Two dice with a difference ... They appear identical, but one die is fair and the other is loaded so that the probability of throwing a 6 is 0.25. Bev picks a die at random and throws it, and it shows a 6. Find the probability that she picked the biased die.
(Hint: Let A be the event 'the die chosen is the biased one' and B be 'a die chosen at random shows a 6 when it is thrown'. Find $P(A|B).)
I know $P(A) = 1/2$, $P(B) = (1/2 \times 0.25) + (1/2 \times 1/6) = 5/24$
Don't know what $P(A|B)$ is and what to do do next.
The basic formula that defines the conditional probability $\Pr(A|B)$ is $$\Pr(A|B)=\frac{\Pr(A\cap B)}{\Pr(B)}.\tag{1}$$ You already computed $\Pr(B)$.
As to $\Pr(A\cap B)$, the probability that you picked the biased die and threw a $6$, you also already computed it as part of the calculation of $\Pr(B)$: It is $(1/2)(0.25)$.
So all that is left to do is the division asked for by Formula (1).