1. **Problem Statement:** In a survey of 600 students, percentages of students liking football, hockey, and basketball are given along with overlaps. We need to find: a) Number of students who like all three games. b) Ratio of students who like only football to those who like only hockey. c) Number of students who like only one of the three games. d) Number of students who like at least two of the games. 2. **Given Data:** - Total students, $N = 600$ - $P(F) = 51\% = 0.51$ - $P(H) = 48\% = 0.48$ - $P(B) = 60\% = 0.60$ - $P(F \cap H) = 27\% = 0.27$ - $P(H \cap B) = 24\% = 0.24$ - $P(F \cap B) = 29\% = 0.29$ - $P(\text{none}) = 5\% = 0.05$ 3. **Formula and Rules:** - Use the Inclusion-Exclusion Principle for three sets: $$P(F \cup H \cup B) = P(F) + P(H) + P(B) - P(F \cap H) - P(H \cap B) - P(F \cap B) + P(F \cap H \cap B)$$ - Since $P(\text{none}) = 0.05$, then $P(F \cup H \cup B) = 1 - 0.05 = 0.95$ 4. **Step 1: Find $P(F \cap H \cap B)$** $$0.95 = 0.51 + 0.48 + 0.60 - 0.27 - 0.24 - 0.29 + P(F \cap H \cap B)$$ $$0.95 = 1.59 - 0.80 + P(F \cap H \cap B)$$ $$0.95 = 0.79 + P(F \cap H \cap B)$$ $$P(F \cap H \cap B) = 0.95 - 0.79 = 0.16$$ Number of students who like all three games: $$600 \times 0.16 = 96$$ 5. **Step 2: Find number of students who like only football and only hockey** - Only football: $$P(F) - P(F \cap H) - P(F \cap B) + P(F \cap H \cap B) = 0.51 - 0.27 - 0.29 + 0.16 = 0.11$$ - Only hockey: $$P(H) - P(F \cap H) - P(H \cap B) + P(F \cap H \cap B) = 0.48 - 0.27 - 0.24 + 0.16 = 0.13$$ Number of students: - Only football: $600 \times 0.11 = 66$ - Only hockey: $600 \times 0.13 = 78$ Ratio of only football to only hockey: $$\frac{66}{78} = \frac{\cancel{6}6}{\cancel{7}8} = \frac{11}{13}$$ 6. **Step 3: Find number of students who like only one game** - Only basketball: $$P(B) - P(F \cap B) - P(H \cap B) + P(F \cap H \cap B) = 0.60 - 0.29 - 0.24 + 0.16 = 0.23$$ Number of students who like only one game: $$600 \times (0.11 + 0.13 + 0.23) = 600 \times 0.47 = 282$$ 7. **Step 4: Find number of students who like at least two games** - Students who like exactly two games: $$P(F \cap H) + P(H \cap B) + P(F \cap B) - 3 \times P(F \cap H \cap B) = 0.27 + 0.24 + 0.29 - 3 \times 0.16 = 0.80 - 0.48 = 0.32$$ - Students who like at least two games (including all three): $$0.32 + 0.16 = 0.48$$ Number of students: $$600 \times 0.48 = 288$$ **Final Answers:** - a) 96 students - b) Ratio $11:13$ - c) 282 students - d) 288 students