The distribution of a random variable X is given below
X = x –2 –1 0 1 2 3
P(X=x) $\frac{1}{10}$ k $\frac{1}{5}$ 2k $\frac{3}{10}$ k

The value of k is

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$\frac{2}{10}$

$\frac{3}{10}$

$\frac{7}{10}$

$\frac{1}{10}$

answer is A.

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Detailed Solution

Given

X = x	–2	–1	0	1	2	3
P(X=x)	$\frac{1}{10}$	k	$\frac{1}{5}$	2k	$\frac{3}{10}$	k

We know that $\sum P (X = x) = 1$

$\Rightarrow \frac{1}{10} + k + \frac{1}{5} + 2 k + \frac{3}{10} + k = 1 \Rightarrow 4 k + \frac{1}{10} + \frac{3}{10} + \frac{1}{5} = 1 \Rightarrow 4 k + \frac{3}{5} = 1 \Rightarrow 4 k = 1 - \frac{3}{5} = \frac{2}{5} \Rightarrow k = \frac{1}{10}$

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