Question

A coin is tossed $m+n$ times $(m>n)$. (i) Show that the probability of at least m consecutive heads come up is $\frac{n+2}{2^{m+1}}$. (ii) Show that the probability of exactly m consecutive heads come up is $\frac{n+1}{2^{m+1}}$.

Answer 1

$1).$ Number of ways of getting $m$ consecutive heads $=2^n$

$\Rightarrow$ When tail comes first and $m$ consecutive heads no. of ways $=2^{n-1}$

$\Rightarrow$ The first position can either be H or T. No. of cares $=\left(\frac{1}{2}\right)2^{n-2}$

$\Rightarrow$ Let $m$ consecutive heads come after $n$ trials $=\frac{2^{n-1}}{2^0}$

$\therefore$ Total $=2^n+2^{n-1}+\left(\frac{1}{2}\right)2^{n-2}+\left(\frac{1}{2^2}\right)2^{n-3}+.....+2^{n-1}.\frac{1}{2^0}$

$=2^n+\frac{(n+1)}{2^{n-1}}$

$\therefore \frac{2^n+\frac{(n+1)}{2^{n-1}}}{2^{m+n}}\frac{n+2}{2^{m+1}}$

$2).$ Probability the exactly $m$ consective heads show up $=\frac{1}{2^{m+1}}$

$\Rightarrow \frac{n+2}{2^{m+1}}-\frac{1}{2^{m+1}}=\frac{n+1}{2^{m+1}}$

Hence, Solved.