MATH SOLVE

3 months ago

Q:
# 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).

Accepted Solution

A:

Answer:a) 0.0659b) 0.1648Step-by-step explanation:6 Freshmen5 Sophomores3 Juniors1 SeniorTotal 15 peoplea) Form a subcommittee with 4 people and all classesshould display this subcommittee. That is, of the 4 categories, allthey 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 considerA: "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]