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QUESTION

A pair of unbiased coins is tossed repeatedly until 2 heads are
   obtained. What is the probability that it will happen on the
   $r$-th toss? What is the probability that it will take more then
   two tosses? Find the mean and variance of the number of tosses.

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ANSWER

 $p$(2 heads with pair of unbiased coins)=$\frac{1}{4}$.

 Hence
   the number of trials, $r$, has a geometric distribution

   $p(r)=(\frac{3}{4})^{r-1}\frac{1}{4}$

   $p$(more than 2 tosses) $=p$(first and second tosses
   fail) $=(\frac{3}{4})^2=\frac{9}{16}$

   \begin{eqnarray*}
    \sum_{r=1}^\infty p^r&=&\frac{p}{1-p} \textrm{
    differentiating}\\
    \sum_{r=1}^\infty rp^{r-1}&=&\frac{1}{(1-p)^2}\\
    \sum_{r=2}^\infty r(r-1) p^{r-2}&=&\frac{2}{(1-p)^3}
   \end{eqnarray*}

   \begin{eqnarray*}
   \mu = \sum_{r=1}^\infty \frac{r}{4}\left(\frac{3}{4}\right)^{r-1}\\
   =\frac{\frac{1}{4}}{\left(1-\frac{3}{4}\right)^2}=4
   \end{eqnarray*}

   \begin{eqnarray*}
    \sigma^2 &=&\sum_{r=1}^\infty
    \frac{r(r-1)}{4}\left(\frac{3}{4}\right)^{r-1}+\sum_{r=1}^\infty
    \frac{r}{4}\left(\frac{3}{4}\right)^{r-1}-\mu^2\\
    &=&\frac{1}{4}\times \frac{2 \times \frac{3}{4}}{\left(1-
    \frac{3}{4}\right)^3}+4-16\\
    &=&24+4-16=12
   \end{eqnarray*}


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