Here’s a problem that I encountered today in my AMC/AIME prep. It came from the 2011 AMC 10A #21:
Two counterfeit coins of equal weight are mixed with 8 identical genuine coins. The weight of each of the counterfeit coins is different from the weight of each of the genuine coins. A pair of coins is selected at random without replacement from the 10 coins. A second pair is selected at random without replacement from the remaining 8 coins. The combined weight of the first pair is equal to the combined weight of the second pair. What is the probability that all 4 selected coins are genuine?
Try this problem out by yourself – it can be solved easily if you know your combinatorics well enough.
Here’s my solution:
Note that this is a conditional probability question. Why? Take a look at this snippet of the problem:
“The combined weight of the first pair is equal to the combined weight of the second pair.“
The problem explicitly states that this must be the case in the first place! So we must determine the probability that the given four coins are all genuine under this condition.
Luckily, we only have two cases that correspond to this constraint: the case where there is one counterfeit and one genuine coin in each pair, and the case where all four coins are genuine. We compute the probability that each of these happens, add them up (because we need the constraint to hold), and divide the probability that the 4 coins are genuine by this constraint-probability.
The probability that the there is a coin of each type in the two pairs is:
Additionally, the probability that all four coins are genuine (without constraints) is:
Using conditional probability, our probability of the two pairs having equal weight is 
If you don’t know how I got the specific probabilities in the problem, I suggest you take the Intro. to Counting/Probability AoPS course, or you can ask another math pro to teach you those skills.