Practicing Success

Target Exam

CUET

Subject

-- Mathematics - Section A

Chapter

Probability

Question:

Probability distribution of random variable X is

X

0

1

2

3

4

5

 P(X) 

 3k 

 3k 

 2k 

 2k2 

 3k2 

 4k2 

The value of k is:

Options:

$k=-1, \frac{1}{9}$

$k=-1$

$k=\frac{1}{9}$

$k=0,-9$

Correct Answer:

$k=\frac{1}{9}$

Explanation:

The correct answer is Option (3) - $k=\frac{1}{9}$