Colored balls

M

MasonRay

New member
Joined
Feb 2, 2021
Messages
1
I have 10 red balls 10 green balls and 10 blue balls
I mix them all together in a big bin. I then have 4 empty bins.
I take 2 balls from the mixed bin and put them into each of the 4 empty bins(8 balls total) What are my chances of at least 2 bins having 2 red balls? Please help.
 
What have you tried, and where are you stuck?

Hint: It would be the same probability if you had 10 red balls and 20 non-red balls
 
You ask for help but seem to be expecting a solution to the problem.
Do you really not realize that getting help and getting the problem solved for you are not the same thing? In fact, in my opinion, they are opposite of one another. In one case you learn something and in the other case you learn absolutely nothing.

Now if you really want help from the forum and I truly hope that you do, then you need to read the posting guidelines and reply back with a post that follows those guidelines.
 
Ah, I see what the real wording of this problem is. Interesting problem!
 
You pick your 8 balls from the first bin and you'll have some collection 8 colored balls.
From that the probability of at least 2 bins having 2 red balls is a relatively straightforward multivariate hypergeometric distribution problem

I would look at it like this

[MATH]\sum \limits_{k=4}^8~P[\text{at least 2 bins having 2 red balls | $k$ of the 8 balls chosen are red}]P[\text{$k$ of the 8 balls are red}][/MATH]
All of those probabilities shouldn't be too difficult to compute. Give it a go.
 
Romsek, the big bin will certainly have 2 red balls after removing 8 balls. So I feel that the problem reduces to the follow.

In a large bucket you have 10 red balls and 20 non-red balls. You take out 8 balls from the large bucket and place two balls into each of 4 bins. What is the probability that at least one bin has 2 red balls?
 
Jomo said:
Romsek, the big bin will certainly have 2 red balls after removing 8 balls. So I feel that the problem reduces to the follow.

In a large bucket you have 10 red balls and 20 non-red balls. You take out 8 balls from the large bucket and place two balls into each of 4 bins. What is the probability that at least one bin has 2 red balls?
Click to expand...

I think you need to look at this again. We can certainly pull 8 balls out of the big bin with none of them being red.

We don't care about the contents of the big bin after the 8 balls have been selected.

Further the problem is asking what is the probability at least 2 of the bins have 2 red balls in them. I don't understand why you've reduced that to 1.
 
Romsek said:
I think you need to look at this again. We can certainly pull 8 balls out of the big bin with none of them being red.

We don't care about the contents of the big bin after the 8 balls have been selected. Where does it say that??

Further the problem is asking what is the probability at least 2 of the bins have 2 red balls in them. I don't understand why you've reduced that to 1. I reduced to one bin since we are guaranteed that the large bin has 2 red balls in it. After removing this large bin from the problem I can now say that the problem is asking for at least one bin having 2 red balls.
Click to expand...
See red comments above.
 
I picture 5 bins with balls in them. Then all the balls are put into one bin and mixed. Then 8 balls are taken out of the bin (the one of course that has the balls) and 2 balls each are put into the 4 empty bins. I honestly see 5 bins at this point. I understand how you see 4 but the problem does not say to ignore the bin that the balls were taken out of.

In the end I think I am correct. However, if I learned that the author of the problem meant the problem to be the way you interrupted it, then I would not be the least surprised at that.
 
Jomo said:
I picture 5 bins with balls in them. Then all the balls are put into one bin and mixed. Then 8 balls are taken out of the bin (the one of course that has the balls) and 2 balls each are put into the 4 empty bins. I honestly see 5 bins at this point. I understand how you see 4 but the problem does not say to ignore the bin that the balls were taken out of.

In the end I think I am correct. However, if I learned that the author of the problem meant the problem to be the way you interrupted it, then I would not be the least surprised at that.
Click to expand...
interrupted ?!!!

I did not see any interruption - only interpretation...
 
one last thing.

You seem to think that the problem includes the original big bin as one of the bins to count red balls in.

Well why on earth would the problem designer be so stupid as to include that when as you say it clearly will have at least two red balls in it.

You look at problems so oddly, that's why we never see them the same. It's OBVIOUS that the problem is only referring to the second set of
4 bins when asking about having 2 balls in them.

Adios.
 
You must log in or register to reply here.
Top