Urn A contains 3 red and 2 black balls; urn B contains 2 red and 4 black balls. if the number is 1 or 2 urn A is chosen, otherwise urn B is chosen.…

Use Bayes' theorem with unequal priors: $P(A)=\tfrac{2}{6}=\tfrac13$ and $P(B)=\tfrac23$. For urn $A$, $P(RR\mid A)=\tfrac{3}{5}\cdot\tfrac{2}{4}=\tfrac{3}{10}$. For urn $B$, $P(RR\mid B)=\tfrac{2}{6}\cdot\tfrac{1}{5}=\tfrac{1}{15}$. Then $P(A\mid RR)=\dfrac{\frac13\cdot\frac{3}{10}}{\frac13\cdot\frac{3}{10}+\frac23\cdot\frac1{15}}=\dfrac{1/10}{1/10+2/45}=\dfrac{9/90}{9/90+4/90}=\dfrac{9}{13}$. The option $9/11$ results from wrongly using equal priors, and the others come from mishandling the without-replacement probabilities. The key idea is that the prior probabilities are not equal because the die gives urn A only two of six faces, so the likelihoods must be weighted before normalising by the total probability of drawing two red balls.