Target Exam

CUET

Subject

-- Applied Mathematics - Section B2

Chapter

Probability Distributions

Question:

A random variable X has the following probability distribution:

X

0

1

2

3

4

5

6

7

 P(X) 

 0 

 k 

 2k 

 2k 

 3k 

 $k^2$ 

 $2k^2$ 

 $7k^2+k$ 

Determine: $P(X <3)$

Options:

$\frac{1}{10}$

$\frac{3}{10}$

$\frac{2}{10}$

$\frac{4}{10}$

Correct Answer:

$\frac{3}{10}$

Explanation:

The correct answer is Option (2) → $\frac{3}{10}$

We know that $Σp_i = 1$

$⇒ 0+ k + 2k + 2k + 3k + k^2 + 2k^2 + 7k^2 + k=1$

$⇒ 10k^2 + 9k-1=0⇒ (10k-1) (k + 1) = 0$

$⇒ k=\frac{1}{10},-1$ but $k$ cannot be negative

$⇒ k=\frac{1}{10}$

$P(X < 3) = P(0) + P(1) + P(2) = 0 + k + 2k$

$=3k=3×\frac{1}{10}=\frac{3}{10}$