TITLE: I dont understand this probability question.
QUESTION [3 upvotes]: Bill, George, and Ross, in order, roll a die. The first one to roll an even
  number wins and the game is ended. What is the probability that Bill will
  win the game?


My Answer: Since Bill is the one to first roll the die, and the probability of him gettin an even number in the die is $\frac{3}{6} = \frac{1}{2}$ . Doesn't that mean that regardless of how George or Ross will have their turn, doesn't that mean that Bill will have a $\frac{1}{2}$ chance of winning this game?
Book's Answer: $\frac{4}{7}$ . (What?)


The Mathematics Department of the University of Louisville consists of
  8 professors, 6 associate professors, 13 assistant professors. In how many of
  all possible samples of size 4, chosen without replacement, will every type of professor be represented?


My Answer: Isn't this just a matter of combination? Let $S$ be the event in which every type of professor be represented then
$N(S)=  \binom {8} {4} + \binom {6} {4} +  \binom {13} {4} = 800$
Book's Answer: $7884$ (What?)

REPLY [1 votes]: For the second question. The 3 groups will have at the least one of their members being chosen to fill in a spot and once one has been chosen, then he/she cannot be chosen once again. This leaves us with a combination: $\binom {8} {2} \binom {13} {1} \binom {6} {1} + \binom {8} {1} \binom {13} {2} \binom {6} {1} + \binom {8} {1} \binom {13} {1} \binom {6} {2} = 7488$. This is because one group cannot be left out when choosing the professors. (The solution 7884 is not correct)
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