Question

Find the probability distribution of the random variable X, which can take only the values 1,2,3 given that P(1)=10/30,P(2)=1/3 and P(3)=12/33.

Answer 1

The Probability Distribution is as follows:

$\begin{matrix} x & 1 & 2 & 3\\ P(x) & \frac{10}{33} & \frac{1}{3} & \frac{12}{33}. \end{matrix}$

The mean of a discrete random variable X is the number given by

$\mu =E(x)=\sum x P(x)=1\left(\frac{10}{33}\right)+2\left(\frac{1}{3}\right)+3\left(\frac{12}{33}\right),\\ =\frac{10}{33}+\frac{2}{3}+\frac{36}{33}=\frac{10}{33}+\frac{22}{33}+\frac{36}{33}=\frac{68}{33}=2\frac{2}{33}$

Therefore, mean of the random variable x is

$=2\frac{2}{33}$