Find the probability of the path "Bag A and black" and the path "Bag B and black" by multiplying the probability of getting the bag by the probability of getting a black marble from that bag.

You are asked for the probability that Bag A was used given that the marble was black. Recall that $P(A \mid B) = \dfrac{P(A \cap B)}{P(B)}$.

Add the probabilities of getting a black marble from each bag (after weighting by how likely each bag is used) to get $P(\text{black})$, then form the fraction $P(A \cap \text{black}) / P(\text{black})$.

In Desmos, type (1/4)*(2/5) to get the probability of using Bag A and then drawing a black marble. On a new line, type (3/4)*(7/12) to get the probability of using Bag B and then drawing a black marble.

On another line, type (1/4)*(2/5) + (3/4)*(7/12) to find the total probability of drawing a black marble from either bag.

Finally, type ((1/4)*(2/5)) / ((1/4)*(2/5) + (3/4)*(7/12)). The value shown is the probability that the marble came from Bag A given that it is black; compare this decimal to the answer choices (convert each fraction to a decimal in Desmos if needed).

The die is fair with 4 sides.

• Probability of rolling 1 (use Bag A): $P(A)=\frac{1}{4}$.
• Probability of rolling 2, 3, or 4 (use Bag B): $P(B)=\frac{3}{4}$.

Bag A has 3 white and 2 black marbles, 5 total.

• $P(\text{black} \mid A)=\frac{2}{5}$.

Bag B has 5 white and 7 black marbles, 12 total.

• $P(\text{black} \mid B)=\frac{7}{12}$.

Use the law of total probability:

Substitute:

Compute each term:

So

Find a common denominator (80):

so

We want $P(A \mid \text{black})$, the probability that Bag A was used given that a black marble was drawn.

By definition of conditional probability,

We already know $P(\text{black}) = \frac{43}{80}$.

The event $A \cap \text{black}$ means: used Bag A and then drew a black marble from A. Its probability is

Now compute

So the correct answer is $\frac{8}{43}$, which corresponds to choice C.