 conditional probability and drawing cards. So, today we're going to discuss probability and we're going to discuss conditional probability. This is different than when we discuss independence. The second event is not affected by the first event. But with conditional probability, this changes. The second event is affected by the outcome of the first event. So, here's the question. What is the probability that two cards drawn at random from a deck of playing cards will both be aces? So, we're drawing two cards. We're drawing card one, first card, and then we're drawing card two, the second card. If you look at a standard deck of playing cards, I have a standard deck right here. You see there are four aces out of 52 cards. So, the probability of randomly drawing an ace would be four out of 52, and you can reduce that to one out of 13, but I'm not going to yet. Now we need to talk about this second card. So, we're going to, with the first card, we're reaching into a deck of cards, pulling one card out, and then we're going to set that card to the side. We're not going to have replacement. So, let's just say I draw an ace. Let's say it's the ace of diamonds. With my first draw, I pull out the ace of diamonds. I then set the ace of diamonds to the side. So, how many total cards do I have now? Instead of having 52 cards, I now have only 51 because the ace of diamonds has been removed, and instead of having four aces, I now have just three. So, I would have three aces. One of the aces was removed in the first set, the first draw, and then I would have 51 cards left. So, you see how this changed, and that's because the second card is conditional on the first card. So, if we do four out of 52 times three out of 51, we would get 12 over 2,652, and then that would reduce to one out of 221. So, it's a very low probability. You're not very likely to draw two aces out of a stutter deck of playing cards.