Can you find E(X) and Var(X) for a biased die without the binomial formula?

[buddingscientist]

binomial distribution

Prob of rolling a 1 = 1/10, rolling a 2 = 2/10, 3 = 3/10, 4 = 4/10
Let X be the value thrown
Calculate E(X) and Var(X)


To do this can't use E(X) = np and can't use Var(X) = npq
is this correct?

 

[matt grime]

This isn't a binomial distribution, so no using those formulae won't help.

 

[HallsofIvy]

You can, however, use the basic definitions:

E(x)= &Sigma(xProb(x))= 1*prob(1)+ 2*prob(2)+ 3*prob(3)+ 4*prob(4).

&sigma(x)= &sqrt((x- E(x))²Prob(x)).

 

[buddingscientist]

thanks a lot,

so to use those formula, we could find that the E(X) amount of 4's, out of 10 rolls, would be

4/10 * 10 = 4

and the variance 4/10 * 6/10 * 10 = 2.4

 

[matt grime]

"E(X) amount of 4's"

E(X) is the expectation of the score. I don't see what the 'amount of 4s' has to do with it.

As was written above the expectation is:

1/10 + 2*2/10 + 3*3/10 +4*4/10 = 3.

 

[buddingscientist]

yes, i know
i wanted to know a use of the E(X) = np formual with respect to that question, the use of it was to find the probability of the amount of 4's out of 10 rolls

 

[matt grime]

In that case why did you use X for two different things? The outcome of one throw and the number of 4s occurring in 10 rolls?