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Toss 6 independent fair coins, and let X be the number that comes up heads. Compute $E(X)$ and $Var(X)$.

Is this the negative geometric distribution? (6-x-1 Choose x-1) (1/2)^x (1/2)^(6-x) ?

Then wouldn't it be $E(X)=x(1-1/2)/1/2 $ ? But it is not an actual number.

Am I misinterpreting the question?

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    $\begingroup$ It's binomial, not negative geometric. $\endgroup$
    – Ian
    Oct 21, 2015 at 2:51
  • $\begingroup$ Then since it's binomial: $E(X)=6*1/2$ and $Var(X)=6*1/2*1/2$ ? $\endgroup$
    – lonelyboy67
    Oct 21, 2015 at 2:53
  • $\begingroup$ That's right. ${}{}$ $\endgroup$
    – Ian
    Oct 21, 2015 at 2:53
  • $\begingroup$ Determining the distribution of X is not neede to compute the expectation and the variance of X. $\endgroup$
    – Did
    Oct 21, 2015 at 7:07

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