A committe of 5 members will be chosen from a group of 10 teachers and 5 students. What is the probability that the committee
will have
A) all teachers?
B) 3 teachers and 2 students?
C) 3 or 4 teachers?
To find the probability in each scenario, we need to use the concept of combinations.
In this case, we need to choose a committee of 5 members from a total of 15 people (10 teachers and 5 students).
A) To find the probability of selecting all teachers, we need to calculate the number of ways to choose 5 teachers from a group of 10 teachers. The probability can be calculated as:
Number of ways to choose 5 teachers from 10 teachers / Total number of ways to choose 5 members from 15 people
The number of ways to choose 5 teachers from 10 teachers can be calculated using the combination formula: C(n, r) = n! / (r!(n-r)!)
C(10, 5) = 10! / (5!(10-5)!) = 252
The total number of ways to choose 5 members from 15 people can also be calculated using the combination formula: C(15, 5) = 15! / (5!(15-5)!)= 3,003
Therefore, the probability of selecting all teachers is:
P(all teachers) = 252 / 3,003 ≈ 0.084
B) To find the probability of selecting 3 teachers and 2 students, we need to calculate the number of ways to choose 3 teachers from 10 teachers and 2 students from 5 students. The probability can be calculated as:
(Number of ways to choose 3 teachers from 10 teachers) * (Number of ways to choose 2 students from 5 students) / Total number of ways to choose 5 members from 15 people
The number of ways to choose 3 teachers from 10 teachers is: C(10, 3) = 10! / (3!(10-3)!) = 120
The number of ways to choose 2 students from 5 students is: C(5, 2) = 5! / (2!(5-2)!) = 10
The total number of ways to choose 5 members from 15 people is: C(15, 5) = 15! / (5!(15-5)!)= 3,003
Therefore, the probability of selecting 3 teachers and 2 students is:
P(3 teachers and 2 students) = (120 * 10) / 3,003 ≈ 0.400
C) To find the probability of selecting 3 or 4 teachers, we need to calculate the number of ways to choose 3 teachers from 10 teachers, the number of ways to choose 4 teachers from 10 teachers, and the number of ways to choose 1 or 0 students from 5 students. The probability can be calculated as:
(Number of ways to choose 3 teachers from 10 teachers) * (Number of ways to choose 2 students from 5 students) + (Number of ways to choose 4 teachers from 10 teachers) * (Number of ways to choose 1 student from 5 students) / Total number of ways to choose 5 members from 15 people
The number of ways to choose 3 teachers from 10 teachers is: C(10, 3) = 10! / (3!(10-3)!) = 120
The number of ways to choose 2 students from 5 students is: C(5, 2) = 5! / (2!(5-2)!) = 10
The number of ways to choose 4 teachers from 10 teachers is: C(10, 4) = 10! / (4!(10-4)!) = 210
The number of ways to choose 1 student from 5 students is: C(5, 1) = 5! / (1!(5-1)!) = 5
The total number of ways to choose 5 members from 15 people is: C(15, 5) = 15! / (5!(15-5)!)= 3,003
Therefore, the probability of selecting 3 or 4 teachers is:
P(3 or 4 teachers) = (120 * 10 + 210 * 5) / 3,003 ≈ 0.362
all you have to do is figure out the probability of each and multiply appropriately.
a) the teachers represent 10/15 or 2/3 of the candidates, we need 5 teachers. imagine someone choosing commitee members randomly. they would have a 2/3 chance of picking the first teacher, a 9/14 chance of picking the second teacher, (there are now 14 candidates and 9 teachers), a 8/13 chance of picking the third teacher, a 7/12 chance of picking the fourth teacher, and a 6/11 chance of picking the fifth teacher.
We now multiply.
(2/3)*(9/14)*(8/13)*(7/12)*(6/11)