The random variable X has probability distribution P(X) of the following form, where k is some number:
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 $\mathrm{P}(\mathrm{X}<2), \mathrm{P}(\mathrm{X} \geq 2), \mathrm{P}(\mathrm{X} \geq 2)$.
(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) $P(X<2)=P(X=0)+P(X=1)$
$=k+2 k$
$=3 k$
$=\frac{3}{6}$
$=\frac{1}{2}$
$\mathrm{P}(\mathrm{X} \leq 2)=\mathrm{P}(\mathrm{X}=0)+\mathrm{P}(\mathrm{X}=1)+\mathrm{P}(\mathrm{X}=2)$
$=k+2 k+3 k$
$=6 k$
$=\frac{6}{6}$
$=1$
$\mathrm{P}(\mathrm{X} \geq 2)=\mathrm{P}(\mathrm{X}=2)+\mathrm{P}(\mathrm{X}>2)$
$=3 k+0$
$=3 k$
$=\frac{3}{6}$
$=\frac{1}{2}$