{10}$

-1

$\frac{-1}{10}$

$\frac{1}{9}$

Correct Answer:

$\frac{1}{10}$

Explanation:

The correct answer is Option (1) → $\frac{1}{10}$

$∑P_i(x_i)=1$

$9k+10k^2=1$

$10k^2+9k-1=0$

$10k^2+10k-k-1=0$

$10k(k+1)-1(k+1)=0$

$(10k-1)(k+1)=0$

so $k=\frac{1}{10}$ as $k≠-1$