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)=

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)=

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

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

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)=

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.

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.

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 variable X takes the values 0,1,2,3,..., with prbability PX(=x)=k(x+1)((1)/(5))^x , where k is a constant, then P(X=0) is.

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

If X is discrete random variable takes the values x_1, x_2, x_3 , ... x_n then sum_(i=1) P(x_i) = ........