$$Question: How many of the 100 smallest positive integers are congruent to 3 modulo 7?
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- \binom{6}{1} \cdot \binom{5}{1} \cdot \binom{4}{1} = 6 \cdot 5 \cdot 4 = 120
- P(\text{at least one from each profession}) = \frac{120}{455} = \frac{24}{91}
- \boxed{\frac{24}{91}}
- Solution: We are asked to count how many of the first 100 positive integers satisfy the congruence:
- n \equiv 3 \pmod{7}
- This means $ n = 7k + 3 $ for some integer $ k $.
