# Thread: Permutations of Combinations question? Help

1. ## Permutations of Combinations question? Help

Just wondering if anyone can point me in the right direction as to how to solve this?....

"In an experiment at Boston's Computer Museum, each of 10 judges communicated with four computers and four other people and was asked to distinguish between them.

1. Assume that the first judge cannot distinguish between the four computers and the four people. If this judge makes random guesses, what is the probability of correctly identifying the four computers and the four people?
2. Assume that all ten judges cannot distinguish between computer and people, so they make random guesses. Based of the previous result, what is the probability that all 10 judges make all correct guesses?

2. Originally Posted by Aquameatwad
Just wondering if anyone can point me in the right direction as to how to solve this?....

"In an experiment at Boston's Computer Museum, each of 10 judges communicated with four computers and four other people and was asked to distinguish between them.

1. Assume that the first judge cannot distinguish between the four computers and the four people. If this judge makes random guesses, what is the probability of correctly identifying the four computers and the four people?
2. Assume that all ten judges cannot distinguish between computer and people, so they make random guesses. Based of the previous result, what is the probability that all 10 judges make all correct guesses?
The problem statement is incomplete. We need to know if a judge will always identify four computers and four people, or if it's possible for a judge to guess for example that there are two computers and six people.

3. i think the question is, whats the chance of the a judge correctly identifying the 4 people and 4 computers on the first try?

The question i posted was all the information given...

4. Originally Posted by Aquameatwad
i think the question is, whats the chance of the a judge correctly identifying the 4 people and 4 computers on the first try?

The question i posted was all the information given...
I'm saying that the question is not specific enough, and since you copied it accurately, this has no reflection on you but rather on the author(s) of the problem.

Assumption (1): The judge knows in advance that there are four computers and four people, therefore the judge (being logical) will always guess four computers and four people.

Assumption (2): The judge does not know how many of each there are, so potentially the judge could guess that they are all computers, or that there are two computers and six people, etc.

Obviously in (2) the probability of guessing correctly is reduced.

You can pick which assumption to work with and we can work from there.

5. Oh I understand. The assumption i think is that a judge already knows there are 4 people and 4 computers but doesn't know which is which. So i I'm assuming assumption 1

6. Originally Posted by Aquameatwad
Oh I understand. The assumption i think is that a judge already knows there are 4 people and 4 computers but doesn't know which is which. So i I'm assuming assumption 1
Okay, so note that when the judge selects the four computers, the people are automatically determined. So there are C(8,4) ways the judge can guess, where C(n,k) is binomial coefficient aka nCk and $\displaystyle \binom{n}{k}$. Exactly one of those guesses is correct. So...

When you have ten judges, they all have the same probability of guessing correctly, so..

7. Ok then using N(D)/N = [(4C4) * (4C4)] / (8C4)? which i got as being 1/70, is this correct? Now with all 10 having the same answer is it just (1/70)^10???

8. Originally Posted by Aquameatwad
Ok then using N(D)/N = [(4C4) * (4C4)] / (8C4)? which i got as being 1/70, is this correct? Now with all 10 having the same answer is it just (1/70)^10???
Well the numerator is just 1, I'm not sure why you would write it as 4C4 * 4C4 (well I can see why but it's unnecessary), but yes that's correct.

9. Thank you very much, i really appreciate the guidance. Prob and stats is not one of my strong points.