Probability distribution using random variable

Question

Let $X$ be a random variable. Suppose $P(X = 1) = \frac{1}{2}$, $P(X = 2)= \frac{1}{3}$, $P(X = 3)= \frac{1}{6}$. Then how can I choose a random number using a probability distribution which takes the above mentioned value?

My work: I could not find any probability distribution function from this point. But using $E[X]$, I can calculate the mean. I think using this one can choose a number.

Kindly help and correct me.

Answer 1

You can pick a randon number $x\in [0,1]$ with uniform distribution and create a roulette: $$f(x)=\begin{cases}1,& x\in \left[0,\frac{1}{2}\right)\\ 2, & x\in \left[\frac{1}{2}, \frac{5}{6}\right)\\ 3, & x\in \left[ \frac{5}{6}, 1\right]\end{cases}$$

Calculating $E[x]$ may give you the expected value, which is $\frac{5}{3}$ - definitely not a number from the set of possible values.

Answer 2

You can toss a dice, if it comes up $\{1, 2, 3\}$, take it as 1; if it comes up $\{4, 5\}$ take it as 2; if $\{6\}$ take it as 3.