Combinatorics problems

[canada324]

Just cant seem to completely wrap my head around some problems involving combinatorics. Here's a question that I think uses combinations and that kind of stuff

There are 10 distinct items to choose from. You pick 6 of the items at random. Another person chooses 8 items at random (after they have all been put back obviously). what is the probability that you picked 3 items in common?

I kind of made this one up, but it's along the lines of the problems that I sometimes struggle with. Does this even require the use of combinations? And if you have any tips for handling these kinds of problems that would be great.
thanks!

[bsd058]

Just cant seem to completely wrap my head around some problems involving combinatorics. Here's a question that I think uses combinations and that kind of stuff

There are 10 distinct items to choose from. You pick 6 of the items at random. Another person chooses 8 items at random (after they have all been put back obviously). what is the probability that you picked 3 items in common?

I kind of made this one up, but it's along the lines of the problems that I sometimes struggle with. Does this even require the use of combinations? And if you have any tips for handling these kinds of problems that would be great.
thanks!

Combinations are not needed for this specific question. Just intuition.

If I take 6 items out and put them back in, and then the second person takes 8 items out (assuming the two that are in the basket are ones that were not originally chosen), then the 2nd person has all 6 items I originally chose. If the two in the basket were both chosen at first, then the 2nd person has four of my original items for sure. The min(same items) = 4, and the max(same items) = 6. Therefore it is impossible to get 3 of the same items exactly and you are certain to get at least 3.

Try doing the problem with you picking out 2 and the second person picking out 4. Then try and see the probability that they picked at least 1 of your items.

Then combinations will be needed.


This would be:

1 - [(number of ways to not pick one of your items by choosing 4 items) / (number of ways to choose 4 items)]

= 1 - (8C4 / 10C4)

= 1 - (70/210) = 2/3.

I think this is correct. It's been a long day, so I could be too tired to do this.

The 8C4 is how many ways there are to choose 4 items without picking one of the 2 that were originally picked.

The 10C4 is how many ways the second person can pick 4 items in general.


Or you could say that the first person that picked out the items picked out 4 and the second picked out 2.

Then the probability would be:

1 - (6C2/10C2) = 1 - 1/3 = 2/3.


These are the same probabilities just solved differently. The first person picked 4 items, and so 6C2 is the number of ways to choose 2 items from the 6 that were not chosen. 10C2 is the number of ways to choose 2 items from 10.

[Happychick90]

my guess: It looks like the probability that you two pick only 3 items in common is zero because even if you minimize the overlap, you'd both still pick 4 of the same items.