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A fair die is tossed twice. About how many times would you expect to roll 3 or greater?
So based on sequence of Bernoulli trials:
P(exactly k successes in n trials) = C(n,k) p^k q^(n-k) where p = probability of success, and q = probability of failure
So in this case, p = 2/3, and q = 1/3
P = C(2,0) p^0 q^2 + C(2,1) p^1 q^1 + C(2,2) p^2 q^0 = 1,
While on the other hand, P is absolutely greater than 1, by common sense. So any ideas? Maybe I mess up with some definitions though.
Since the successes $X$ are binomially distributed with parameters $p=2/3$ and $n=2$ as you correctly have, then the expected number of sucesses $E[X]$ is given by $$E[X]=np=2\cdot\dfrac{2}{3}=\dfrac{4}{3}=1.333>1$$
You have calculated probabilities and you have forgotten to multiply by $k$ in order to find the expected value. Since you have all three probabilities they just add up to 1. Your formula would be correct as follows $$E[X]=0\cdot P(X=0)+1\cdot P(X=1)+2\cdot P(X=2)$$ (you missed the $0,1$ and $2$ in front of the probabilities).
