If the probability distribution of a ran

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Target Exam

CUET

Subject

-- Mathematics - Section A

Chapter

Determinants

Question:

If the probability distribution of a random variable X is as given below :

X	-1	0	1	2	3
P(X)	K	$\frac{1}{5}$	2K	$\frac{3}{10}$	K

Then the value of K is :

Options:

$\frac{3}{8}$

$\frac{1}{4}$

$\frac{5}{8}$

$\frac{1}{8}$

Correct Answer:

$\frac{1}{8}$

Explanation:

X	-1	0	1	2	3
P(X)	K	$\frac{1}{5}$	2K	$\frac{3}{10}$	K

we know that $\sum P\left(x_i\right)=1$

$\Rightarrow K+\frac{1}{5}+2 K+\frac{3}{10}+K=1$

So $4K + \frac{2+3}{10} = 1$

$4K + \frac{5}{10} = 1$

$K = \frac{6}{10}$

$K = \frac{1}{2×4}$

$K = \frac{1}{8}$