Socks in a drawer conundrum

sirellwood · #1 $Old$ October 25th, 2009, 06:12 PM

A drawer contains 8 Red socks, 4 green socks and 4 yellow socks. Two socks
are chosen randomly from the drawer, without replacement. Let R be the number of red socks chosen and Y be the number of yellow socks chosen.

(a) Derive the joint probability mass function of R and Y , i.e. P(R = r, Y = y).
(b) Calculate the marginal probability distributions of R and Y , i.e. P(R = r) and P(Y = y), respectively.
(c) Are R and Y independent? Justify your answer.
(d) What is the probability that a pair of the same colour is chosen?
(e) Obtain the conditional distribution of R|Y = 1.
(f) Evaluate E[R], Var(R) and Cov(R, Y).

ANDS! · #2 $Old$ October 25th, 2009, 11:42 PM

A. Your PMF is not a function perse, in the sense you are thinking (perhaps) of it as. It is simply a chart with R and Y being as you say. To set this up you need to know the possible values of R and Y. You are told that you are choosing two socks without replacement, and so your values of R and Y can range from 0 to 2 (do you see why?)

You then set up a grid, but how you set up the grid is the important part. In this scenario, both the R and the Y are columns and they can take on values of 0, 1, 2 since we are selecting two socks. P(R=0) would be the probability you select no red socks, which means the probability that on the first and second draw you get a green or a yellow - (12/16)(11/16). P(R=1) means you got a red on the first or second draw - (8/16)(12/15)+(12/16)(8/15). And so on and so forth.

Once you set up your probability mass function, figuring out the rest should be easy (assuming you know how to figure out the rest).

matheagle · #3 $Old$ October 26th, 2009, 12:50 AM

This is a hypergeo.

$P(R=r,Y=y)={{8\choose r}{4\choose 2-r-y}{4\choose y}\over {16\choose 2}}$

$P(R=r)={{8\choose r}{8\choose 2-r}\over {16\choose 2}}$

$P(Y=y)={{12\choose 2-y}{4\choose y}\over {16\choose 2}}$

I doubt that they are independent. Just show $P(R=1,Y=1)\ne P(R=1)P(Y=1)$
I picked (1,1), try any spot you wish and hope they are not equal.

ANDS! · #4 $Old$ October 26th, 2009, 10:23 AM

Isn't a hypergeometric distribution only used for when we can classify each trial as a success or a failure? Wouldn't the above denote three possible outcomes?

matheagle · #5 $Old$ October 26th, 2009, 12:04 PM

If you have a fixed chance of success and failure, that's p and q from a binomial OR multinomial.
This is just a multi-hypergeometric.

ANDS! · #6 $Old$ October 26th, 2009, 01:24 PM

Apologies, I was referring to success being we draw one item, and failure being we draw the other. The words I were using were not correct. I meant, isn't a hyper-geometric used when we have only two classes of objects.

In any case I understand where you're coming from with it being a multiple hypergeometric distribution. I assumed that based on the questions being asked, perhaps the instructor didn't expect their students to use that particular method.