The probability mass function of a random X is zero except at the points X=0,1,2 . At these points it has the values $P(0)=3c^3 ,P(1)=4c-10c^2 ,P(2)=5c-1$.

(i) find c

(ii) Find P(X<1) , P(1 < X ≤ 2) , P(0 < X ≤2)

(i) We know that ∑p$_i$ = 1, we have $p(0) + p(1) +p(2) = 1$

i.e. $3c^3 + 4c-10c^2 + 5c-1 =1 \,\,\ or \,\,\ 3c^3 -10c^2 + 9c-2 = 0$

$(3c -1 )(c -2)(c – 1)=0$

∴ c = $\frac{1}{3}$

By putting the value of c, the probability distribution is

(ii) $P(X \lt 1) = P(X=0)$ = $\frac{1}{9}$; $P(1\lt X ≤ 2) = P(2)$ = $\frac{2}{3}$

$P(0 \lt X ≤ 2) = P(1) +P(2) = \frac{2}{9} + \frac{2}{3} = \frac{8}{9}$