91Ó°ÊÓ

Coin's probability $=p$

Number of flipping times$=n,$

Define random variable $I_j$that marks whether the run of Heads of size $k$started from the trial number $j$or not, $j=1,2,…n-k+1$.

Now define an indicator variable as follows:

$I_j=\left\{\begin{array}{l}1â€\dots â\P Ä\dots â\P Ä\dots â\P Ä\dots \text{(if a}\mathrm{k}\text{sized run begins at}{\mathrm{j}}^{\text{th}}\text{position}\\ 0â€\dots â\P Ä\dots â\P Ä\dots â\P Ä\dots \text{(otherwise)}\end{array}\right\}$

So, if we define $X$as the random variable that marks the total number of $k-$runs of Heads, So we have that:

$X=\underset{j=1}{\overset{n-k+1}{∑}}{I}_j$

Using the linearity of the expectation, so:

$E\left(X\right)=\underset{j=1}{\overset{n-k+1}{∑}}E\left(I_j\right)$

$=E\left(\underset{j=1}{\overset{n-k+1}{∑}}{I}_j\right)$

$=P\left(I_1=1\right)+\underset{j=2}{\overset{n-k}{∑}}P\left(I_j=1\right)+P\left(I_{n-k+1}\right)$

Now solving above expression:

$=P\left(I_1=1\right)+\underset{j=2}{\overset{n-k}{∑}}P\left(I_j=1\right)+P\left(I_{n-k+1}\right)$

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

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

$=(n-k)p^k(1-p{)}^2+p^k\left(1-p^2\right)$

Hence, the expected number of runs of heads of size as per question is:

$(n-k)p^k(1-p{)}^2+p^k\left(1-p^2\right)$