#27 **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) = 0.72. 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) = 0.23. 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(\

From given,P(X) = 0.3P(Y) = 0.7P( Room | X ) = 0.8P( Room | Y ) = 0.4We have to find If Brianna gets…

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

A First Course in Probability (10th Edition)

10th Edition

ISBN:9780134753119

Author:Sheldon Ross

Publisher:Sheldon Ross

Chapter1: Combinatorial Analysis

Section: Chapter Questions

Problem 1.1P: a. How many different 7-place license plates are possible if the first 2 places are for letters and...

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