Probability - Combination

a) For forming the 5 groups in such manner:

In formint the 1st group, we have to choose 3 randomly out of 15 students, so we have 15C3 combinations.

In formint the 2nd group, we have to choose 3 randomly out of 12 students, so we have 12C3 combinations.

Continuing the process, we have a total of:

15C3 x 12C3 x 6C3 x 9C3 x 3C3

combinations.

Then for the requirement "at least one study group consists of students from 3 different classes", we can consider in this manner:

When forming the first group, the 3 students must be from different classes, then don't care in forming the remaining 4 groups since at least one group already has 3 students from different classes.

So when forming the 1st group, we can choose 3 classes from 5, giving 5C3 combiniations and then for each selected class, we have 3C1 selections.

Hence when forming the first group, we have 5C3 x 3C1 x 3C1 x 3C1 choices.

Then for the remaining 4 groups, no. of selections = 12C3 x 6C3 x 9C3 x 3C3
since we don't care anymore.

So the required probability is:

(5C3 x 3C1 x 3C1 x 3C1 x 12C3 x 6C3 x 9C3 x 3C3)/(15C3 x 12C3 x 6C3 x 9C3 x 3C3)

= (5C3 x 3 x 3 x 3)/(15C3)

= 54/91

b) This event is the complementary event of that in (a), so probability is 1 - 54/91 = 37/91