Suppose that a state university has to form a committee of 5 members from a list of 20 candidates out of whom 12 are teachers and 8 are students.If members of the committee are selected at random,what is the probability that the majority of the committee members are students?

I8tufufhdu

suppose that a state university has to form committee of 5 members from a list of 20

We can do it by binomial taking n=5 p=8/20 and q=12/20

P=2/5

Q=3/5
N=5
P(X>=3)= P(X=0)+P(X=1)+P(X=2)+P(X=3)
=0.07776+0.2592+0.3456+0.2304
=0.9129

To find the probability that the majority of the committee members are students, we need to calculate the probability of different scenarios where the majority of the committee members are students.

First, let's determine all possible scenarios where the majority of committee members are students:

1. Selecting 3 students and 2 teachers: There are 8 students and 12 teachers to choose from. The number of ways to select 3 students from 8 is "8 choose 3" or C(8,3), which is equal to 8! / (3! * (8-3)!) = 56. Similarly, the number of ways to select 2 teachers from 12 is "12 choose 2" or C(12,2), which is equal to 12! / (2! * (12-2)!) = 66. Therefore, the number of ways to form the committee with 3 students and 2 teachers is 56 * 66 = 3,696.

2. Selecting 4 students and 1 teacher: Similar to the previous scenario, there are 8 students and 12 teachers to choose from. The number of ways to select 4 students from 8 is "8 choose 4" or C(8,4), which is equal to 8! / (4! * (8-4)!) = 70. The number of ways to select 1 teacher from 12 is "12 choose 1" or C(12,1), which is equal to 12! / (1! * (12-1)!) = 12. Therefore, the number of ways to form the committee with 4 students and 1 teacher is 70 * 12 = 840.

3. Selecting all 5 students: There are 8 students to choose from. The number of ways to select 5 students from 8 is "8 choose 5" or C(8,5), which is equal to 8! / (5! * (8-5)!) = 56.

Now, let's calculate the total number of ways to form a committee of 5 members from a list of 20 candidates. There are 20 candidates to choose from, so the number of ways to select 5 members is "20 choose 5" or C(20,5), which is equal to 20! / (5! * (20-5)!) = 15,504.

Finally, we can calculate the probability by summing up the number of ways for each scenario and dividing it by the total number of possible committees.

Probability = (Number of ways to have a majority of students) / (Total number of possible committees)
= (Number of ways to have 3 students and 2 teachers + Number of ways to have 4 students and 1 teacher + Number of ways to have all 5 students) / (Total number of possible committees)
= (3,696 + 840 + 56) / 15,504
= 4,592 / 15,504
= 0.2968

Therefore, the probability that the majority of the committee members are students is approximately 0.2968 or 29.68%.

Suppose that a SRM University, Sikkim has to form a committee of 5 members from a list of 20 candidate out of which 12 are faculty members and 8 are students. If the members of committee are selected at random. What is the probability that the majority of the committee members are students?