Bag A contains 3 red and 2 black balls and bag B contains 2 red and 4 black balls. A fair die is thrown; if the number is 1 or 2 a bag A is chosen…

Use Bayes' theorem. $P(A)=\frac13$, $P(B)=\frac23$.
Same colour from $A$: $\dfrac{\binom32+\binom22}{\binom52}=\dfrac{4}{10}=\dfrac25$.
Same colour from $B$: $\dfrac{\binom22+\binom42}{\binom62}=\dfrac{7}{15}$.
$P(\text{same})=\frac13\cdot\frac25+\frac23\cdot\frac7{15}=\frac{6}{45}+\frac{14}{45}=\frac{20}{45}=\frac49$.
$P(B\mid\text{same})=\dfrac{14/45}{20/45}=\dfrac{7}{10}$.
The value $\frac3{10}$ is $P(A\mid\text{same})$, $\frac{14}{45}$ is the joint probability $P(B\cap\text{same})$ with no normalisation, and $\frac7{15}$ is only the likelihood $P(\text{same}\mid B)$. Each of these is a typical slip when the conditioning event is ignored or only partly used. The problem is solved cleanly by writing the total probability of the conditioning event first and then dividing the relevant joint probability by it.