probability of sequence of exactly 3 heads A fair coin is tossed 5 times. What is the probability of getting a sequence of exactly 3 heads? 
Attempt:
Sample space = $2^5 = 32$
\number of possible positions of first head = $5 - 3 + 1 = 3$
\number of ways the other 2 outcomes could occur = $2^2 = 4$
Therefore prob of exactly 3 heads in sequence = $\frac {3 × 4}{32} $
Should i count the ways that the other two outcomes could occur?

Let us use your idea. If the sequence of consecutive heads begins at Position 1, then the next term must be T, and the last term can be chosen freely, $2$ choices.
If the sequence begins at  Position 2, everything is forced, we only have THHHT, $1$ choice. 
And if the consecutive heads start at Position 3, our sequence must be of shape XTHHH, $2$ possibilities.
