Show clear work for all problems. Give answers as simplified fractions or decimals rounded to four places unless stated otherwise. No formulas or hints are provided on the exam.
Part A (Chapter 2) — Bayes’ Theorem: Dorm Problems
1) Three Dorms (A/B/C).
The campus has Dorm A, Dorm B, and Dorm C. The fractions of students in each dorm are: Dorm A 25%, Dorm B 25%, Dorm C 30%. The probability a randomly chosen student is female within each dorm is: Dorm A 60%, Dorm B 45%, Dorm C 35%. Given that a randomly chosen student is female, compute the probability she is in each dorm: P(Dorm A | Female), P(Dorm B | Female), and P(Dorm C | Female).
Bayes by “match then normalize” for each dorm d:
matchA=0.60×0.25=0.1500, matchB=0.45×0.25=0.1125, matchC=0.35×0.30=0.1050.
Evidence P(Female)=0.1500+0.1125+0.1050=0.3675.
Excel: put P(d) and P(F|d) in columns and compute =P(d)*P(F|d), sum, then divide each match by the sum.
2) Four Dorms (A/B/C/D).
Dorm populations: A 25%, B 25%, C 30%, D 20%. The probability a student is female within each dorm: A 60%, B 45%, C 35%, D 50%. Given that a student is female, find the probability she is in each dorm: P(A | Female), P(B | Female), P(C | Female), P(D | Female).
Excel: column of matches =P(d)*P(F|d) → sum → divide each by the sum.
3) Five Dorms (A/B/C/D/E).
Dorm populations: A 15%, B 20%, C 25%, D 20%, E 20%. The probability a student is female within each dorm: A 70%, B 60%, C 50%, D 40%, E 30%. Given that a student is female, find the probability she is in each dorm: P(A | Female), P(B | Female), P(C | Female), P(D | Female), P(E | Female).