
#1
Mar612, 09:12 PM

P: 5

A box contains 12 marbles. Six of the marbles are red, four are green, and two are yellow. Suppose that you choose three marbles at random. let x be the number of red marbles in the sample, y be the number of green marbles, and z be the number of yellow marbles.
a. Give the precise formulas for the probability mass functions of the three random variables, f(x), f(y), and f(z). b. Suppose you win $1 for each green marble in your sample and lose $1 for each red marble (yellow has no financial consequences). Let w be your net winnings or losses. Construct the P.M.F of w as a table. (obviously the table part cannot be shown in this answer) c. Find the expected value of w and explain its significance. d. Find the standard deviation of w. Any help is greatly appreciated! If anyone can point me in the right direction, that'll be great! 



#2
Mar612, 09:38 PM

P: 40

Look up the hypergeometric distribution.




#3
Mar612, 09:43 PM

P: 5

Yes, thank you. I put it into the formula and it gave me the correct answer :)




#4
Mar612, 10:04 PM

HW Helper
Thanks
P: 4,677

Probability Mass Function(Hint: you need a covariance between two random variables, so you need a bivariate distribution.) RGV 



#5
Mar612, 11:11 PM

P: 5

This is what I got for part (a)
F(x)= 3Ck * 0.5^3 F(y)=3Ck * 0.667^3 F(z)=3Ck * 0.167^3 Now for part (b) I have some problems, I don't know if I got all the possible outcomes RRR RRG RGR GRR GGG GGR GRG RGG RGY RYG GYR GRY YRG YGR YYR YYG YRY RYY YGY GYY YYY I got the probability of each and how much cash you would have. 



#6
Mar612, 11:29 PM

HW Helper
Thanks
P: 4,677

Let's look at choosing two reds in 3 draws (starting with a box of 6 Red, 4 green, 2 yellow) Suppose we label the outcomes as R (red) and N (not red). What is the probability of the specific outcome RRN? Look at it in pieces: what is the probability of R1: the first draw is red? There are 6 reds in 12 balls, so P{R1} = 6/12. Now you have 11 balls left and 5 are red, so what now is the probability of the second draw = red (R2)? Remember, we have already observed R1, so we are really asking for P{R2R1} = 5/11. Now we have 10 balls left, of which 4 are red and 6 are nonred. So, P{N3R1,R2} = 6/10. Altogether, P{R1 R2 N3} = P{RRN} = (6/12)(5/11)(6/10). You can work out similar probabilities for RNR and NRR, and add them up. That will give you P{2 red}. Once you have done this a few times you will realize there are slick shortcut formulas allowing you to write down the answer quickly, but first you need to realize what is involved. RGV 


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