1. The problem involves calculating the probability of drawing marbles from two bags, X and Y, with given probabilities for each draw. 2. The first step is to understand the tree diagram structure: the first level splits into Bag X and Bag Y, each with probability $\frac{1}{2}$. 3. The second level shows the probability of drawing a white (W) or red (R) marble from each bag. For Bag X, $P(W|X) = \frac{3}{4}$ and $P(R|X) = \frac{1}{4}$. For Bag Y, $P(W|Y) = \frac{1}{5}$ and $P(R|Y) = \frac{4}{5}$. 4. The third level shows the probability of drawing the second marble given the first draw. For example, if the first marble from Bag X was white, then $P(W_2|X, W_1) = \frac{2}{3}$ and $P(R_2|X, W_1) = \frac{1}{3}$. This is because after removing one white marble, the bag now has fewer white marbles left. 5. Specifically, for Bag X, initially there are 3 white and 1 red marbles (total 4). Drawing one white marble reduces white marbles to 2 and total marbles to 3, so the probability of drawing another white marble is $\frac{2}{3}$. 6. The fraction $\frac{3}{4}$ is the probability of drawing the first white marble from Bag X, and $\frac{2}{3}$ is the probability of drawing the second white marble given the first was white. These probabilities multiply along the branches of the tree to find combined probabilities. 7. Therefore, the $\frac{3}{4}$ comes from the initial composition of Bag X, and the $\frac{2}{3}$ comes from the updated composition after one white marble is removed. Final answer: The $\frac{3}{4}$ is the probability of drawing the first white marble from Bag X, and $\frac{2}{3}$ is the probability of drawing the second white marble from Bag X after one white marble has already been drawn.