Using Karthikeyan's hint.

I misunderstood the hint as no "n" tails in succession. The hint actually asks us to think about recursion for no two tails in succession in "n" tosses.

Let the number of sequences with no two tails in succession in " n " tosses be f(n) .

Let the number of sequences with no two tails in succession in " n " tosses, ending in a T be g(n) .

Let the number of sequences with no two tails in succession in " n " tosses, ending in an H be h(n) .

Then f(n) = g(n) + h(n) .

Is there a way to find a relationship between f(n+1), g(n+1), h(n+1) and f(n), g(n), h(n) ? Yes!

If the sequence of length "n" ends in a T, then we extend it by appending it with an H.

If the sequence of length "n" ends in an H, then we extended by appending it with both an H and a T.

That means g(n+1) = h(n) + g(n) = f(n) and h(n+1) = g(n) = f(n-1)

That means f(n+1) = g(n+1) + h(n+1) = f(n) + f(n-1) .

We have our recursion and we want to find f(10) .

f(1) = 2 (either "H" or "T")

f(2) = 3 (either "HH", "TH, or "HT")

f(3) = f(2) + f(1) = 5

f(4) = f(3) + f(2) = 8

f(5) = f(4) + f(3) = 13

f(6) = f(5) + f(4) = 21

f(7) = f(6) + f(5) = 34

f(8) = f(7) + f(6) = 55

f(9) = f(8) + f(7) = 89

f(10) = f(9) + f(8) = 144

Therefore, the probability = \frac{144}{1024} . Same as the other answer.