Theory of Probability and Apples Text of the task:

From a basket containing 5 red apples and 7 green apples, take out one by one all except one. Find the probability that the last apple left in the basket will be green.

I tried to solve through the classical definition of probability, and this is what I got:
1. c(5,5)*c(6,7) / c(11,12)
2. 1*7/12
3. 7/12

The question is the following: 1. Is this solution correct? 2. Is it possible to solve the problem somehow using conditional probabilities?
 A: Your answer is right. Here's another way to see why: note that the problem is equivalent to drawing just one apple and calling it the "last" apple.
It is possible to solve the problem using conditional probabilities, but I can't think of a way to do it that would be either wise or practical. The issue is that there are far too many cases to enuerate; you'd need to consider the possibility that your first apple is green / red, then given that, that your second apple is green / red, and so forth. You could summarize this work with a diagram, perhaps; you could have levels indicating what possibilities are on the table after each draw. After one draw you'd have two levels (either 4 red / 7 green apples, or 5 red / 6 green apples), then after two draws you'd have three levels, and so on. You could carefully consider the number of possibilities after each step and the probability of going from each state to each state. This would be a Markov chain approach to the problem, which is essentially equivalent to considering conditional probabilities; it would be much more work than the solution you originally posed.
Of course, maybe someone else can think of a better conditional probability argument than I did.
