Let X be a random variable which assumes values `x_1, x_2, x_3,\ x_4` such that `2P(X=x_1)=3P(X=x_2)=P(X=x_3)=5P(X=x_4)dot` Find the probability distribution of X.

Let X be a random variable which assumes values x_(1),x_(2),x_(3),x_(4) such that 2P(X=x_(1))=3P(X=x_(2))=P(X=x_(3))=5P(X=x_(4)) Find the probability distribution of X .

For the p.m.f. P (X = x) of discrete random variable X which takes values 1, 2, 3, 4 such that 2P(X=1)=3P(X=2)=P(X=3)=4P(X=4) , then E (X)=

A random variable X takes the value 1,2,3 and 4 such that 2P(X=1)=3P(X=2)=P(X=3)=5P(X=4) . If sigma^(2) is the variance and mu is the mean of X then sigma^(2)+mu^(2)=

A random variable X takes the value 1,2,3 and 4 such that 2P(X=1)=3P(X=2)=P(X=3)=5P(X=4) . If sigma^(2) is the variance and mu is the mean of X then sigma^(2)+mu^(2)=

Let X be a random varibale such that P(X = -2) = P(X = -1) = P(X = 2) = P(X = 1) = (1)/(6) and P(X = 0) = (1)/(3) .

A random variable X takes the values 0,1,2,3 and its mean = 1,3 .If P(X=3) =2P(X=1) and P(X=2) = 0.3 then P(X=0) is