$\begin{array}{|c|c|c|c|} \hline X& 0 & 1 & 2 \\ \hline P(X=x)& 3c^3& 4c-10c^2&5c-1\\ \hline \end{array}$

find C and determine $P(X \lt 1)$

Since ,

$\begin{aligned} \sum p_i &= 1 \\ We \ have\ p(0) +p(1) + p(2) &= 1\\ 3c^3+4c-10c^2+5c-1&= 1\\ 3c^3-10c^2+9c-1 &=1\\ 3c^3-10c^2+9c-1-1&=0\\ 3c^3-10c^2+9c-2&=0\\ \end{aligned}$

c=2,1,$\frac{1}{3}$

Consider only smallest number

$c=\frac{1}{3}$

probability distribution is given by

$\begin{array}{|c|c|c|c|} \hline X& 0 & 1 & 2 \\ \hline P(X=x)& 3(\frac{1}{3})^2& 4(\frac{1}{3})-10(\frac{1}{3})^2&5(\frac{1}{3})-1\\ \hline \end{array}$

$\begin{array}{|c|c|c|c|} \hline X& 0 & 1 & 2 \\ \hline P(X=x)& (\frac{1}{9})& (\frac{2}{9})&(\frac{2}{3})\\ \hline \end{array}$

p(x<1)=p(x=0) = $\frac{1}{9}$