Combinatorics problem

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aybala0

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I'm trying to calculate a possibility but I don't know how to...

There are 200 balls/cards whatever you call them
We mark 20 of these, let's say there are 20 cards in a bag and the rest of the cards are black (180 black cards)
These 200 cards are going to be distributed randomly to 10 different bags, each bag will contain 20 random cards
how can I calculate the possibility of 2 red cards getting into the same bag?
My friends said I have to calculate the possibilities this not happening and subtract them from 1 but I'm not sure how
 
aybala0 said:
I'm trying to calculate a possibility but I don't know how to...
There are 200 balls/cards whatever you call them
We mark 20 of these, let's say there are 20 cards in a bag and the rest of the cards are black (180 black cards)
These 200 cards are going to be distributed randomly to 10 different bags, each bag will contain 20 random cards
how can I calculate the possibility of 2 red cards getting into the same bag?
My friends said I have to calculate the possibilities this not happening and subtract them from 1 but I'm not sure how
Click to expand...
The quoted question contains many miss-statements and contradictions.
Please review and re-post.
 
pka said:
The quoted question contains many miss-statements and contradictions.
Please review and re-post.
Click to expand...
I asked a question how can it have MANY miss-statements and contradictions. Are you sure you speak english?
 
There are 200 balls/cards whatever you call them."

It is quite unhelpful to leave the objects with multiplicity of definition or a measure of ambiguity. Pick one. "objects"?

"We mark 20 of these, "

This appears to have no relationship to the rest of the oddly-constructed sentence.

"let's say there are 20 cards in a bag"

Which 20? Are these the marked ones? Typically, cards and bags don't mix well. Maybe a hat or a deck or a box? You can have a bag if you like; it's just a bit unusual.

"and the rest of the cards are black (180 black cards)"

Did you also put these in the bag?

"These 200 cards are going to be distributed randomly to 10 different bags,"

How? Will you first distribute the 20 marked cards from the bag. This is an easy problem. 100% probability by the pigeon hole principle.
Will you also be distributing the 180 cards not in the bag at the same time? How will you decide bag or no bag on each selection?

"each bag will contain 20 random cards"

There is no such thing as a "random card". Each card you put in a distribution bag will be a real, identifiable card.

"how can I calculate the possibility of 2 red cards getting into the same bag? My friends said I have to calculate the possibilities this not happening and subtract them from 1 but I'm not sure how "

Your friends are functioning on limited knowledge. Perhaps they saw this method once and now they believe it is the only way to proceed. This is most unlikely. Please refer to my signature for additional insight on this matter.

These are also troubling sentences due to vocabulary. "possibility" and "probability" don't mean the same thing. Is there a possibility that 2 red will be in the same bag? Sure. 100% possible. The antithesis of possibility is impossibility. Typically, "possibility" refers to whether or not something can happen at all, where "probability" refers to how likely something that can happen will be realized.

"I asked a question how can it have MANY miss-statements and contradictions."

See above.

"Are you sure you speak english?"

Perhaps you should focus on answering legitimate questions, rather than responding with invective alone.

Also, the word "English" should be capitalized.

We volunteers are delighted to help if we can understand the question and if we are provided with a sample of your personal efforts.

Please review and re-post.
 
aybala0 said:
I'm trying to calculate a possibility but I don't know how to...

There are 200 balls/cards whatever you call them
We mark 20 of these, let's say there are 20 cards in a bag and the rest of the cards are black (180 black cards)
These 200 cards are going to be distributed randomly to 10 different bags, each bag will contain 20 random cards
how can I calculate the possibility of 2 red cards getting into the same bag?
My friends said I have to calculate the possibilities this not happening and subtract them from 1 but I'm not sure how
Click to expand...
Let's be a little generous and translate this into consistent English so we can get to the math. Here is the question, corrected:

I'm trying to calculate a probability but I don't know how to...​
​
There are 200 balls.​
Let's say there are 20 red balls and the rest of the balls are black (180 black balls).​
These 200 balls are going to be distributed randomly to 10 different bags, so that each bag will contain 20 balls.​
How can I calculate the probability of 2 red balls being in the same bag?​
My friends said I have to calculate the probability of this not happening and subtract from 1 but I'm not sure how.​

Now the problem makes sense. The friends' hint also makes sense, though of course it is not the only way.

So, what is the probability that no two red balls are in the same bag? That is, that there is one red ball in each bag.

One approach is to count all the ways to distribute 200 distinct balls into 10 sets of 20, and then the ways to distribute them with 1 red ball in each bag.

Please show what you've tried.
 
It seems to me that if there are 10 bags, and 20 red balls, then it isn't possible to have ALL bags with only one red ball (if each bag has one red ball then there would be 10 red balls left over that remain to be placed into bags). So there is 100% chance of one of the bags ending up with two (or more) red balls.

OP, could your question actually be:- what is the probability that at least one bag contains exactly two red balls ?
Or maybe the question should actually be 20 bags with 10 balls each?

Please let us know.
 
Yes, I misread the numbers even as I was trying (too hard?) to make it make sense.

This may be a trick question that's really about pigeonholes rather than probability/combinatorics. And not even that.
 
okay I'm rewriting the question, I thought it was understandable, sorry
I just want to understand the question and the logic
there are 200 balls in total.
20 of these balls are red
180 of them are black.
So they make up a total of 180+20= 200 balls, they are all in a BIG bag that can contain all 200.
This should be okay.
Now, I take 10 other smaller bags. These bags can contain 20 balls maximum.
Remember the big bag that had all the balls, 180 of them black and the rest 20 are red?
Now, I'm taking the big bag and randomly distributing every one of the 200 balls to these smaller bags (10 smaller bags)
After the distribution, I no longer have a big bag that has 200 balls, I now have 10 smaller bags and each contain 20 balls
For example, the first bag may have 18 black balls and 2 red balls, that makes up 20 in total.
In the final situation, we have 10 small bags and each contain some black and some red balls

The question is: What is the probability of at least 2 red balls being in the same bag?
so there can also be 3 or 4 in a bag that we will count as a possibility.

If I still can't be clear please reply
 
OK, then it is an absolute certainty that at least one of the small bags will contain 2 or more red balls.

An equivalent way of thinking about this problem is to ignore the black balls, and simply place the 20 red balls randomly into the 10 bags. Can this be done so that each of the 10 bags has fewer than two balls?
 
aybala0 said:
okay I'm rewriting the question, I thought it was understandable, sorry
I just want to understand the question and the logic
there are 200 balls in total.
20 of these balls are red
180 of them are black.
So they make up a total of 180+20= 200 balls, they are all in a BIG bag that can contain all 200.
This should be okay.
Now, I take 10 other smaller bags. These bags can contain 20 balls maximum.
Remember the big bag that had all the balls, 180 of them black and the rest 20 are red?
Now, I'm taking the big bag and randomly distributing every one of the 200 balls to these smaller bags (10 smaller bags)
After the distribution, I no longer have a big bag that has 200 balls, I now have 10 smaller bags and each contain 20 balls
For example, the first bag may have 18 black balls and 2 red balls, that makes up 20 in total.
In the final situation, we have 10 small bags and each contain some black and some red balls

The question is: What is the probability of at least 2 red balls being in the same bag?
so there can also be 3 or 4 in a bag that we will count as a possibility.
Click to expand...
As written two of the smaller bags could contain all twenty red balls. You do write that "each contain some black and some red balls" but do not state that is an requirement. Is it? Must each small bag contain at least one red and one black?
 
Excuse me but you, not pka, has the English problem!

There are 200 balls/cards whatever you call them OK, no problem here.

We mark 20 of these, let's say there are 20 cards in a bag and the rest of the cards are black (180 black cards). So you marked 20. There are 20 in a bag (the 20 marked ones? You did not say!). The remains 180 are black. You never said what the color of the 20 marked cards are!

These 200 cards are going to be distributed randomly to 10 different bags, each bag will contain 20 random cards 200 cards are going to be put in 10 different bags with 20 each. Does this include the bag with the 20 marked ones. Did you put the 20 marked ones in a single bag?
how can I calculate the possibility of 2 red cards getting into the same bag? You never mentioned anything about red cards? Did you notice that? How many are red?

What does the marked cards have to do with this problem?

In my opinion, absolutely nothing you wrote made any sense. Next time you want to attack a helper on this site I suggest that you keep quiet.
 
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