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

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

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

(a) Determine the value of $k$.

(b) Find $\mathrm{P}(\mathrm{X}<2), \mathrm{P}(\mathrm{X} \geq 2), \mathrm{P}(\mathrm{X} \geq 2)$.

Solution:

(a) It is known that the sum of probabilities of a probability distribution of random variables is one.

$\therefore k+2 k+3 k+0=1$

$\Rightarrow 6 k=1$

$\Rightarrow k=\frac{1}{6}$

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

$=k+2 k$

$=3 k$

$=\frac{3}{6}$

$=\frac{1}{2}$

$\mathrm{P}(\mathrm{X} \leq 2)=\mathrm{P}(\mathrm{X}=0)+\mathrm{P}(\mathrm{X}=1)+\mathrm{P}(\mathrm{X}=2)$

$=k+2 k+3 k$

$=6 k$

$=\frac{6}{6}$

$=1$

$\mathrm{P}(\mathrm{X} \geq 2)=\mathrm{P}(\mathrm{X}=2)+\mathrm{P}(\mathrm{X}>2)$

$=3 k+0$

$=3 k$

$=\frac{3}{6}$

$=\frac{1}{2}$

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