1. (2004 College Entrance Examination, Chongqing Paper) In a speech contest held by a senior high school grade, there are 10 students participating, among whom 3 are from Class 1, 2 are from Class 2, and 5 are from other classes. If the order of their speeches is determined by drawing lots, then the probability that the 3 students from Class 1 are exactly grouped together (meaning their speaking order numbers are consecutive), while the 2 students from Class 2 are not grouped together is ( )
A. $\frac{1}{10}$
B. $\frac{1}{20}$
C. $\frac{1}{40}$
D. $\frac{1}{120}$
8. Let the first box contain $n$ balls, $b$ of which are black; then the second box contains $(25-n)$ balls, with $B$ being black. Then $\frac{b B}{n(25-n)}=\frac{27}{50} . \therefore n(25-n)$ is a multiple of 5, $\therefore n$ is a multiple of 5. Let $n=5$ or 10 (otherwise swap the boxes), if $n=5$ then $b B=54, \therefore b \mid 54$ and $b \leqslant 5 . \therefore b=1,2,3$. Verification: $B=54,27,18$. Only $B=18 \leqslant 20$ meets the condition. $\therefore b=3, B=18 \quad P=\frac{2}{5} \cdot \frac{1}{10}=\frac{1}{25}$. If $n=10, b B=81, \therefore b=B=9, \therefore P=\frac{1}{10} \cdot \frac{2}{5}=\frac{1}{25}$. In summary: $P=\frac{1}{25}$.
1. B $P=\frac{6!\times 3!\times 7 \times 6}{10!}=\frac{1}{20}$