Problem 5

For 5 consecutive heads, if you start with a head then there is a probability of success of $\frac{1}{2}$ each time, but you also need a tail next to avoid falling into the 6 consecutive heads group. You also need to consider that if you don't start or end with a heads then you must get a tail just before and just after the 5 consecutive heads. Using this principle with casework (grouping): $\large \ P(5 \ consecutive \ heads ) = 2(\frac{1}{2^6}) +3(\frac{1}{2^7}) = \frac{1}{2^5} +3(\frac{1}{2^7})$ $\large \ P(6 \ consecutive \ heads ) = 2(\frac{1}{2^7} +\frac{1}{2^8}) = \frac{1}{2^6} + \frac{1}{2^7}$ $\large \ P(7 \ consecutive \ heads ) = 2(\frac{1}{2^8}) + \frac{1}{2^9} = \frac{1}{2^7} + \frac{1}{2^9}$ $\large \ P(8 \ consecutive \ heads ) = 2(\frac{1}{2^9})= \frac{1}{2^8}$ $\large \ P(9 \ consecutive \ heads) = \frac{1}{2^9}$ Adding these together gives: $\therefore\huge\color{#3D99F6}{\ P( \geq\ 5 \ consecutive \ heads ) = \boxed{\frac{3}{32}}}$

$4\times \frac{1}{2^5}\times \frac{1}{2}+\frac{1}{2^5}$

= $\frac{3}{32}$