A random variable X has the following pr

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Target Exam

CUET

Subject

-- Applied Mathematics - Section B2

Chapter

Probability Distributions

Question:

A random variable X has the following probability distribution :

x	1	2	3	4	5
P(X=x)	$K$	$K$	$3K$	$2K^2$	$4K^2$

Based on the above information, the value of 'K' is :

Options:

$\frac{1}{6}$

$\frac{1}{5}$

$\frac{1}{3}$

$\frac{5}{6}$

Correct Answer:

$\frac{1}{6}$

Explanation:

$\text{Total probability}=1.$

$K+K+3K+2K^2+4K^2=1.$

$5K+6K^2=1.$

$6K^2+5K-1=0.$

$K=\frac{-5+\sqrt{25+24}}{12}=\frac{-5+7}{12}=\frac{1}{6}.$

$K=\frac{1}{6}.$