Brianna is taking part in a lottery for a room in one of two new dormitories at her college. She will draw a card randomly to determine which room she will have. Each card has the name of a dormitory, (X)avier or (Y)oung, and also a two-person room number or an apartment number. Thirty percent of the available spaces are in X. Eighty percent of the available spaces in X are rooms and 40% of the spaces in Y are rooms. If Brianna gets a room, what is the probability that she is in dorm Y? 68 ... The probability is (Type an integer or decimal rounded to three decimal places as needed.) P(Room | X)=0.8 0.3 P(Y) = 0.7 Room = 0.24 0.2 Apartment = 0.06 0.4 Room= 0.28 P(Apartment | Y)=0.6 Apartment = 0.42 icorre
Brianna is taking part in a lottery for a room in one of two new dormitories at her college. She will draw a card randomly to determine which room she will have. Each card has the name of a dormitory, (X)avier or (Y)oung, and also a two-person room number or an apartment number. Thirty percent of the available spaces are in X. Eighty percent of the available spaces in X are rooms and 40% of the spaces in Y are rooms. If Brianna gets a room, what is the probability that she is in dorm Y? 68 ... The probability is (Type an integer or decimal rounded to three decimal places as needed.) P(Room | X)=0.8 0.3 P(Y) = 0.7 Room = 0.24 0.2 Apartment = 0.06 0.4 Room= 0.28 P(Apartment | Y)=0.6 Apartment = 0.42 icorre
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![**Brianna's Dormitory Lottery Probability Explanation**
Brianna is taking part in a lottery for a room in one of two new dormitories at her college. She will draw a card randomly to determine which room she will have. Each card has the name of a dormitory, (X)avier or (Y)oung, and also a two-person room number or an apartment number. Thirty percent of the available spaces are in X. Eighty percent of the available spaces in X are rooms and 40% of the spaces in Y are rooms. If Brianna gets a room, what is the probability that she is in dorm Y?
**Diagram Explanation:**
The diagram represents probabilities branching from two primary dorm choices, X and Y.
1. Brianna can either get a dorm in X or Y:
- P(X) = 0.3
- P(Y) = 0.7
2. If she gets dorm X, she can get:
- A room: P(Room | X) = 0.8 (which is 80% of the spaces in X)
- An apartment: P(Apartment | X) = 0.2
3. If she gets dorm Y, she can get:
- A room: P(Room | Y) = 0.4 (which is 40% of the spaces in Y)
- An apartment: P(Apartment | Y) = 0.6 (which is 60% of the spaces in Y)
**Probabilities Calculated:**
- For dorm X:
- P(Room ∩ X) = P(X) * P(Room | X) = 0.3 * 0.8 = 0.24
- P(Apartment ∩ X) = P(X) * P(Apartment | X) = 0.3 * 0.2 = 0.06
- For dorm Y:
- P(Room ∩ Y) = P(Y) * P(Room | Y) = 0.7 * 0.4 = 0.28
- P(Apartment ∩ Y) = P(Y) * P(Apartment | Y) = 0.7 * 0.6 = 0.42
**Final Probability Calculation:**
The total probability of Brianna getting a room (either in X or Y):
\[ P(\text{Room}) = P(\](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F128593f5-7ea4-4e3b-94cd-55abf91c7165%2F8f8049c6-8f19-433f-9fad-05e70ff2e338%2F43f6jk_processed.jpeg&w=3840&q=75)
Transcribed Image Text:**Brianna's Dormitory Lottery Probability Explanation**
Brianna is taking part in a lottery for a room in one of two new dormitories at her college. She will draw a card randomly to determine which room she will have. Each card has the name of a dormitory, (X)avier or (Y)oung, and also a two-person room number or an apartment number. Thirty percent of the available spaces are in X. Eighty percent of the available spaces in X are rooms and 40% of the spaces in Y are rooms. If Brianna gets a room, what is the probability that she is in dorm Y?
**Diagram Explanation:**
The diagram represents probabilities branching from two primary dorm choices, X and Y.
1. Brianna can either get a dorm in X or Y:
- P(X) = 0.3
- P(Y) = 0.7
2. If she gets dorm X, she can get:
- A room: P(Room | X) = 0.8 (which is 80% of the spaces in X)
- An apartment: P(Apartment | X) = 0.2
3. If she gets dorm Y, she can get:
- A room: P(Room | Y) = 0.4 (which is 40% of the spaces in Y)
- An apartment: P(Apartment | Y) = 0.6 (which is 60% of the spaces in Y)
**Probabilities Calculated:**
- For dorm X:
- P(Room ∩ X) = P(X) * P(Room | X) = 0.3 * 0.8 = 0.24
- P(Apartment ∩ X) = P(X) * P(Apartment | X) = 0.3 * 0.2 = 0.06
- For dorm Y:
- P(Room ∩ Y) = P(Y) * P(Room | Y) = 0.7 * 0.4 = 0.28
- P(Apartment ∩ Y) = P(Y) * P(Apartment | Y) = 0.7 * 0.6 = 0.42
**Final Probability Calculation:**
The total probability of Brianna getting a room (either in X or Y):
\[ P(\text{Room}) = P(\
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