Find k for the above probability distribution , also find P(X<4).

Solution: We know that

$\sum P(X=x)=1$

$i=0,1,2,3$,

$\therefore \quad P(x=1)+P(x=2)+P(x=3)+P(x=4)+P(x=5)+P(x=6)=1$

$\therefore \quad 3 k+5 k+7 k+9 k+11 k+13 k=1$

$\therefore \quad 48 k=1$

$\therefore \quad k=\frac{1}{48}$

$\begin{aligned} & \begin{aligned} P(x\lt4) &=P(x=1)+P(x=2)+P(x=3) \\ &=3 k+5 k+7 k \\ &=15 k \\ &=15 \times \frac{1}{48} \\ p(x\lt4) &=\frac{15}{48} \end{aligned} \end{aligned}$