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: $P(X>6)$

Options:

$\frac{1}{100}$

$\frac{13}{100}$

$\frac{17}{100}$

$\frac{11}{100}$

Correct Answer:

$\frac{17}{100}$

Explanation:

The correct answer is Option (3) → $\frac{17}{100}$

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

$⇒ k=\frac{1}{10}$

$P(X > 6) = P(7) = 7k^2 + k = 7.\frac{1}{100}+\frac{1}{10}=\frac{17}{100}$