We have two identical baskets. In both of them there are 10 balls. In basket 1 there are 7 blue balls and 3 red balls. In basket 2 there are 3 blue balls and 7 red balls.
One chooses one of these baskets, picks 5 balls from there and finds that 2 of them were red. What is the probability that the chosen basket is the basket 2?
Let's denote A = "The basket is basket 1.", B = "The basket is basket 2." and C = "2 of the picked 5 balls are red.".
Then (I guess) $$\mathbb{P}\left(A\right)=\mathbb{P}\left(B\right)=1/2 ,$$ $$\mathbb{P}\left(C\right)=\frac{\binom{7}{3}\binom{3}{2}}{\binom{10}{5}}+\frac{\binom{3}{3}\binom{7}{2}}{\binom{10}{5}}=\frac{1}{2} ,$$ and therefore the asked probability would be $$ \mathbb{P}\left(B\mid C\right)=\frac{\mathbb{P}\left(B\cap C\right)}{\mathbb{P}\left(C\right)}=\frac{\frac{\binom{3}{3}\binom{7}{2}}{\binom{10}{5}}}{1/2}=\frac{1/12}{1/2}=\frac{1}{6}.$$
Especially I am not sure about the probability of C. Is this logic correct? I guess that Bayesian theorem could be applied at some point.
The marginal probability $\Pr[C]$ you computed is not correct, because the law of total probability is $$\Pr[C] = \Pr[C \mid A]\Pr[A] + \Pr[C \mid B]\Pr[B] = \frac{\binom{7}{3}\binom{3}{2}}{\binom{10}{5}} \cdot \frac{1}{2} + \frac{\binom{3}{3}\binom{7}{2}}{\binom{10}{5}} \cdot \frac{1}{2} = \frac{1}{4}.$$ Just because $\Pr[A] = \Pr[B] = 1/2$ prior to observing the result, does not mean it can be ignored.
Then the posterior calculation of $\Pr[B \mid C]$ is $$\Pr[B \mid C] = \frac{\Pr[C \mid B]\Pr[B]}{\Pr[C]} = \frac{\frac{\binom{3}{3}\binom{7}{2}}{\binom{10}{5}} \cdot \frac{1}{2}}{\frac{1}{4}} = \frac{1}{6}.$$
In this case, you obtained the correct answer, but had the prior probability of choosing Basket 1 versus Basket 2 not been equal, you'd get a wrong answer.