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 :

$\frac{3}{8}$ $\frac{1}{4}$ $\frac{5}{8}$ $\frac{1}{8}$

$\frac{1}{8}$

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  $4K + \frac{2+3}{10} = 1$ $4K + \frac{5}{10} = 1$ $K = \frac{6}{10}$ $K = \frac{1}{2×4}$ $K = \frac{1}{8}$