I dont understand this probability question. 
  
*
  
*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?)
 A: Bill will ultimately win if he rolls an even number before the other two do so.


*

*If he immediately rolls an even number he wins.

*If he immediately rolls an odd number and both his opponents also do, then he's back in the same situation.

*If any other result, he looses.
So by partitioning on these three cases, the probability of ultimately winning $p$ is defined recursively by the Law of Total Probability. 
$$\begin{align}p ~=~& \mathsf P(\textsf{E}.^+) + \mathsf P(\textsf{OOO}.^+) p + 0 
\\[1ex] ~=~& \tfrac 1 2 + \tfrac 1 8 p
\\[2ex] \therefore ~ p = 4/7
\end{align}$$


$N(S)=  \binom {8} {4} + \binom {6} {4} +  \binom {13} {4} = 800$

You have counted ways to select four professors, or four associate professors, or four assistant professors.   That is not the ways to select a group of four with at least one of each type.   This is:

 $$N(S_{\text{true}})=  \binom{8}{2}\binom{6}{1}\binom{13}{1} + \binom{8}{1}\binom{6}{2}\binom{13}{1}+\binom{8}{1}\binom{6}{1}\binom{13}{2}$$

A: 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)
A: 1) No. Bill can win by rolling an even number, but he can also win by rolling an odd number: for instance, if George and Ross also roll odd numbers and then Bill rolls an even number.
2) No. A sample of size 4 contains at least one of each kind of professor if and only if it contains a professor, an associate professor, an assistant professor, and any one of the remaining 7+5+12 people. (You will have to make sure you don't overcount if you do it like this: if we have Professor 1, Associate Professor 4, Assistant Professor 5, and Professor 2, we need to make sure we don't count the duplicate Professor 2, Associate Professor 4, Assistant Professor 5, and Professor 1.)
A: 
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?

The probability that Bill wins on the first throw is $1/2$.  If Bill does not win on the first throw, he can still win if George and Ross both toss odd numbers until Bill tosses an even number.  You did not take into account the possibility that Bill could win after the first round.
Since the probability of rolling an odd number is $1/2$, the probability that nobody wins during the first round is $(1/2)^3 = 1/8$.  Bill can then win in the second round by rolling an even number with probability $(1/8)(1/2)$, where $1/8$ is the probability that nobody won during the first round and $1/2$ is the probability that Bill wins on his throw during the second round.  
If nobody wins during the second round, Bill can win in the third round with probability $(1/8)^2(1/2)$, where $(1/8)^2$ represents the probability that all three men toss odd numbers during the first two rounds and $1/2$ represents the probability that Bill tosses an even number in the third round.
More generally, Bill can win in the $k$th round if everybody tosses odd numbers for the first $k - 1$ rounds and Bill tosses an even number in the $k$th round.  Thus, the probability that Bill wins in the $k$th round is $$\left(\frac{1}{8}\right)^{k - 1}\left(\frac{1}{2}\right)$$
The probability that Bill wins is the sum over all $k$ of his winning in the $k$th round.
$$P(\text{Bill wins}) = \sum_{k = 1}^{\infty} \left(\frac{1}{8}\right)^{k - 1}\left(\frac{1}{2}\right) = \frac{1}{2}\sum_{k = 1}^{\infty} \left(\frac{1}{8}\right)^{k - 1}$$
You can calculate the probability by finding the sum of the geometric series.

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?

You added up the number of ways of selecting four professors, four associate professors, and four assistant professors.  In each case, you only considered samples with one type of professor.  Instead, you were supposed to consider samples with at least one representative of each type of professor.  Since the four selected people must contain at least one professor of each type, one type of professor must be represented by two people while each of the other types is represented by one person.  

  The number of ways this can be done is $$\binom{8}{2}\binom{6}{1}\binom{13}{1} + \binom{8}{1}\binom{6}{2}\binom{13}{1} + \binom{8}{1}\binom{6}{1}\binom{13}{2}$$

