The random variable X has a probability distribution P(X) of the following form, where k is some number:

$P(X=x)=\left\{\begin{array}{ccc}2 k, & \text { if } & x=0 \\ 3 k, & \text { if } & x=1 \\ 4 k, & \text { if } & x=2 \\ 0, & \text { if } & \text { otherwise }\end{array}\right.$

Then P(X ≤ 1) is equal to:

$\frac{5}{9}$

The correct answer is Option (3) → $\frac{5}{9}$

$P(X=0)+P(X=1)+P(X=2)=1$

$2k + 3k + 4k = 1$

$9k = 1 \;\Rightarrow\; k = \frac{1}{9}$

$P(X \leq 1) = P(X=0) + P(X=1)$

$= 2k + 3k = 5k$

$= 5 \cdot \frac{1}{9} = \frac{5}{9}$

$P(X \leq 1) = \frac{5}{9}$