Statistics (probability) problem

[Lateral]

Homework Statement

Basically the garbage information summed up: a truck is transporting a bunch of batteries, with lots of different brands. The truck crashes and the batteries and scrambled.

Suppose the number of pairs of batteries is n. Each pair is a different brand. If 2r batteries are chosen at random, and 2r < n, show that the probability there is no matching pair, by brand, is:

2$$^{}2r$$ n! (2n - 2r)! / (2n)! (n - 2r)!

Homework Equations

I don't know, but I can see that the expectation formula E(x) = $$\Sigma$$ (x P(X=x)) and the Permutation formula nPr = n! / (n - r)! and Combination formula nCr = n! / (n - r)! r! may be useful as they seem to be of a similar format to the given equation.

The Attempt at a Solution

The part is struggle most with is why they have taken 2 to the power of 2r. (2n - 2r) is obviously total batteries less chosen batteries. The denominator is total batteries by pairs less batteries, but again I'm at a loss as to why they have done this.

Can anyone shed some light?

 

[EnumaElish]

2^(2r) = # of all subsets of a set with 2r elements.

n! / [(2r)! (n - 2r)!] = nC(2r)

(2n)! / [(2r)! (2n - 2r)!] = (2n)C(2r)

Does this help?

 

[Architeuthis Dux]

I understand your confusion and I will try to explain the reasoning behind this equation. Firstly, the problem is asking for the probability that there are no matching pairs of batteries after choosing 2r batteries out of the total n pairs. This can be thought of as a combination problem, where we are choosing 2r objects out of n objects without replacement.

Now, let's break down the numerator of the equation. We have 2^{}2r, which represents the number of ways to choose 2r batteries out of 2r pairs. This is equivalent to (2r)! as each pair can be arranged in 2 different ways (e.g. A-B or B-A). Next, we have n! which represents the number of ways to choose the remaining n-2r pairs out of the total n pairs. Finally, we have (2n-2r)! which represents the number of ways to arrange the remaining batteries within the chosen pairs.

Moving on to the denominator, we have (2n)! which represents the total number of ways to arrange all the batteries. And (n-2r)! which represents the number of ways to arrange the remaining n-2r pairs.

Putting it all together, we can see that the numerator and denominator represent the total number of ways to choose and arrange the batteries without any matching pairs, divided by the total number of ways to choose and arrange all the batteries. This gives us the probability of no matching pairs.

I hope this helps to clarify the reasoning behind the equation. It may be helpful to think of it in terms of combinations and permutations, as you mentioned, and also to draw out a few examples to see the pattern.