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) = 1/33, and P(3) = 12/33

Answer 1

Check

$$\sum _ip(x_i)=10/33+1/3+12/33=1$$

$$mean=E(X)=\sum_ix_ip(x_i)$$

$$=1(10/33)+2(1/3)+3(12/33)=58/33$$

The mean of the probability distribution of the random variable $x$ is $58/33.$