If a discrete variable X has the following probability distribution : X $\frac{2}{3}$ 1 $\frac{4}{3}$ P(X) $c^2$ $c^2$ $c$ then c is :

$\frac{1}{2}$ $\frac{1}{3}$ $\frac{1}{4}$ $\frac{1}{5}$

$\frac{1}{2}$

From the table $∑p_i=1⇒c^2+c^2+c=1$ so $2c^2+c-1=0$ $2c^2+2c-c-1=0⇒2c(c+1)-1(c+1)=0$ so $c=\frac{1}{2},-1$ (Not possible) so $c=\frac{1}{2}$