1. **Problem Statement:** We have 500 students surveyed about their preferences for watching football, hockey, and basketball. Given percentages for each sport and their pairwise intersections, plus the percentage who like none, we need to find how many students like all three sports. 2. **Given Data:** - Total students, $N = 500$ - $P(F) = 49\% = 0.49$ - $P(H) = 53\% = 0.53$ - $P(B) = 62\% = 0.62$ - $P(F \cap H) = 27\% = 0.27$ - $P(F \cap B) = 28\% = 0.28$ - $P(B \cap H) = 26\% = 0.26$ - $P(\text{none}) = 0.5\% = 0.005$ 3. **Formula Used:** For three sets, the inclusion-exclusion principle states: $$ P(F \cup H \cup B) = P(F) + P(H) + P(B) - P(F \cap H) - P(F \cap B) - P(H \cap B) + P(F \cap H \cap B) $$ Since $P(\text{none}) = 0.005$, the probability of liking at least one sport is: $$ P(F \cup H \cup B) = 1 - P(\text{none}) = 1 - 0.005 = 0.995 $$ 4. **Calculate $P(F \cap H \cap B)$:** Rearranging the inclusion-exclusion formula: $$ P(F \cap H \cap B) = P(F \cup H \cup B) - [P(F) + P(H) + P(B)] + [P(F \cap H) + P(F \cap B) + P(H \cap B)] $$ Substitute values: $$ P(F \cap H \cap B) = 0.995 - (0.49 + 0.53 + 0.62) + (0.27 + 0.28 + 0.26) $$ Calculate step-by-step: $$ 0.49 + 0.53 + 0.62 = 1.64 $$ $$ 0.27 + 0.28 + 0.26 = 0.81 $$ So, $$ P(F \cap H \cap B) = 0.995 - 1.64 + 0.81 = 0.165 $$ 5. **Find the number of students who like all three sports:** $$ \text{Number} = P(F \cap H \cap B) \times N = 0.165 \times 500 = 82.5 $$ Since number of students must be whole, we round to 83. **Final answer:** 83 students like watching all three sports.