Probability of getting H from tossing a coin is $p$. We toss a coin $n$ times. What is probability that number of H will be odd.

Asked Oct 28 '22 at 17:19

Active Oct 28 '22 at 17:42

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My work.

If we toss one time then it's just $p$. P(#2)=$p(1-p)$, (#$2$) - $2$ tosses

And so if n is even then

$P$(#n)$=p(1-p)^{n-1} + p^3(1-p)^{n-3} + \cdots p^{n-1}(1-p)$

If n is odd.

$P$(#n)$=p(1-p)^{n-1} + p^3(1-p)^{n-3} + \cdots p^{n}$

How combine this two and get some compact form?

I can see geometric progression pattern but can't make use of that.

edited Oct 28 '22 at 17:42

asked Oct 28 '22 at 17:19

unit 1991

Can you recognise some kind of geometric series? – jlammy Oct 28 '22 at 17:22
Well hold on a moment. Why is there no binomial coefficient here? Do I misunderstand the problem? – Brian Tung Oct 28 '22 at 17:44
You forgot the binomial coefficients. The number of heads follows a binomial distribution: $$P(\text{# heads }=k)=\binom nkp^k(1-p)^{n-k}$$ – Mike Earnest Oct 28 '22 at 17:45
@MikeEarnest Ok but still how compute sum? – unit 1991 Oct 28 '22 at 17:50
Duplicate question does not answer my question? It starts with guess which seems pretty unintutive. – unit 1991 Oct 28 '22 at 17:57
I just added an answer to that question which is more intuitive. – Mike Earnest Oct 28 '22 at 18:07

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