Suppose we have $n$ coins and we flip them at the same time. if the probability of getting head of the $i$ th coin is $\frac{1}{i}$. prove that the probability of getting even number of heads is $\frac{1}{2}$.
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3Please edit to include your efforts. Working the problem for small $n$ would be a natural way to start. – lulu Mar 27 '21 at 16:56
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Welcome to MSE. You'll get a lot more help if you show that you have made a real effort to solve the problem yourself, even if you haven't made much progress. What are your thoughts? What have you tried? How far could you get? Where are you stuck? This question will likely be closed if you don't add more context. Please respond by editing the question body. Clarifications don't belong in the comments. – saulspatz Mar 27 '21 at 17:10
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Does this answer your question? If nine coins are tossed, what is the probability that the number of heads is even? – Ethan Bolker Mar 27 '21 at 19:20
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1If $n=1$, isn't the probability of an even number of heads equal to $0$? – paw88789 Mar 27 '21 at 19:33
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Let $e_n,o_n$ denote the events that an even/odd number of heads were thrown. Then $P(e_n)+P(o_n)=1$, and \begin{align*} P(e_{n+1})&=P(e_n)(1-1/(n+1)) + P(o_{n})(1/(n+1))\\ P(o_{n+1})&=P(e_n)(1/(n+1)) + P(o_{n})(1-1/(n+1)). \end{align*}
Subtracting these gives:
$$P(e_{n+1})-P(o_{n+1})=-\frac{2}{n+1}(P(e_n)-P(o_n))=\cdots=\frac{(-2)^n}{(n+1)!}(P(e_1)-P(o_1))=\frac{(-2)^n}{(n+1)!}(1-1)$$
So $P(e_n)=P(o_n)=1/2$ (assuming 0 is even).
Alex R.
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