See there are 6 boxes to fill.

For such probs. always imagine things..try to visualize

If you fill first three by H (which stands for Head), only the possibility of 2 Hs exist that too after a gap of 1 box: HHHTHH== Max. 5 Hs expected

Now, similarly second way in which 3 Hs can come in a row: THHHTH==Thus only 4 Hs expected at max

similarly Third way: HTHHHT=4 Hs again at max

now Fourth way: HHTHHH= 5 Hs at max.

So do sum of individual possibilities of H in all 4 instances:

1: (1/2)^3 * (1/2)^2

2: (1/2)^3 * (1/2)

3: (1/2)^3 * (1/2)

4: (1/2)^3*(1/2)^2

Calculating===> (1/2)^4 {1/2+1+1+1/2}

===> 1/16*6/2===>3/16

Hope that Helps.

PS in probablities, there r no shortcuts please dont go by formulas rather try visualize imho