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Toss a die 10 times. If you get six 1's, find the expected number of 2's.

The answer given is $\frac{4}{5}$. I'm trying to understand where the following method of solving the problem goes wrong.

$X = 1, 2, 3, 4$

$E(1) = 1 * [\frac{1}{5}* (\frac{4}{5})^3]$

$E(2) = 2 * [(\frac{1}{5})^2 * (\frac{4}{5})^2]$

$E(3) = 3 * [(\frac{1}{5})^3*\frac{4}{5}]$

$E(4) = 4 * (\frac{1}{5})^4$

$E(1) + E(2) + E(3) + E(4) = \frac{112}{625} = 0.1792$

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1 Answer 1

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You have to account for the positions of 2's, i.e. $$E_i=i\times{4 \choose i}(1/5)^i(4/5)^{4-i}$$ Then, you'll obtain $4/5$.

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