Target Exam

CUET

Subject

-- Mathematics - Section A

Chapter

Probability

Question:

The probability distribution of a random variable X is given by

X

0

1

2

3

4

 P(X) 

 0.1 

 k 

 2k 

 2k 

 k 

The value of k is:

Options:

0

0.1

1

$\frac{3}{20}$

Correct Answer:

$\frac{3}{20}$

Explanation:

The correct answer is Option (4) → $\frac{3}{20}$

Since the sum of all probability be 1.

$0.1+k+2k+2k+k=1$

$6k=0.9$

$k=\frac{9}{60}=\frac{3}{20}$