# Probability Example Question

I'm reading my book on probability and I don't understand the example problem:

Question: Suppose that n + m balls, of which n are red and m are blue, are arranged in a linear order in such a way that all (n + m)! possible orderings are equally likely. If we record the result of this experiment by listing only the colors of the successive balls, show that all the possible results remain equally likely.

My approach: I thought about doing my own example. Assume n= 2 red and m = 1 blue. Therefore, we have (n+m)! = 3! = 6 total permutations, namely:

$b \ r_1 \ r_2 \\r_1 \ b \ r_2 \\r_1 \ r_2 \ b$

$b \ r_2 \ r_1 \\r_2 \ b \ r_1 \\r_2 \ r_1 \ b$

So I understand that each one of these permutations will have a probability of 1/6 of occurring as mentioned in the question stem. But since we only are showing colors of successive balls, we remove the repeats by dividing by 2!: $\frac{3!}{2!} = 3$, which is easily seen from the list above.

This is where I am confused. What does it want me to do from here? The book says the following: "As a result, every ordering of colorings corresponds to n! m! different orderings of the n + m balls, so every ordering of the colors has probability $\frac{n!m!}{(n+m)!}$ of occurring."

Where are they getting n!m!? Shouldn't it be $\frac{(n+m)!}{n!m!}$ as I show above?

• Here is how I would explain the solution for others to see. I understand this from Andre's help: The total number of outcomes is (n+m)!. However, since we are only doing listings by color, we must divide by the number of repeats which gives us the following: $\frac{(n+m)!}{n!m!}$. So this value gives us a new set which we calculate our probability. The question asks us the probability of choosing a value from this set. It's simply 1 divided by that value which is the reciprocal: $\frac{n!m!}{(n+m)!}$ – user1527227 Apr 25 '13 at 19:48

You are saying exactly the same thing as the text.

You are saying that all the $\frac{(m+n)!}{m!n!}$ choices are equally likely.

The text is saying that each choice has probability $\frac{m!n!}{(m+n)!}$.

We have $N$ equally likely choices if and only if each choice has probability $\frac{1}{N}$.

• I guess I'm confused what n!*m! represents. I thought n!m! represents the number of ways to order the red balls and blue balls separately. Also if $\frac{(m+n)!}{m!n!}$, then does this mean that the probability of each of these is 1/3 or 2/3? – user1527227 Apr 25 '13 at 18:43
• Suppose the red balls and blue balls each have labels. Then $(m+n)!$ is the number of arrangements. When we strip the labels off the $n$ reds, then a family of $n!$ arrangements collapses into one, so the number of arrangements becomes $(m+n)!/n!$. When we then strip off the labels from the blues, there is a further collapse by a factor of $m!$. It seemed clear from your post that you knew that. But $1/3$, $2/3$ have no connection to the problem. There are $\frac{(m+n)!}{m!n!}$ equally likely possibilities. To find the probability of each, flip this number over (take the reciprocal). – André Nicolas Apr 25 '13 at 18:54
• Ok this is exactly where I'm confused. Why do you say to take the reciprocal to find the probability? For a sample space of equally likely outcomes, Probability = {# outcomes in event}/{# outcomes in S} right? So what my intuition was telling me was the following: $\frac{\frac{(m+n)!}{m!n!}}{(m+n)!}$ – user1527227 Apr 25 '13 at 19:17
• the 1/3 and 2/3 I was referring to my example of n=2, m=1. – user1527227 Apr 25 '13 at 19:18
• A $24$-sided die has all its faces equally likely, and numbered $1$ to $24$. If we toss it, what is the probability that an $8$ turns up? In your $1$ and $2$ example, each of the $3$ possibilities has probability $\frac{1}{3}$. – André Nicolas Apr 25 '13 at 19:19