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Flip a fair coin 491 times. What is the probability that tail occurs even number of times and why?

A six-sided die is thrown 10 times. What is the probability for that you get an odd number of sixes?!

My attempt for the first question is:

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The second one:

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Are these answers correct? If the first question is correct who can I prove it, because it is hard to calculate I want to prove instead in sample way.

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    $\begingroup$ @Adam what did you do on this question? what is it that you find hard when trying to solve it? $\endgroup$ – Math-fun Feb 21 '15 at 14:17
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    $\begingroup$ Please post context and an attempted solution, rather than just expecting an answer to what's clearly homework. $\endgroup$ – David Etler Feb 21 '15 at 14:17
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    $\begingroup$ A friendly message from the friendly community moderator: Don't delete significant parts of the question, please. Keeping even incorrect attempts visible helps the answerers. $\endgroup$ – Jyrki Lahtonen Feb 23 '15 at 10:51
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The questions can be done without any advanced formulae.

Here's a hint for the first part:

Would the probability be any different if we wanted an even number of heads?

Hint for the second part:

$\mathrm{Answer }=P(\mathrm{One\ six})+P(\mathrm{Three\ sixes})+P(\mathrm{Five\ sixes})+P(\mathrm{Seven\ sixes})+P(\mathrm{Nine\ sixes}).$

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The probability that a binomial random variable $X$ with success probability $p$ and has an even number of successes after $n$ trials, is $\frac{1}{2}(1+(1-2p)^n)$. For a proof, see my answer here: probability that a random variable is even

So for the first question we get: $$P(X_1 \text{ is even})=\frac{1}{2}(1+(1-2\frac{1}{2})^{491})=\frac{1}{2}$$ For the second question: $$P(X_2 \text{ is odd})=1-\frac{1}{2}(1+(1-2\frac{1}{6})^{10})=\frac{1}{2}-\left(\frac{2}{3}\right)^{10}=\frac{57001}{118098}\approx 0.483$$

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