Probability question related to coin tosses In an exam I gave recently, the following question was asked:
A fair coin is tossed $10$ times and the outcomes are listed. let $H_i$ be the event that the $i^{th}$ outcome is a head and $A_m$ be the event that the list contains exactly m heads, then are $H_2$ and $A_5$ independent events ?
The Official solution to this question was as follows:
$$p(H_i) = \frac{1}{2},\qquad p(A_m)=\frac{^{10}C_m}{2^{10}}\\p(H_i\cap{A_m})=\frac{^9C_{m-1}}{2^{10}}.\\\text{For}\;H_i\;\text{and}\;A_m\;\text{to be independent},\;\frac{^9C_{m-1}}{2^{10}}=\frac{1}{2}.\frac{^{10}C_m}{2^{10}}\\ \Rightarrow1=\frac{1}{2}.\frac{10}{m}\Rightarrow m=5\\ \Rightarrow H_2\;\text{and}\;A_5\;\text{are independent events} $$
Now while I understand the mathematics behind this answer, I find it logically confusing that $p(A_5)$ does not get affected whether or not the $2^{nd}$ outcome is heads. Any ideas ?

The big thing here is this:
If the 2nd toss is heads, you need to get exactly 4 other heads in the other nine tosses. There are $\binom{9}{4}$ equally likely ways to do this.
If the 2nd toss is tails, you need to get exactly 4 other tails in the other nine tosses.  Again, there are $\binom{9}{4}$ equally likely ways to do this.
So, regardless of the outcome of the second toss, conditioned on it you have the same probability of getting exactly 5 heads.
(Of course, you could also have said that you need to get exactly 5 heads in the latter case; but, given that $\binom{9}{5}=\binom{9}{4}$, this isn't an issue.)
