A coin that when flipped comes up heads with probability $p$ is flipped until either heads or tails has occurred twice. Find the expected number of flips

Given in the question that the coin was flipped comes up heads with probability $p$is flipped until either heads or tails has occurred twice.

Consider $X$is a number of tosses of the coin.

For $X=0$and $p_0=0$as two heads or two tails are not possible.

For $X=1$and $p_1=0$as two heads or two tails are not possible.

For $X=2$and $p_2=0$as two heads or two tails are possible.

Therefore,

$p_2=HH$or $TT$

$=pp+(1-p)(1-p)$

We get,

$=p^2+(1-p{)}^2$

Since if head occurs then the probability is $p$otherwise if tail occurs then the probability is $(1-p)$

For the $X=3,$probability of two heads or two tails is surely one.

But this has not occurred with $X=2$

It means that if $X=1$does not occur with$X=3,$

Outcome surely occurs

Thus, we find out changes outcome does not happen with$X=2$

That is,

$p_3=1-p_2$

$=1-p^2-(1-p{)}^2$

The Distribution Table is shown as below:

$\begin{array}{|cc|}\hline X_i& p_i\\ 0& 0\\ 1& 0\\ 2& p^2+(1-p{)}^2\\ 3& 1-p^2-(1-p{)}^2\\ \hline\end{array}$

Find the value

$E\left(X\right)=\sum xp\left(x\right)$

$=(0\times 0)+(1\times 0)+2\times \left(p^2+(1-p{)}^2\right)+3\times \left(1-p^2-(1-p{)}^2\right)$

$=0+0+2p^2+2(1-p{)}^2+3-3p^2-3(1-p{)}^2$

Simplifying the equation

$=-p^2-(1-p{)}^2+3$

$=-p^2-1-p^2+2p+3$

We get,

$=-2\left(p^2+p-2\right)$.

The expected number of flips $E\left(X\right)$is$-2\left(p^2+p-2\right).$