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 ?

Let $P\left(H_{2}\mid A_{5}\right)$ stand for the probability that
outcome $2$ is a head under the condition that $5$ of the outcomes
are heads. 
Then $P\left(H_{2}\mid A_{5}\right)=\frac{5}{10}=\frac{1}{2}$.
We know that $P\left(H_{2}\right)=\frac{1}{2}$ and we observe that
$P\left(H_{2}\right)=P\left(H_{2}\mid A_{5}\right)$. 
This justifies
the conclusion that event $A_{5}$ does not 'affect' this probability,
which means: independence. 
Things become different when you compare $H_2$ and $A_i$ where $i\neq 5$
Especially in case $A_5$ the number of heads agrees with what was to be 'expected' ($5$ out of $10$). 
