The random variable $X$ has a probability distribution $P(X)$ of the following form,where $k$ is some number:
$P(X) = \begin{cases} k, & \text{if } x=0 \\ 2k, & \text{if } x=1 \\ 3k, & \text{if } x=2 \\ 0, & \text{otherwise} \end{cases}$
Find $P(X < 2)$,$P(X \leq 2)$,and $P(X \geq 2)$.
$P(X) = \begin{cases} k, & \text{if } x=0 \\ 2k, & \text{if } x=1 \\ 3k, & \text{if } x=2 \\ 0, & \text{otherwise} \end{cases}$
Find $P(X < 2)$,$P(X \leq 2)$,and $P(X \geq 2)$.
For any probability distribution,the sum of all probabilities must be $1$.
$\sum P(X) = k + 2k + 3k = 6k = 1 \implies k = \frac{1}{6}$.
$P(X<2) = P(X=0) + P(X=1) = k + 2k = 3k = 3 \times \frac{1}{6} = \frac{1}{2}$.
$P(X \leq 2) = P(X=0) + P(X=1) + P(X=2) = k + 2k + 3k = 6k = 6 \times \frac{1}{6} = 1$.
$P(X \geq 2) = P(X=2) + P(X>2) = 3k + 0 = 3k = 3 \times \frac{1}{6} = \frac{1}{2}$.
$\sum P(X) = k + 2k + 3k = 6k = 1 \implies k = \frac{1}{6}$.
$P(X<2) = P(X=0) + P(X=1) = k + 2k = 3k = 3 \times \frac{1}{6} = \frac{1}{2}$.
$P(X \leq 2) = P(X=0) + P(X=1) + P(X=2) = k + 2k + 3k = 6k = 6 \times \frac{1}{6} = 1$.
$P(X \geq 2) = P(X=2) + P(X>2) = 3k + 0 = 3k = 3 \times \frac{1}{6} = \frac{1}{2}$.
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