Question on sampling without replacement

#1
Here is a question regarding sampling without replacement.

Say there is a class size of 24, and the teacher is breaking students into groups of (4) for project work. The teacher selects the groups by putting all (24) names into a hat, then drawing (4) at a time. What is the probability that (2) friends, Adam and Bob, are chosen at random for the same group?

Q1: does order matter in selecting the groups? For the first group, there are P(24,6) ordered choices selected from the hat, but only C(24,6) group combinations. I have included a chart showing the math for using both.

Q2: do we simply add the combinations (or permutations) for each group as I’ve shown?

Q3: is the final probability (using combinations) simply 7,462/17,857 as shown? This is fairly close to the result using permutations as noted.

Q5: Here’s where my head starts to spin. Let’s say the teacher decided to make (2) large groups of 12. Using the same analysis (see 2nd attachment), the probability for Adam and Bob in the same group is now 646,647/2,704,157 or approx. 24%.

Does this make sense? The sample space is one of (4) possibilities for tables X&Y:

1: X(Bob & others), Y(Adam & others): NO
2: X(Adam & others), Y(Bob & others): NO
3: X(Adam & Bob together), Y(others): YES
4: X(others), Y(Adam & Bob together): YES

This “intuitive” analysis suggests there is a 50% chance Adam and Bob are in the same group!

But I think there is a disconnect between looking at the apparent sample space and in how the groups are chosen. Any thoughts? I’d originally thought about using the hypergeometric distribution, which gives a probability for [C(2,2)xC(22,2)]/C(24,4) = 2.17%.

Which is correct?

I’ve simplified this example from something I worked on this weekend, based on an offsite meeting. There were 140 of us, split into tables of (6), with 1 leftover table of 2. I shared a table with another co-worker, and she wondered what the odds were of us sitting together if the tables were chosen randomly, as above (see attachment for calculations). In this case, using the same analysis I calculated the probability as 73.7%, which makes no sense at all. This is why I scaled things down with the classroom example, to get a better feel for the numbers; it appears to be more reasonable. Again, using the hypergeometric distribution (I’ve fudged out the last group of 2, I know), I get approximately: [C(2,2)xC(138,4)]/C(140,6) = 0.154%, which makes much more sense.

Any thoughts/ideas where my method has run aground? Thanks! /Steve
 

katxt

Active Member
#2
Look at the group Adam is in. There are three remaining places and 23 remaining students, one of which is BoB. So there are 3 chances in 23 of Bob being in the same group.