Why is there the same probability to choose an object after arbitrary attempts?

Question

Here's an example problem with which I can explain what I'm asking:

"In a bag there are $3$ white balls, $4$ green balls and $5$ red balls, given you pull out a ball out of the bag and not return it, what is the probability that the second ball you pull out is white?" Using The full probability formula, the solution to it is $1/4$, however if we calculate the probability that the first pulled out ball is white we still get probability $1/4$, I did several problems like this and this property remains.

I think the crux of proving the property is that if you pull out $n$ balls and put them in a line the $k$th ball has the same probability of being a specific ball (e.g. a white one) as pulling out $k$ balls one by one, and the last one being a ball of the same type (white).

Here's my attempt at giving a generalized problem which when proven proves the property:

Given a bag and $n$ type of balls (e.g. for $n=3$ there could be specifically at least $1$ white, green and red one). Let $m$ be the total number of balls

Let the event $A_{i,j} =$ {After pulling out $i$ balls, the last one I got was of type $j$}.

Prove that

$P(A_{1,1}) = P(A_{2,1}) = P(A_{3,1}) ... = P(A_{m, 1})$

$P(A_{1,2}) = P(A_{2,2}) = P(A_{3,2}) ... = P(A_{m, 2})$

$P(A_{1,3}) = P(A_{2,3}) = P(A_{3,3}) ... = P(A_{m, 3})$

$...$

$P(A_{1,n}) = P(A_{2,n}) = P(A_{3,n}) ... = P(A_{m, n})$

I think the best way to prove it would be using induction, but I wasn't able to "inductionize" the problem.

Edit: rephrased example problem

Answer 1

I am assuming that by the "full probability formula", you mean considering all possible permutations of the balls in a row, else your statement is certainly not true.

You need to understand this carefully.

When we have no information, since balls have no preference for position, a white ball has a probability of $\frac{3}{3+4+5}=\frac{3}{12}=\frac14$ of being anywhere
Your statement "...given you pull out a ball out of the bag and not return it...the solution is $\frac14$" is true only if you are not told what ball was drawn.

But suppose we know that the first ball is green, and the rest unknown, the probability of a white ball being in any particular place in the rest of the row is $\frac3{11}$

In other words, the unconditional probability of finding a white ball anywhere in the row is $\frac14$, but the conditional probability given some information changes the probability.

Answer 2

As others have pointed out, the "easy way" is to observe that since we have no information regarding the color of the first ball, all balls are equally likely to be the second ball. But sometimes it pays to do things the hard way, both to improve understanding and as a check on intuition. So here's another way.

Suppose we draw all $12$ balls, one at a time. There are $12!$ possible sequences, all of which we assume are equally likely. How many of those have a white ball in the second position? There are $3$ choices for the white ball, and then the other $11$ balls can be arranged in $11!$ ways in the remaining positions; so there are $3 \times 11!$ ways in all. So the probability that the second ball is white is $$\frac{3 \times 11!}{12!} = \frac{3}{12}$$