Question

committee consists of six Freshmen, five Sophomores, three Juniors, and one Senior.a.) A subcommittee of four is to be chosen at random. What is the probability that all fourclasses are represented on the subcommittee?b.) Answer the question in part (a) if a subcommittee of five is chosen (instead of a sub-committee of four).

Answer 1

a) 0.0659

b) 0.1648

Explanation:

6 Freshmen

5 Sophomores

3 Juniors

1 Senior

Total 15 people

a) Form a subcommittee with 4 people and all classes

should display this subcommittee. That is, of the 4 categories, all

they must participate.

For a freshman to participate we have to:

`{6 \choose 1} = 6`

For a sophomore to participate, we have to:

`{5 \choose 1} = 5`

For a junior to participate, we have to:

`{3 \choose 1} = 3`

For a senior to participate, we have to:

`{1 \choose 1} = 1`

The number of possible subcommittees is

`{15 \choose 4} = 1365`

So consider

A: "All 4 classes are represented"

`\mathbb{P}(A) = (6\cdot 5\cdot 3\cdot 1)/(1365) \approx 0.0659`

b) Taking the last class, we can choose 2 people from the other groups.

Therefore, we vary the number of people as follows:

Choosing two freshmen and 1 from the other classes

`{6 \choose 2}\cdot {5 \choose 1}\cdot {3 \choose 1}\cdot {1 \choose 1} = 225`

Choosing two from sophomore and one from other classes

`{6 \choose 1}\cdot {5 \choose 2}\cdot {3 \choose 1}\cdot {1 \choose 1} = 180`

Choosing two juniors and 1 from the other classes

`{6 \choose 1}\cdot {5 \choose 1}\cdot {3 \choose 2}\cdot {1 \choose 1} = 90`

The number of possible subcommittees is

`{15 \choose 5} = 3003 \\\mathbb{P}(A) = ((225+180+90))/(3003) \ \approx 0.1648`