MHB Probability of a particular item not being assigned

  • Thread starter Thread starter ATroelstein
  • Start date Start date
  • Tags Tags
    Probability
AI Thread Summary
The discussion revolves around calculating the probability that none of the M cups receive any of the N green marbles when K total marbles are used. The initial calculations correctly identify the probabilities for each cup, leading to the conclusion that the probability for the Mth cup should be based on K - (M - 1) marbles remaining. However, the book presents a different formula, suggesting the last term should be based on K - M, which raises confusion. Participants in the discussion agree that the original reasoning appears accurate, indicating a potential error in the book's approach. The consensus is that the correct calculation should reflect the removal of marbles as cups are filled.
ATroelstein
Messages
15
Reaction score
0
Lets say I have M cups and K marbles. All these marbles are blue, except for N of them that are green. At random, I will select a marble and drop it in a cup. The marble will not be returned to the original set of marbles before the next marble is selected. I would like to know the probability that these M cups don't have any of the N green marbles after I drop a marble into each cup. For this, I know the probability that the first cup does get a green marble is $\frac{N}{K}$. Therefore the probability that the cup does not get a green marble is $1 - \frac{N}{K}$. The second cup now has a probability of $\frac{N}{K - 1}$ of getting a green marble, since we have one less marble to randomly select now, which gives if a probability of $1 - \frac{N}{K-1}$ of not having a green marble. Now if I put this all together, I would assume the probability that none of the M cups have a green marbles is the product of all these probabilities for each individual cup not having a green marble.

$(1 - \frac{N}{K}) * (1 - \frac{N}{K - 1}) * (1 - \frac{N}{K - 2}) * ... * (1 - \frac{N}{K - (M-1)})$

I have $K - (M-1)$ as the denominator in the last probability because for the Mth cup, $M - 1$ marbles have been removed from the original set of K marbles. My issue is, in the book I am looking at, it states the probability being

$(1 - \frac{N}{K}) * (1 - \frac{N}{K - 1}) * (1 - \frac{N}{K - 2}) * ... * (1 - \frac{N}{K - M})$

I'm confused as to why their last probability term is $(1 - \frac{N}{K - M})$ as to me it seems like this would be correct if we have $M + 1$ cups. Is there something that I'm missing about how this probability should be calculated? Thanks.
 
Mathematics news on Phys.org
ATroelstein said:
Lets say I have M cups and K marbles. All these marbles are blue, except for N of them that are green. At random, I will select a marble and drop it in a cup. The marble will not be returned to the original set of marbles before the next marble is selected. I would like to know the probability that these M cups don't have any of the N green marbles after I drop a marble into each cup. For this, I know the probability that the first cup does get a green marble is $\frac{N}{K}$. Therefore the probability that the cup does not get a green marble is $1 - \frac{N}{K}$. The second cup now has a probability of $\frac{N}{K - 1}$ of getting a green marble, since we have one less marble to randomly select now, which gives if a probability of $1 - \frac{N}{K-1}$ of not having a green marble. Now if I put this all together, I would assume the probability that none of the M cups have a green marbles is the product of all these probabilities for each individual cup not having a green marble.

$(1 - \frac{N}{K}) * (1 - \frac{N}{K - 1}) * (1 - \frac{N}{K - 2}) * ... * (1 - \frac{N}{K - (M-1)})$

I have $K - (M-1)$ as the denominator in the last probability because for the Mth cup, $M - 1$ marbles have been removed from the original set of K marbles. My issue is, in the book I am looking at, it states the probability being

$(1 - \frac{N}{K}) * (1 - \frac{N}{K - 1}) * (1 - \frac{N}{K - 2}) * ... * (1 - \frac{N}{K - M})$

I'm confused as to why their last probability term is $(1 - \frac{N}{K - M})$ as to me it seems like this would be correct if we have $M + 1$ cups. Is there something that I'm missing about how this probability should be calculated? Thanks.
I think it's the book that is wrong, not you.
 
Suppose ,instead of the usual x,y coordinate system with an I basis vector along the x -axis and a corresponding j basis vector along the y-axis we instead have a different pair of basis vectors ,call them e and f along their respective axes. I have seen that this is an important subject in maths My question is what physical applications does such a model apply to? I am asking here because I have devoted quite a lot of time in the past to understanding convectors and the dual...
Insights auto threads is broken atm, so I'm manually creating these for new Insight articles. In Dirac’s Principles of Quantum Mechanics published in 1930 he introduced a “convenient notation” he referred to as a “delta function” which he treated as a continuum analog to the discrete Kronecker delta. The Kronecker delta is simply the indexed components of the identity operator in matrix algebra Source: https://www.physicsforums.com/insights/what-exactly-is-diracs-delta-function/ by...
Back
Top