An urn contains $4$ blue and $4$ red marbles. At first a marble is drawn (without looking) and removed from the urn. Then, a marble is drawn from the urn, its color recorded and put back in the urn. This process is repeated $1000$ times. Let $A$ be the event that the removed marble is blue. Let $B$ the event exactly $503$ draws out of $1000$ draws are red. What is $P(A\vert B)$?
My attempt:
$$P(A\vert B)=\frac{P(A \cap B )}{P(B)}$$
If red removed: $\implies 3 \, \text{red},4 \, \text{blue} \implies P(\text{red})=\frac{3}{7}$
If blue removed: $\implies 4 \, \text{red},3 \, \text{blue} \implies P(\text{red})=\frac{4}{7}$
Probability of choosing exactly $503$ out of $1000$ is: $${1000 \choose 503}p^{503}(1-p)^{497}$$
Therefore: $$P(B)=\frac{1}{2}{1000 \choose 503}\left( \frac{3}{7}\right)^{503}\left( \frac{4}{7}\right)^{497}+\frac{1}{2}{1000 \choose 503}\left( \frac{4}{7}\right)^{503}\left( \frac{3}{7}\right)^{497}=1.15 \cdot10^{-6}$$
$$P(A\cap B)= \frac{1}{2}{1000 \choose 503}\left( \frac{4}{7}\right)^{503}\left( \frac{3}{7}\right)^{497}=9.78 \cdot 10^{-7}$$
$$P(A \vert B)=\frac{P(A \cap B )}{P(B)}=\frac{9.78 \cdot10^{-7}}{1.15 \cdot 10^{-6}}=0.8504 \approx 85 \%$$
Is my reasoning correct or did I make some mistake somewhere?
Your analysis is spot on, except for the minor mistake of incorrectly adding a leading factor of $\displaystyle \left(\frac{1}{2}\right)$ to the computation of $p(A\cap B).$
An intuitive visualization that you made a mistake is by noticing that
$$p(B) = p(A\cap B) + p([\neg A] \cap B).$$
Therefore,
$$p(A|B) = \frac{p(A\cap B)}{p(A\cap B) + p([\neg A] \cap B)}.$$
Edit
Also, just noticed, you have a typo in your very last equation, where you switched the numerator and denominator. In effect, the typo proved harmless, because your common sense kicked in, to force you to construe the larger number as the denominator.