Balls in Boxes, Probability Question

In summary, the probability that box #1 still has a black ball and b white balls after a ball is randomly chosen from box #1 and placed in box #2, and then a ball is randomly chosen from box #2 and placed in box #1, can be determined using conditional probabilities. The probability of choosing a black ball from box #1 is a/(a+b) and the probability of choosing a black ball from box #2 given a black ball was chosen from box #1 is (c+1)/(c+d+1). This can be used to calculate the probability of the correct allocation of balls in both boxes. Alternatively, one can use the hypergeometric distribution to get the same result.
  • #1
dogma
35
0
Hello one and all. I could use a little guidance here on a probability problem.

Box #1 contains a black balls and b white balls while box #2 contains c black balls and d white balls. A ball is chosen randomly from box #1 and placed in box #2. A ball is then randomly chosen from box #2 and placed in box #1. What is the probability that box #1 still has a black balls and b white balls?

Okay, from that I come up with the following:

Let Random Variable X1 = a black ball is transferred to box #2 from box #1
Let R.V. X2 = a white ball is transferred to box #2 from box #1

Let R.V. Y1 = a black ball is transferred to box #1 from box #2
Let R.V. Y2 = a white ball is transferred to box #1 from box #2

[tex]P(X_1) = \frac{a}{a+b}[/tex] and [tex]P(X_2) = \frac{b}{a+b}[/tex]

[tex]P(Y_1 \mid X_1) = \frac{c+1}{c+d+1}[/tex] and [tex]P(Y_1 \mid X_2) = \frac{c}{c+d+1}[/tex]

[tex]P(Y_2 \mid X_1) = \frac{d}{c+d+1}[/tex] and [tex]P(Y_2 \mid X_2) = \frac{d+1}{c+d+1}[/tex]

This is where I get stuck.

I know (for example) that I can define another R.V. to represent, say, a black ball was selected from box #2 (I'll call it R.V. A), and...

[tex]P(A) = P(X_1) \cdot P(Y_1 \mid X_1) + P(X_2) \cdot P(Y_1 \mid X_2)[/tex]

Assuming I'm somewhat on the right track and haven't screwed things up, how would I go about determining the probability that box #1 still has a black balls and b white balls?

Thanks in advance for your enlightenment (and do I need it).

dogma
 
Last edited:
Physics news on Phys.org
  • #2
You're on the right track by thinking conditionally.


Can I make some notational changes and abuses?

P(correct allocation of balls) = P(correct number given a white transferred or correct n umber given a black transferred)

mutuall exclusive

=P(correct number given white transferred) +P(correct number given black transferred)

=P(white put back given white taken) + P(balck put back given black taken)

= somethings you've worked out.
 
  • #3
First of all, thanks for your response. I greatly appreciate your help.

My mind is swimming…so hopefully I'm not going to make this worse. If I'm correctly utilizing the info you provided:

P(correct allocation of balls) = P(a black balls and b white balls), since box #1 started off with a black balls and b white balls [and box #2 started off with c black balls and d white balls].

P(white ball chosen from box #1) =[tex]\frac{b}{a+b}[/tex]

P(black ball chosen from box #1) =[tex]\frac{a}{a+b}[/tex]

and

P(white ball chosen from box #2 given a white ball chosen from box #1) =[tex]\frac{d+1}{c+d+1}[/tex]

P(black ball chosen from box #2 given a black ball chosen from box #1) =[tex]\frac{c+1}{c+d+1}[/tex]

and finally,the correct allocation of balls:

P(a black balls and b white balls) = [tex]\frac{b}{a+b} \cdot \frac{d+1}{c+d+1}+\frac{a}{a+b} \cdot \frac{c+1}{c+d+1}[/tex]

I'm still a little fuzzy about this…then again, using Playdoh is challenging for me.

Am I warmer, colder, or way out in left field?

Thanks again!

dogma
 
  • #4
yp, that seems to be about right (and correcting any errors i may have made).
 
  • #5
thank you for your guidance and wisdom.

best of wishes,

dogma
 
  • #6
Follow on question:

Could I have used a hypergeometric distribution to figure this out? I guess I would just have to figure out how to set it up that way.

dogma
 
  • #7
Hypergeometric

One can use the hypergeometic distribution to get to the same result. Using the conditional probability approach, the probabilities (for choosing the black or white ball from box#1 )and the conditional probabilities (of choosing the black or white balls from box#2) are the same as the one you got earlier.

Lets see for one case:
Box#1: choosing a black ball
Probability = (a choose 1) * (b choose 0) / (a+b choose 1)
= a/(a+b)
(a choose 1 : ways of choosing 1 ball from a black balls)
Similarly the probability of choosing a black ball from box# given black ball chosen from box#1 = (c+1)/ (c+d+1)

This is exactly the first term in your equation.

I would appreciate if someone could let me know how to handle mathematical notation in this text editor.
 
  • #8
Lele said:
I would appreciate if someone could let me know how to handle mathematical notation in this text editor.
Lele,

Check out this link on LaTex (it's in the General Physics forum): https://www.physicsforums.com/showthread.php?t=8997

It is a thread containing info about LaTex typesetting in a message. It's pretty easy to do.

Thanks and good luck.

dogma
 

1. What is the "Balls in Boxes" problem?

The "Balls in Boxes" problem is a common probability question in which a certain number of balls are randomly distributed into a certain number of boxes. The problem typically asks for the probability of a specific outcome, such as a certain number of balls being in a certain box.

2. How do you approach solving a "Balls in Boxes" problem?

To solve a "Balls in Boxes" problem, you first need to determine the total number of possible outcomes. This can be done by calculating the total number of ways the balls can be distributed into the boxes. Then, you need to determine the number of favorable outcomes, or the number of ways the desired outcome can occur. Finally, you can calculate the probability by dividing the number of favorable outcomes by the total number of possible outcomes.

3. What is the formula for calculating the probability in a "Balls in Boxes" problem?

The formula for calculating the probability in a "Balls in Boxes" problem is:

P(desired outcome) = number of favorable outcomes / total number of possible outcomes

4. Can the "Balls in Boxes" problem be solved using different types of distributions?

Yes, the "Balls in Boxes" problem can be solved using different types of distributions, such as the binomial distribution or the hypergeometric distribution. The distribution used will depend on the specific details of the problem, such as if the balls are replaced after each selection or if they are drawn without replacement.

5. Are there any real-world applications of the "Balls in Boxes" problem?

Yes, the "Balls in Boxes" problem has many real-world applications, such as in genetics, where it can be used to calculate the probability of inheriting certain traits based on the distribution of genes from parents. It is also commonly used in statistics and probability theory to model various scenarios, such as stock market fluctuations or disease outbreaks.

Similar threads

  • Set Theory, Logic, Probability, Statistics
Replies
3
Views
1K
  • Set Theory, Logic, Probability, Statistics
Replies
6
Views
747
  • Set Theory, Logic, Probability, Statistics
Replies
5
Views
473
  • Set Theory, Logic, Probability, Statistics
Replies
10
Views
2K
  • Set Theory, Logic, Probability, Statistics
2
Replies
36
Views
3K
  • Set Theory, Logic, Probability, Statistics
Replies
7
Views
1K
  • Precalculus Mathematics Homework Help
Replies
10
Views
968
  • Set Theory, Logic, Probability, Statistics
Replies
2
Views
1K
  • Set Theory, Logic, Probability, Statistics
Replies
10
Views
807
  • Set Theory, Logic, Probability, Statistics
Replies
8
Views
1K
Back
Top