A fair coin is tossed 15 times. The probability of getting atleast 9 consecutive heads is

The 9 consecutive heads may start with $1^{\text{st }},2^{\text{nd }}$$3^{\text{rd }},\dots ,7^{\text{th }}$ toss.

H=Getting head, T=Getting tail, K= getting either  head or tail

P(H)=P(T)=1/2,P(k)=1

$$\begin{gathered} HHHHHHHHHKK\dots K\Rightarrow P(first head in first toss)=\frac{1}{2^9} \\ THHHHHHHHHkk...k\Rightarrow P(first head in second toss)=\frac{1}{2^{10}} \\ similarly \\ kTHHHHHHHHHkkkk\Rightarrow P(first head in third toss)=\frac{1}{2^{10}} \\ kkkkkTHHHHHHH\Rightarrow P(first head in seventh toss)=\frac{1}{2^{10}} \end{gathered}$$

$\therefore$The probability of getting 9 consecutive heads $=\frac{1}{2^9}+\frac{6}{2^{10}}=\frac{8}{2^{10}}=\frac{2^3}{2^{10}}=\frac{1}{2^7}$