Question

find the mean of the probability distribution of the random variable x, which can take only the values 1,2 and 3 given that p(1)=10/33,p(2)= 1/3, and p(3)= 12/33

Answer 1

The probability of each of these events and the corresponding value of X, can be summarized below.

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

The mean (also known as the expected value) 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 $2\frac{2}{33}.$