combinatorics probability question.

[usmanqd]

Question.
A certain club has 10 members. 6 elite and 4 standard members. If 6 members are to be selected, what is the probability that 3 or more of the members are elite?

High Level Solution:
One way to solve this question is to calculate number of ways to select each of the following and then divide by number of ways of selecting 6 out of 10.

-3 elite + 3 standard
-4 elite + 2 standard
-5 elite + 1 standard
-6 elite

Specific issue:
My question is how do we calculate the number of ways in which 3 elite and 3 standard members can be selected?

[nt2011]

You would get the 3 elite and 3 standard using the slot method

6/10 * 5/9 * 4/8 * 4/7*3/6*2/5*1/4 = 1/210

Similarly you would calculate other scenarios as well and add them all.

[jlucero]

nt2011-

2 things-

1) you ended up with seven different fractions, which is probably just a mistake from you rushing through the problem

2) you calculated the probability that the first person would be elite, the second would be elite, the third would be elite, the fourth would be standard... etc. But there are other ways that you could achieve the same outcome, but without having selected the first, second, and third as elite, but rather the first, third, and sixth. Once you calculated your probability, you would need to multiply by any other possible arrangement of EEESSS = 6!/3!3! = 20 unique orders for selecting exactly 3 elite and 3 standard memebers.

[jlucero]

usmanqd-

Since you are looking for the number of possible arrangements, start by selecting your elite and standard members separately. If you wanted to select 3 elite members, you could select 6x5x4 members, but since it doesn't matter what order you select them in, you would divide this by 3! to eliminate identical groups. As an equation, you would say there are 6!/3!3! = 6! total / 3! in x 3! out = 20 ways of selecting 3 elite members.

By the same method, you would say there are 4!/3!1! = 4! total / 3! in x 1! out = 4 ways of selecting 3 standard members (common sense check- if I'm selecting three out of four people, I can leave out person A, B, C, or D).

Since there are 20 ways to select elite members and 4 ways to select standard members, there are 20 x 4 = 80 ways to select exactly 3e & 3s members.

*Note that this 80 is different than the 20 posted above b/c 80 represents the total # of groups with 3e & 3s members (group ABCWYZ), while 20 represents the total # of different orders that I could have selected those members (order eseess)

[usmanqd]

Thanks Joe. This makes sense.

[tim]

glad to hear it!