Voice lottery program

Solution: From the meaning of the question, we know that this problem is a classical probability.

∫ All the events involved in the experiment are in the order of 10 * * A10/0;

Meet the event conditions to get "three students in class one are arranged together, and two students in class two are not arranged together" through the following steps:

① There are 33 ways to "bind" three students in Class One;

(2) There are A66 ways to treat "one child" in class one as an object and queue up with five students and six objects in other classes;

③ There are 72 ways to insert two students from Class Two into the seven gaps (including the positions at both ends) where the above six objects are arranged.

According to the step counting principle (multiplication principle), * * * has A33? A66？ A72 method.

There are three students in class one who just line up together (referring to the serial number of the speech).

The probability that two students in Class Two don't line up together is: P=

A33？ A66？ A27A 10 10

=

120

.

So choose B.