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.