What is the probability that there will be three or more heads in a row? You are to toss a coin five times.
I have a few questions about this.
a) Does Three + heads in a row equate to P(X=3)+P(X=4)+P(X=5) (and thus we could solve with exactly 3 *.... * exactly 5)?
b) Is the solution 8/32?
My prof listed this but does 'in a row' change the simple answer of "counting" the # of 3 heads in 32 total combinations?
 A: Three heads in a row occur for the first time in position 123 or 234 or 345. Case 123 corresponds to the set of sequences HHH?? which has probability 1/8. Case 234 corresponds to the set of sequences  THHH? which has probability 1/16. Case 345 corresponds to the set of sequences  ?THHH which has probability 1/16. These cases are disjoint hence three heads in a row occur with probability 1/8+1/16+1/16=1/4.
A: Part (a) is correct. Visualize:
$\begin{array} {c | c c c c c | c} \\
0 & T &T &T &T &T & \\
1 & H &T &T &T &T & \\
2 & T &H &T &T &T & \\
3 & H &H &T &T &T & \\
4 & T &T &H &T &T & \\
5 & H &T &H &T &T & \\
6 & T &H &H &T &T & \\
7 & H &H &H &T &T & 3 \text{ heads in a row }  \\
8 & T &T &T &H &T & \\
9 & H &T &T &H &T & \\
10 & T &H &T &H &T & \\
11 & H &H &T &H &T & \\
12 & T &T &H &H &T & \\
13 & H &T &H &H &T & \\
14 & T &H &H &H &T & 3 \text{ heads in a row }  \\
15 & H &H &H &H &T & 4 \text{ heads in a row }  \\
16 & T &T &T &T &H & \\
17 & H &T &T &T &H & \\
18 & T &H &T &T &H & \\
19 & H &H &T &T &H & \\
20 & T &T &H &T &H & \\
21 & H &T &H &T &H & \\
22 & T &H &H &T &H & \\
23 & H &H &H &T &H & 3 \text{ heads in a row }  \\
24 & T &T &T &H &H & \\
25 & H &T &T &H &H & \\
26 & T &H &T &H &H & \\
27 & H &H &T &H &H & \\
28 & T &T &H &H &H & 3 \text{ heads in a row }  \\
29 & H &T &H &H &H & 3 \text{ heads in a row }  \\
30 & T &H &H &H &H & 4 \text{ heads in a row }  \\
31 & H &H &H &H &H & 5 \text{ heads in a row }  \\
\end{array}$
So the answer is $\frac 8 {32}$, which is the sum of the cases where there are exactly 3 in a row, exactly 4 in a row, and exactly 5 in a row.
For part (b), notice how if you didn't specify "in a row", then we would also have to count rows 11, 13, 19, etc.  So $\frac {5 \choose 3} {32} $ would overcount solutions.
