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 -1,0,1. It's mean is 0.6. If P(X=0)=0.2 then P(X=1)=

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 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

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

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. Giventhat P(X=0)=P(X=1)=p and that E(X^(2))=E(X); find the value of p.

A random variable X has its range (1 2 3} If P(X=1)=alpha P(X=2)=beta, p(X=3)=(1)/(6) and the mean is (5)/(3) then alpha , beta=

If the random variable X takes the values x_1,x_2,x_3….,x_(10) with probablilities P (X=x_i)=ki , then the value of k is equal to