A coin is tossed $m+n$ times $(m>n)$.Show that the probability of atleast $m$ consecutive heads is $\frac{n+2}{2^{m+1}}$.
I could not attempt this question,except few initial steps.Let $H$ and $T$ denote turning up of the head and tail.$\therefore P(H)=P(T)=\frac{1}{2}$
Please help me.
Best Answer
The probability the first $m$ tosses are heads is $\dfrac1{2^m}$.
The probability that the $j$th is tails and the next $m$ are heads is $\dfrac1{2^{m+1}}$, providing that $1 \le j \le n$.
If $n \le m$ then these are disjoint events, so the probability any of them occurs is $\dfrac1{2^m}+ n \times \dfrac1{2^{m+1}} = \dfrac{n+2}{2^{m+1}}.$