4 balls and 4 cells probabillity

• diracdelta
In summary: As an example, it's like you can count the number of ways a deck of cards can be shuffled and so you can count the number of ways a hand in poker can be made. But you soon realize that there are patterns and this is different than that. So you learn the rules of probability, you learn how to apply them and you learn how to apply them to problems like this one. This is a very basic problem and I would say that the counting method is over-complicating it. Just like counting the number of ways a deck of cards can be shuffled is over-complicating the problem of counting the number of ways a hand in poker can be
diracdelta

Homework Statement

Four balls are distributed randomly and independently into 4 cells.
What is probability that:
a) In every cell there is exactly one ball?
b) All four balls be in one cell?
c)Two balls in one cell ( any) , other two balls in other cell (any)?
(assume that the balls are different one from each other)

The Attempt at a Solution

[/B]
For a)
P=4! / (4^4)
4! because there is 4! ways that we could distribute them, and 4^4 is all possible ways to distribute balls.
b)
$$\frac{\binom{4}{1} \cdot 4!}{4^{4}}$$
Out of four cells i choose one where i will put them in, and since balls are not the same 4! to permutate them all

c)
$$\frac{\binom{4}{1} \cdot \frac{4!}{2!}\cdot \binom{3}{1}\frac{2!}{2!}}{4^{4}}$$
Out of four cells choos one where i will put two balls and out of three cells choose one where i will put two balls.

Is this ok?

diracdelta said:
Out of four cells i choose one where i will put them in, and since balls are not the same 4! to permutate them all

No, this is wrong. You have 4 possibilities to chose the cell. The order in which the balls enter is irrelevant (there is only one order - all balls must go into the same cell).

diracdelta said:
Out of four cells choos one where i will put two balls and out of three cells choose one where i will put two balls.

No, you need to chose two cells in which to put two balls. Then you need to consider in how many different ways you can put 4 different balls into two different cells such that each cell has two balls (i.e., you need to pick which two balls goes into the first cell and which two goes into the second).

diracdelta
diracdelta said:

Homework Statement

Four balls are distributed randomly and independently into 4 cells.
What is probability that:
a) In every cell there is exactly one ball?
b) All four balls be in one cell?
c)Two balls in one cell ( any) , other two balls in other cell (any)?
(assume that the balls are different one from each other)

The Attempt at a Solution

[/B]
For a)
P=4! / (4^4)
4! because there is 4! ways that we could distribute them, and 4^4 is all possible ways to distribute balls.
b)
$$\frac{\binom{4}{1} \cdot 4!}{4^{4}}$$
Out of four cells i choose one where i will put them in, and since balls are not the same 4! to permutate them all

c)
$$\frac{\binom{4}{1} \cdot \frac{4!}{2!}\cdot \binom{3}{1}\frac{2!}{2!}}{4^{4}}$$
Out of four cells choos one where i will put two balls and out of three cells choose one where i will put two balls.

Is this ok?

Using the counting method doesn't seem like the best way to tackle this problem. Can you think of a more direct approach?

diracdelta
diracdelta said:

b)
$$\frac{\binom{4}{1} \cdot 4!}{4^{4}}$$
Out of four cells i choose one where i will put them in, and since balls are not the same 4! to permutate them all

By this logic, there are ##4! * 4^4## ways of distributing the balls. I can put any ball anywhere but I can also permute the order that I pick them.

You should assume that the balls are numbered 1-4 and you pick them in number order. Why, because it's simpler and that's usually what these questions want.

diracdelta
Orodruin said:
No, this is wrong. You have 4 possibilities to chose the cell. The order in which the balls enter is irrelevant (there is only one order - all balls must go into the same cell).No, you need to chose two cells in which to put two balls. Then you need to consider in how many different ways you can put 4 different balls into two different cells such that each cell has two balls (i.e., you need to pick which two balls goes into the first cell and which two goes into the second).

Alright. So I only choose one out of 4 cells, which yields p= 4/256

As far as this, 4 choose 2 (cells) and (4 choose 2) for balls?

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diracdelta said:
Alright. So I only choose one out of 4 cells, which yields p= 4/256

As far as this, 4 choose 2 (cells) and (4 choose 2) for balls?
Yes.

diracdelta
diracdelta said:
As far as this, 4 choose 2 (cells) and (4 choose 2) for balls?

You need to be super careful to handle the permutations correctly. This answer is correct though (as the number of ways).

haruspex said:
Yes.
Ok. Thanks for help!
verty said:
You need to be super careful to handle the permutations correctly. This answer is correct though (as the number of ways).
I noticed.
By the way, could someone help me with PeroK's suggestion?
I have no clue what other acces should I approach this problem with. Only thing that pops into my mind is number of number of ways "A" occurs / number of all possible ways.

The more direct way is to learn probability and the rules of probability and then to apply them to the problem. If you know the rules, you just apply them and it's more direct. For example, you might have heard about inclusion-exclusion, that is one of the rules. I think there are about 5 rules, you learn them, you apply them. Any good probability book will help you to do that (they usually cover them in the early chapters).

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verty said:
The more direct way is to learn probability and the rules of probability and then to apply them to the problem. If you know the rules, you just apply them and it's more direct. For example, you might have heard about inclusion-exclusion, that is one of the rules. I think there are about 5 rules, you learn them, you apply them. Any good probability book will help you to do that (they usually cover them in the early chapters).
I'm intrigued. I cannot think of a more direct way (only equally direct ways) to do (c) than that in the OP. Would you be so kind as to elaborate?

haruspex said:
I'm intrigued. I cannot think of a more direct way (only equally direct ways) to do (c) than that in the OP. Would you be so kind as to elaborate?

This is as much as I'll say on this topic.

The balls being distributed are independent trials. We can think of it as a 4-sided die being thrown 4 times. Call a throw acceptable if the desired outcome is still possible. Let ##P(i)## be the probability that throw ##i## was acceptable. We want to find ##P(\cap_i)##.

Now if you know rules like the multiplication rule and complimentary events, you will be able to find a formula for the answer.

verty said:
This is as much as I'll say on this topic.

The balls being distributed are independent trials. We can think of it as a 4-sided die being thrown 4 times. Call a throw acceptable if the desired outcome is still possible. Let ##P(i)## be the probability that throw ##i## was acceptable. We want to find ##P(\cap_i)##.

Now if you know rules like the multiplication rule and complimentary events, you will be able to find a formula for the answer.
That's more direct!?

haruspex said:
That's more direct!?

Perhaps not in this problem but I'll give another problem here and perhaps it'll be more obvious that it is more direct (as a method to find the answer).

A town council of 7 members contains a steering committee of size 3. New ideas for legislation go first to the steering committee and then on to the council as a whole if at least 2 of the 3 committee members approve the legislation. Once at the full council, the legislation requires a majority vote (of at least 4) to pass.

Consider a new piece of legislation, and suppose that each town council member will approve it, independently, with probability p. What is the probability that a given steering committee member's vote is decisive in the sense that if that person's vote were reversed, then the final fate of the legislation would be reversed? What is the corresponding probability for a given council member not on the steering committee?

This is from the book by Sheldon Ross, it's a very good book actually. Most methods fail here but my approach works.

verty said:
Perhaps not in this problem but I'll give another problem here and perhaps it'll be more obvious that it is more direct (as a method to find the answer).
This is from the book by Sheldon Ross, it's a very good book actually. Most methods fail here but my approach works.
The key difference with this 'council' question is that the process is inherently serialised, so requires the serial analysis you propose.

haruspex said:
The key difference with this 'council' question is that the process is inherently serialised, so requires the serial analysis you propose.

Hmm, that is a good point. And I'm starting to see that the counting method can also work here, by which I mean, thinking of it as a set and counting the elements. And that is much easier for 99% of problems than what was my approach. Like who is ever going to want to know how decisive a vote is? It's a silly question.

I think I was too critical of the other methods. There is definitely merit in them. My approach of course works but it is complicated and why shouldn't easy questions have easy solutions?

1. What is the probability of getting exactly 2 balls in each of the 4 cells?

The probability of getting exactly 2 balls in each of the 4 cells is 1/6, or about 16.67%. This can be calculated by using the formula for combinations: C(n,r) = n! / (r!(n-r)!), where n is the total number of objects (4 balls and 4 cells in this case) and r is the number of objects we are choosing (2 balls for each cell). So, C(8,2) = 8! / (2!(8-2)!) = 8! / (2!6!) = 8*7/2 = 28/2 = 14. Then, we divide this by the total number of possible outcomes, which is 4^4 (since there are 4 choices for each ball and 4 balls), giving us a probability of 14/256 = 1/6.

2. What is the probability of getting all 4 balls in one cell?

The probability of getting all 4 balls in one cell is 1/64, or about 1.56%. This can be calculated by using the formula for combinations: C(n,r) = n! / (r!(n-r)!), where n is the total number of objects (4 balls and 4 cells in this case) and r is the number of objects we are choosing (4 balls for one cell). So, C(8,4) = 8! / (4!(8-4)!) = 8! / (4!4!) = 8*7*6*5/(4*3*2*1) = 70. Then, we divide this by the total number of possible outcomes, which is 4^4 (since there are 4 choices for each ball and 4 balls), giving us a probability of 70/256 = 1/64.

3. What is the probability of getting at least 1 ball in each cell?

The probability of getting at least 1 ball in each cell is 1/1, or 100%. This is because it is impossible to not have at least 1 ball in each cell with 4 balls and 4 cells. The only way this could happen is if all 4 balls landed in the same cell, which would be counted in the previous question. Therefore, the probability is 1, or 100%.

4. What is the probability of getting exactly 3 balls in one cell and 1 ball in another cell?

The probability of getting exactly 3 balls in one cell and 1 ball in another cell is 4/64, or about 6.25%. This can be calculated by using the formula for combinations: C(n,r) = n! / (r!(n-r)!), where n is the total number of objects (4 balls and 4 cells in this case) and r is the number of objects we are choosing (3 balls for one cell and 1 ball for another cell). So, C(8,3) * C(5,1) = (8! / (3!(8-3)!) * (5! / (1!(5-1)!)) = (8*7*6 / (3*2)) * (5 / 1) = 56 * 5 = 280. Then, we divide this by the total number of possible outcomes, which is 4^4 (since there are 4 choices for each ball and 4 balls), giving us a probability of 280/256 = 4/64.

5. What is the probability of getting all 4 balls in different cells?

The probability of getting all 4 balls in different cells is 4/256, or about 1.56%. This can be calculated by using the formula for permutations: P(n,r) = n! / (n-r)!, where n is the total number of objects (4 balls and 4 cells in this case) and r is the number of objects we are choosing (4 balls in different cells). So, P(8,4) = 8! / (8-4)! = 8! / 4! = 8*7*6*5 = 1680. Then, we divide this by the total number of possible outcomes, which is 4^4 (since there are 4 choices for each ball and 4 balls), giving us a probability of 1680/256 = 4/256.

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