Register to reply 
Probability Mass Function 
Share this thread: 
#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

Sci Advisor
HW Helper
Thanks
P: 5,086

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

Sci Advisor
HW Helper
Thanks
P: 5,086

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 


Register to reply 
Related Discussions  
Probability Mass Function vs Probability Measure  Set Theory, Logic, Probability, Statistics  2  
Probability Mass Function/Moment Generating Function  Precalculus Mathematics Homework  3  
Probability (probability mass function,pmf)  Calculus & Beyond Homework  5  
Probability mass function  Calculus & Beyond Homework  8  
Probability mass function  Calculus & Beyond Homework  3 