Calculate the probability of $P(A|B')$ given that $A$ and $B$ are independent and $P(A)=\frac{1}{3}$
My intuition tells me since $P(A)= \frac{1}{3}$ and $A$ does not depend on $B$ the answer should just be $\frac{1}{3}$ and even more since I looked at this answer.
Is this really just a trick question or is there more to it?
You need to use Bayes theorem for these kind of exercises :
$$\begin{align}P(A \mid B) &= \frac{P(A \cap B)}{P(B)} \\&= P(A)& \text{since A et B are independent}\end{align}$$
And you are right here intuition is quite clear, yet be carefull with intuition when working with conditional porbabilities since some situations can be somehow tricky. But yes that's it.