Problem summary:
99 fair coins
1 unfair coin with a probability of getting heads $P(H)=0.9$
A coin is selected and flipped 10 times.
If the coin lands on head 10 times out of 10 flips, what is the probability that the unfair coin was selected?
My approach:
Event $A$: unfair coin was picked
Event $B$: Getting 10 heads from 10 flips
Event $A^{c}$: did not pick unfair coin
Problem: find $P(A|B)$
$$P(A|B)=\displaystyle \frac{P(A \cap B)}{P(B)}$$
$=\displaystyle \frac{P(A \cap B) }{P(A \cap B) \cup P(A^{c} \cap B)} \label{a}\tag{1}$ $$P(A \cap B)={10 \choose 10} 0.9^{10}0.1^{0}=0.3487 \label{b}\tag{2}$$ $$P(A^{c} \cap B)={10 \choose 10}0.5^{10}0.5^{0}=0.00097656$$
$$P(B)=P(A \cap B)+P(A^{c} \cap B)$$ $$=0.3487+0.00097656=0.34967656 \label{c}\tag{3}$$
Subbing (2) and (3) into (1) gives
$$P(A|B)=\frac{0.3487}{0.34967656}=0.997207248$$
Does this seem correct?
I agree with your setup, but I think your computations for $P(A\cap B)$ and $P(A^C\cap B)$ are off. We have $P(A\cap B) = P(B\mid A)P(A) = \frac{0.9^{10}}{100} \approx 0.0034868$, and $P(A^C\cap B) = P(B\mid A^C)P(A^C) = 2^{-10}\left(\frac{99}{100}\right) \approx 0.0009668$. Thus $P(B) \approx 0.0044536$, and so $P(A\mid B) = P(A\cap B)/P(B) \approx 0.78292$.
Intuitively, this makes sense: the probability of a fair coin getting 10 heads in a row is about 1/1000, so if only 1/100 coins are loaded, there should be a sizeable (~10%) chance that the coin is fair, regardless of how often loaded coins would landheads. The other ~12% come from the fact that a loaded coin getting 10 heads in a row is also not a guarantee.