The random variable X can take only the values `0; 1;2`. Giventhat `P(X=0) = P(X=1) = p` and that `E(X^2) = E(X)`; find the value of `p`.

The random variable X can take only the values 0,1,2. If P(X=0)=P(X=1)=p and E(X^(2))=E[X] , then find valu of p.

A random variable X takes the value -1,0,1. It's mean is 0.6. If P(X=0)=0.2 then P(X=1)=

The random variable X can the values 0,1,2,3, Give P(X=0)=P(X=1)=p and P(X=2)=P(X=3) such that sum p_(i)x_(i)^(2)=2sum p_(i)x_(i) then find the value of p

Gien that X is discrete random variable which take the values, 0,1,2 and P(X=0)=(144)/(169),P(X=1)=(1)/(169) , then the value of P(X=2) is

A random variate X takes the values 0,1,2,3 and its mean is 1.3. If P(X=3)=2P(X=1) and P(X=2)=0.3, then P(X=0) is equal to

Given X~B(n,p) if E(X) = 6, Var (X) = 4.2 , find the value of n and p.

A random variable X takes the values 0,1 and 2 .If P(X=1)=P(X=2) and P(X=0)=0.4 ,then the mean value of the random variable X is

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 has its range {-1,0,1} . If its mean is 0.2 and P(X=0)=0.2 ,then P(X=1)=