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I am currently stuck with this question, as I prepare for my exam. I'm not sure what kind of answer it is looking for:

A biased coin is tossed until a head appears for the first time. Let $p$ be the probability of head, $0 < p < 1$.

a) What is the probability that the number of tosses required is odd?

b) How about even?

Here's what I have so far: \begin{align*} P(1) &= p\\ P(2) &= (1-p)p\\ P(3) &= (1-p)(1-p)p\\ P(4) &= (1-p)(1-p)(1-p)p\\ \end{align*} and so on. I understand that it is multiplying by $(1-p)$ for each extra toss. How would I put this in terms of a probability?

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    $\begingroup$ To find the probability that it is odd, you should add $P(1) + P(3) + P(5) + ...$. You then have an infinite geometric series with first term $p$ and ratio $(1-p)^2$. $\endgroup$ – user2825632 Sep 22 '16 at 0:03
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What are the probabilities that we have to keep in mind when solving this question? You have got the gist, namely that $(1-p)$ represents failure and $p$ a success.

Now, suppose there is success the first time itself. The probability of this happening is $p$.

Now, suppose there is success the third time . The probability of this happening is $p(1-p)^2$, because you have to fail twice before succeeding.

Now, suppose there is success the fifth time . The probability of this happening is $p(1-p)^4$, because you have to fail four times before succeeding.

Can you see what the pattern is? The probability of then getting a head on an odd turn is $p(1 + (1-p)^2 + (1-p)^4 + \ldots)$. $$ p(1 + (1-p)^2 + (1-p)^4 + \ldots) = p\displaystyle\sum_{i=0}^\infty ((1-p)^2)^i = \frac{p}{1-(1-p)^2} = \frac{1}{2-p}. $$ Similarly, the probability that you will get heads on an even turn is $1-\frac{1}{2-p} = \frac{1-p}{2-p}$

Note: $0<p<1$, otherwise the geometric series convergence is questionable.

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You have $P(N=n)=(1-p)^{n-1}p$. $\quad\color{seagreen}{\checkmark}$

And have noticed that $P(N=2k)=(1-p) P(N=2k-1)$ so $$\begin{align}\sum_{k=1}^\infty P(N=2k)~=~&(1-p)~\sum_{k=1}^\infty P(N=2k-1) \\[3ex] P(N\in2\Bbb N^+)~=~&(1-p)~P(N\in 2\Bbb N^+-1)\end{align}$$ So, since $P(N\in2\Bbb N^+)+P(N\in 2\Bbb N^+-1)=1$ therefore...

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