Welcome to our community

Be a part of something great, join today!

Probability of a particular item not being assigned

ATroelstein

New member
Jun 30, 2012
15
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.
 

Opalg

MHB Oldtimer
Staff member
Feb 7, 2012
2,702
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.