Answer:
a) 0.0659
b) 0.1648
Step-by-step 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:
[tex]{6 \choose 1} = 6[/tex]
For a sophomore to participate, we have to:
[tex]{5 \choose 1} = 5[/tex]
For a junior to participate, we have to:
[tex]{3 \choose 1} = 3[/tex]
For a senior to participate, we have to:
[tex]{1 \choose 1} = 1[/tex]
The number of possible subcommittees is
[tex]{15 \choose 4} = 1365[/tex]
So consider
A: "All 4 classes are represented"
[tex]\mathbb{P}(A) = \frac{6\cdot 5\cdot 3\cdot 1}{1365} \approx 0.0659[/tex]
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
[tex]{6 \choose 2}\cdot {5 \choose 1}\cdot {3 \choose 1}\cdot {1 \choose 1} = 225[/tex]
Choosing two from sophomore and one from other classes
[tex]{6 \choose 1}\cdot {5 \choose 2}\cdot {3 \choose 1}\cdot {1 \choose 1} = 180[/tex]
Choosing two juniors and 1 from the other classes
[tex]{6 \choose 1}\cdot {5 \choose 1}\cdot {3 \choose 2}\cdot {1 \choose 1} = 90[/tex]
The number of possible subcommittees is
[tex]{15 \choose 5} = 3003 \\\mathbb{P}(A) = \dfrac{(225+180+90)}{3003} \ \approx 0.1648[/tex]