Understanding the Bias in Binomial Distribution for Probability Calculations

buddingscientist
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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?
 
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This isn't a binomial distribution, so no using those formulae won't help.
 
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)).
 
thanks alot,

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
 
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"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.
 
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
 
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?
 
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