Probability question

Virum

Hey guys, my girlfriend is taking this probability course and I am trying to help her out with this problem. It is hard for her since she is ESL and she gets confused easily, so I try to read the book and interpret, but I can't sit in on the lecture so I don't really know what is going on. I don't know much about probability and after looking at class examples and lecture notes and reading the book I can't figure out a starting point for the problem:

An urn contains numbered chips from 1 to 10. A sample of 3 is chosen at random. Let Y be the maximum of the 3 numbers in the sample. Find the cdf of Y, i.e F(y) = P(Y<=y), and use this to evaluate p(y) = P(Y=y).

Now that notation is greek to me, but by questions asked and looking over notes I think that we need to find the probability of each number being the maximum in the selection of three before we can find the cdf.

Now obviously the probability that 1 being the max, if selected, is 0, regardless of the other two numbers is 0 as well. For 10 it'd always have to be the maximum. However, I have no idea how to find the other 7 numbers.

For a number like 4, I am thinking

(3 #s less than 4 choose 2 / 9 numbers choose 2)

(3 choose 2) / (9 choose 2) = 0.0833333333

Hmm, sounds okay. What about 8?

(7 choose 2) / (9 choose 2) = 0.583333333

Seems a little low to me.

Lets test 10:

(9 choose 2) / (9 choose 2) = 1... seems to work.

Am I way off base? What do I need to start doing?

Tinuz

Simply put (and I never use this stuff, so I may be wrong):

you draw 3 numbers, so, if you draw the 3 lowest numbers then 3 is the max....this is the lowest max (which is what you re getting at). So our cdf starts at 3.

Then, I'd say the probabilities of what we are getting is completely indepent and equal for each and ever chip. so 1/8th, the maximum probability is 1, hence we get:

P(Y<=y)

1 : 0
2 : 0
3 : 1/8
4 : 2/8
5 : 3/8
6 : 4/8
7 : 5/8
8 : 6/8
9 : 7/8
10 : 8/8

That is, if we draw 3 chips, the first two of which are 1 and 2, then I know that the next chip will be the maximum. At that point the probability of it being less than, say, 7 is 4 (for the number of chips that are smaller than 7) out of the total of 8 chips, hence 4/8.

Finding the pdf is done by integration.

EDIT: I don't reeallt get what you mean by your notation, so if I repeated what you just said, I'm sorry.

Savant

If I'm reading you right, then for this problem it is easier to find the CDF directly rather than trying to cobble it together from the probability distribution function. Then, depending on what the problem says, you can find the PDF from that.

Since you are dealing with discrete probability where it is equally likely to draw any number out of the bag, one approach (which you seem to have hit on) is to count the number of ways to satisfy an event and divide that by the total number of possible combinations. Note that this only works directly for distributions where the outcomes of the drawings are random and equally likely.

So you want to count the events (events are subsets of the total set of outcomes) that correspond to the random variable Y having the range of values Y <= y for a particular y. This event occurs when all 3 numbers drawn are between 1 and y, inclusive. Since you don't care about the order the numbers are drawn in and there is no replacement, the number of possible ways to draw 3 numbers between 1 and y is (y choose 3). The total number of possible combinations to draw 3 numbers is (10 choose 3).

So P[Y <= y] = (y choose 3) / (10 choose 3), which is the number of combinations satisfying Y <= y divided by the total number of possible combinations. You can then use this to determine the PDF.

Savant

One way to think of this sort of probability is to think about poker hands. For example, the probability of drawing four of a kind is equal to the number of different hands that are four of a kind (4 of one value of card plus any other card) divided by the total number of different poker hands by combining together 5 cards there are.

Tinuz

Yeap, what he said.

I just Monte Carlo'ed it, and when he wrote it a did the whole combinations thing and they match up.