1. 12/51. One way to think of this is as: If we start with a deck of cards that is missing one spade, and we draw a card at random from it, what is the probability that you draw a spade? Then there are 12 spades and 51 cards. A more careful way to think about this is: We start with a deck of 52 cards, draw one card at random, then draw another card at random. Let A denote the event that the first card is a spade, and B the event that the second card is a spade. We want to calculate Pr[B|A]. The sample space is the set of all 52! ways to order a deck of cards. Now Pr[A] = 13*51!/52! = 13/52, since there are 13*51! ways to order a deck of cards while leaving a spade on top. Also Pr[A and B] = 13*12*50!/52! = 13*12/52*51, since there are 13*12*50! ways to order a deck of cards while leaving two spades on top. Finally Pr[B|A] = Pr[B and A]/Pr[A] = (13*12/52*51)/(13/52) = 12/51. 2. a but not b. Pr[not raining | Apr 8] = 1 - Pr[raining | Apr 8] = 0.7. However, Pr[raining | not Apr 8] could be anything. If we live in Seattle (where it's very rainy for much of the rest of the year), it might be very large. If we live some place where April is about as rainy as it gets, and it is pretty dry much of the rest of the year, Pr[raining | not Apr 8] might be fairly small. We don't have enough information to distinguish.