The probability distribution of a random variable X is given under :

$P(X=x)=\left\{\begin{matrix}kx^2 & for \, x =1, 2,3\\2kx& for \, x=4, 5, 6\\0 & for \, \, otherwise \end{matrix}\right.$

Then $P(X≥4)$ is :

$\frac{15}{22}$

The correct answer is Option (4) → $\frac{15}{22}$

$∑P(X)=1$ always

$=k(1)^2+k(2)^2+k(3)^2+2k(4)+2k(5)+2k(6)$

$=k+4k+9k+8k+10k+12k=1$

$44k=1⇒k=\frac{1}{44}$

$P(X≥4)=2k(4+5+6)=2×\frac{1}{44}×15=\frac{15}{22}$