Tossing a coin

Let $a_1, a_2 \dots a_{17}$ represent the outcome of the 17 coin tosses. To enumerate all the possibilities of at least 11 consecutive heads we consider the following two cases :-

Case 1: The first 11 coin tosses are all heads, i.e. $a_1 = a_2 = a_3 \dots a_{11} = Heads$ . The number of such cases are $2^6$ , corresponding to the number of ways to assign the variables $a_{12} \dots a_{17}$ .

Case 2: The first 11 coin tosses are not consecutive 11 heads. To count the number of such cases let's assume that $a_i, a_{i+1} \dots a_{i+10}$ be the first 11 consecutive heads for some $2 \leq i \leq 7$ . Since $a_i, a_{i+1} \dots a_{i+10}$ are first consecutive heads $a_{i-1}$ must be Tails. It means that ( $a_{i-1},a_i, a_{i+1} \dots a_{i+10}$ ) can be considered as a bundle which corresponds to (Tails, Heads, Heads . . . Heads). Apart from the bundle there are 5 other elements (say $b_1, b_2, b_3, b_4, b_5$ ) which can be assigned in $2^5$ ways. Now the bundle can be placed at 6 places (amongst the remaining elements, $b_i$ 's) which means that the total possibilities are $6\times 2^5$ .

The total number of possibilities thus are $2^6 + 6\times 2^5 = 2^8$ . Which means that the probability is $2^8/2^{17} = 1/512$ . Thus $a=1,b=512$ and $a+b = 513$ .