Answer

This is a classical application of Bayes' conditional probability, where given two events A and B: $$P(A|B) = \frac{P(B|A)\cdot P(A)}{P(B)}$$In our case we want P(d4 picked | rolled 2) - where d4 will stand for 4-sided die and d6 is the six-sided die. Applying the formula, we know that $$P(\text{d4 picked | rolled 2}) = \frac{P(\text{rolled 2 | d4 picked}) \cdot P(\text{d4 picked})}{P(\text{rolled 2})}$$Now let's take these one by one. If we rolled the 4-sided die, the chance of rolling a 2 would be $\frac{1}{4}$ - as 2 is one of 4 possible outcomes. We can further assumed the dice was picked at random, so that P(d4 picked) = P(d6 picked) = 0.5. Then, we know the probability of rolling a 2 depends on whether we selected d4 or d6, so we can decompose: $$P(\text{rolled2})=P(\text{rolled 2 | d4 picked})\cdot P(\text{d4 picked}) + P(\text{rolled 2 | d6}) \cdot P(\text{d6 picked})$$We've said before that P(d4 picked) = P(d6 picked)=0.5. We know that P(rolled 2 | d4 picked) = $\frac{1}{4}$ and that if d6 was picked, 2 is one of 6 possible outcomes, so P(rolled 2 | d6 picked) = $\frac{1}{6}$.

Bringing it all together, we have $$P(\text{d4 picked | rolled 2})= \frac{\frac{1}{4} \cdot \frac{1}{2}}{\frac{1}{6}\cdot \frac{1}{2} + \frac{1}{4} \cdot \frac{1}{2}}=0.6$$