Explanation

(a)
Let $'H'$denote the head and any of them come is $'A'$.Suppose the sequence of $m$ consecutive heads start with first throw, then
$P\lbrack(HH\ldots.m\ times)(AA\ldots n\ times)\rbrack$
$= \left( \frac{1}{2}.\frac{1}{2}\ldots.m\ times \right)(1..1.\ldots.n\ times)$
$= \left( \frac{1}{2} \right)^{m}$
Now, suppose the sequence of $m$ consecutive heads start with second throw, the first must be a tail,
$\therefore P\lbrack T(H.H\ldots m\ times)(A.A\ldots n - 1\ times)\rbrack$
$= \frac{1}{2}.\frac{1}{2^{m}} \times (1)^{n - 1} = \frac{1}{2^{m + 1}}$
Similarly, as above
$\therefore$Required probability
$= \frac{1}{2^{m}} + \left( \frac{1}{2^{m + 1}} + \frac{1}{2^{m + 1}} + \ldots n\ times \right)$
$= \frac{1}{2^{m}} + \frac{n}{2^{m + 1}} = \frac{n + 2}{2^{m + 1}}$