Reason why these two probabilities are equal (Picking Balls in exact order without Replacement) I realize the probability of the following two events are equal. I am curious: is there a reason, besides coincidence, that the probabilities are equal?
Suppose there are five balls in a bucket. 3 of the balls are labelled A, and 2 of the balls are labelled B. There is no way to distinguish between balls labelled A. There is no way to distinguish between balls labelled B.
Suppose I draw balls at random, without replacement. The event $\{AAABB\}$ means I pick 3 balls labelled A, then 2 balls labelled B (in that exact order). Then,
$$ P(\{AAABB\}) = \frac{3}{5} \times \frac{2}{4} \times \frac{1}{3} \times 1 = 0.1 $$
Also,
$$ P(\{AABAB\}) = \frac{3}{5} \times \frac{2}{4} \times \frac{2}{3} \times \frac{1}{2} \times 1 = 0.1$$
As you can see, the probability of the events $\{AAABB\}$ and $\{AABAB\}$ are exactly the same. I have seen the claim that any possible order of 3 A's and 2 B's is the same. Why is this true (if it indeed true)? If the claim is true, then I don't have to multiply out individually for every conceivable event.
Thanks.

The claim indeed seems true.
For a way to see this.
Consider the first probability
$P(\{AAABB\}) = \frac{3}{5} \times \frac{2}{4} \times \frac{1}{3} \times 1 = 0.1$
This can actually be written as
$P(\{AAABB\}) = \frac{3}{5} \times \frac{2}{4} \times \frac{1}{3} \times \frac{2}{2} \times \frac{1}{1} = 0.1$
Now notice the denominators: This will always be $5 \times 4 \times 3 \times 2 \times 1$.
Now the numerator will just be $3 \times 2 \times 1 \times 2 \times 1$, in some order.
The $3$ will appear when you pick the first A, one of the $2$ will appear when you pick the second A or first B etc.
Thus the probability is $\frac{12}{120} = 0.1$.
