Let's break down the problem using the information provided and the tree diagram. We need to find the probability that Brianna is in dorm Y, given that she has a room. This is a conditional probability problem.

We are given:

• P(X) = 0.4 (Probability that Brianna is in dorm X)
• P(Y) = 0.6 (Probability that Brianna is in dorm Y)
• P(Room | X) = 0.7 (Probability of a room in X)
• P(Room | Y) = 0.2 (Probability of a room in Y)

We are asked to find:
P(Y | Room) (Probability that Brianna is in dorm Y, given that she has a room).

We can use Bayes' Theorem for this:

$P(Y | Room) = \frac{P(Room | Y) \cdot P(Y)}{P(Room)}$

Where P(Room) is the total probability of getting a room, which is calculated as:

$P(Room) = P(Room | X) \cdot P(X) + P(Room | Y) \cdot P(Y)$

Now, let's calculate each term:

1. P(Room | X) \cdot P(X) = 0.7 * 0.4 = 0.28
2. P(Room | Y) \cdot P(Y) = 0.2 * 0.6 = 0.12
3. P(Room) = 0.28 + 0.12 = 0.4

Now, applying Bayes' Theorem:

$P(Y | Room) = \frac{0.2 \cdot 0.6}{0.4} = \frac{0.12}{0.4} = 0.3$

So, the probability that Brianna is in dorm Y, given that she has a room, is 0.3.

Let me know if you need any further explanation or details! Would you like me to expand on the steps involved? Here are some related questions for you:

1. How would the probability change if the probabilities for rooms in X and Y were reversed?
2. How can we apply Bayes' Theorem to other types of conditional probability problems?
3. What would happen if Brianna was more likely to get an apartment instead of a room in dorm Y?
4. How do you calculate the total probability for events with more than two possibilities?
5. How could this type of problem be extended to multiple dorms and room types?

Tip: Always verify your total probability (P(Room)) before applying Bayes' Theorem to ensure correctness in calculations!