What Are the Probability Mass Functions for Randomly Selected Marbles?

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Homework Help Overview

The discussion revolves around a probability problem involving a box of marbles, specifically focusing on the probability mass functions (PMFs) for the number of red, green, and yellow marbles drawn from the box. The problem also explores the financial implications of drawing these marbles based on their colors.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the application of the hypergeometric distribution to derive the PMFs for the random variables representing the colors of the marbles. There are attempts to formulate the PMFs and to identify all possible outcomes for the net winnings based on the drawn marbles. Some participants question the correctness of the initial formulas provided for the PMFs.

Discussion Status

Some participants have provided guidance on using the hypergeometric distribution, while others are working through the implications of their findings. There is an ongoing exploration of the relationships between the random variables involved, particularly in relation to calculating expected values and standard deviations.

Contextual Notes

Participants are navigating the complexities of calculating probabilities and outcomes based on a finite sample of marbles, with some expressing uncertainty about the completeness of their outcome lists and the correctness of their formulas.

Jimerd
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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!
 
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Look up the hypergeometric distribution.
 
Yes, thank you. I put it into the formula and it gave me the correct answer :)
 
Jimerd said:
Yes, thank you. I put it into the formula and it gave me the correct answer :)

More to the point: do you know *why* the appropriate distribution is the hypergeometric? BTW: can you see why using the simple hypergeometric alone will not help you do part (d)?
(Hint: you need a covariance between two random variables, so you need a bivariate distribution.)

RGV
 
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.
 
Jimerd said:
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.

Your formulas for F(x), F(y) and F(z) are all incorrect. You have NOT used the hypergoemetric distributions here.

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{R2|R1} = 5/11. Now we have 10 balls left, of which 4 are red and 6 are non-red. So, P{N3|R1,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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