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Conditional Probability Problem

  • Thread starter Frank69
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  • #1
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Homework Statement



4 balls are placed successively into 4 cells, all 4^4 arrangements being equally probable. Given that the first 2 balls are in different cells (event B), what is the probability that one cell contains exactly 3 balls (event A)?

Homework Equations



P(A|B) = P(A intersection B) / P(B), given P(B) != 0 (Conditional Probability Thereom)

The Attempt at a Solution



My reasoning is: The first ball will be placed for sure in an empty cell, the second ball has a probability of 3/4 to be placed in an empty cell (because there are only 3 cells empty), so P(B) = 3/4. Since we know B has occurred, the 3rd ball has 2/4 probability to be placed in one of the cells having already 1 ball, and the 4th ball has 1/4 probability to hit the cell with 2 balls. So we have P(A|B) = 1/8.

I don't really know if it is correct or not, the problem is that I m supposed to solve it using Conditional Probability. How can I apply it to solve this problem?
 

Answers and Replies

  • #2
118
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You have the right answer, but I think the way they want you to do it is to actually compute P(A and B) and P(B) and divide to get the answer. You've found P(B). To find P(A and B) you need to find the number of possibilities of having the first two balls be in different cells and having one cell contain three balls, and divide that by the number of ways 4 balls can go into 4 cells, which is 4^4. Then divide P(A and B) by P(B) to get P(A|B).
 
  • #3
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I have problems computing this: P(A and B)

how can I find the intersection of the events A and B in this case? we have studied this but for simpler event spaces (like a die toss etc). In this case it seems rather complicated. For example we could have B = { 1111, 1210, 1300, 2110, ... } (considering 1111 as cell1 contains 1 ball, cell2 contains 1 ball and so on) and A = { 3100, 3010, 0310, 0031, ...} (i.e. all the sample events having 3 balls in one cell). Wouldn't A intersection of B be just the sample events of A? I really do not know how to simplify the calculation, any idea?
 
  • #4
118
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Well, the easiest way to find out the number of ways A and B can happen is to label the balls ball 1, ball 2, ball 3, and ball 4, where the first two balls are balls 1 and 2. One possibility for both A and B to happen is if cell 1 contains balls 1,3,and 4, and cell 2 contains ball 2. Another possibility is if cell 1 contains balls 1, 3, and 4, and cell 3 contains ball 2. Basically you want to figure out the total number of possibilities, and divide that by 4^4. That will give you P(A and B).

You might want to ask your teacher if your earlier answer is good enough. The problem isn't that hard when you think about it intuitively, and it's more difficult when you actually need to compute P(A and B). If your teacher actually wants you to use the theorem you gave, then basically you have to compute P(A and B).
 
  • #5
27
0

Homework Statement



4 balls are placed successively into 4 cells, all 4^4 arrangements being equally probable. Given that the first 2 balls are in different cells (event B), what is the probability that one cell contains exactly 3 balls (event A)?

Homework Equations



P(A|B) = P(A intersection B) / P(B), given P(B) != 0 (Conditional Probability Thereom)

The Attempt at a Solution



My reasoning is: The first ball will be placed for sure in an empty cell, the second ball has a probability of 3/4 to be placed in an empty cell (because there are only 3 cells empty), so P(B) = 3/4. Since we know B has occurred, the 3rd ball has 2/4 probability to be placed in one of the cells having already 1 ball, and the 4th ball has 1/4 probability to hit the cell with 2 balls. So we have P(A|B) = 1/8.

I don't really know if it is correct or not, the problem is that I m supposed to solve it using Conditional Probability. How can I apply it to solve this problem?
I think you're right on Pr(B)=3/4. Now to find Pr(A and B), consider that "A and B" means "3 balls in one cell and 1 ball in another cell". This has to be 3/64. It would be 1/(4^3)=1/64 if the balls were labeled as to their order, but apparently they are not, so Pr(A and B)=3/64.

At least that's my take on the problem, for what it's worth.
 
  • #6
118
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I think you're right on Pr(B)=3/4. Now to find Pr(A and B), consider that "A and B" means "3 balls in one cell and 1 ball in another cell". This has to be 3/64. It would be 1/(4^3)=1/64 if the balls were labeled as to their order, but apparently they are not, so Pr(A and B)=3/64.

At least that's my take on the problem, for what it's worth.
Wait, I don't think that's correct. I don't think the event of A and B happening is the same as the event of 3 balls in one cell and one ball in the other. And I don't understand how you got 3/64. I found that Pr(A and B)=6/64, after reducing.
 
  • #7
8
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Trying to find A and B I got the following:

A and B = {
(134)(2)()(), (134)()(2)(), (134)()()(2),
(2)(134)()(), ()(134)(2)(), ()(134)()(2),
(2)()(134)(), ()(2)(134)(), ()()(134)(2),
(2)()()(134), ()(2)()(134), ()()(2)(134)
}

Where () is a cell and 1,2,3,4 are the balls labeled according to the order they have been placed.

Since we have 12 elements in the event (A and B) and the total number of arrangements is 4^4 = 256, we have:

P(A and B) = 12/256 = 3/64

We know that P(B) = 3/4 (are we sure about this?)

Therefore P(A|B) = (3/64) / (3/4) = 1/16

Is this correct?

If yes, this is somehow different from the first answer which I computed without using the Conditional Probability Theorem.
Which is the correct answer then, 1/16 or 1/8? (I guess there might be errors in the first reasoning, but I'm not sure).
 
  • #8
118
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What about (1)(234)(0)(0)? In this event, balls 1 and 2 are in different cells (event B) and one cell contains exactly three balls (event A).
 
  • #9
8
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Oh yes, you are rite. Then there are 24 elements in (A and B), giving P(A and B) = 24/256

Therefore P(A|B) = P(A and B) / P(B) = (24/256) / (3/4) = 1/8

Thank you for the help guys
 

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