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

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Given that P(1)= 10/33 P (2) =11/33 and P(3)= 12/33, the mean is given as,

$E(x)=\sum xp(x)=(1\times {10\over33})+(2\times{11\over33})+(3\times {12\over33})={10\over33}+{22\over33}+{36\over33}={68\over33}=2.060606$

The mean of the probability distribution is 2.060606