Explain whether or not the three probabilities add to unity. Combinations with replacement? Problem (1):
A storage contains 200 computers; 5 are defective and 195 are fine. Two computers are selected randomly with replacement.


*

*Calculate the probability that neither computer is defective (correct?)
$$\frac {\binom {195+2-1}{2}}{\binom {200+2-1}{2}}$$

*Calculate the probability that exactly one computer is defective (correct?)
$$\frac {\binom {5+1-1}{1}\binom {195+1-1}{1}}{\binom {200+2-1}{2}}$$

*Calculate the probability that both computers are defective (correct?)
$$\frac {\binom {5+2-1}{2}}{\binom {200+2-1}{2}}$$

*Explain whether or not the three probabilities you have just calculated should add to unity


Question 4 I don't really understand.

Problem (2): ...without replacement.


*

*Calculate the probability that neither computer is defective
$$\frac {\binom {195}{2}}{\binom {200}{2}}$$

*Calculate the probability that exactly one computer is defective
$$\frac {\binom {5}{1}\binom {195}{1}}{\binom {200}{2}}$$

*Calculate the probability that both computers are defective
$$\frac {\binom {5}{2}}{\binom {200}{2}}$$

*Explain whether or not the three probabilities you have just calculated should add to unity


All three add to 1. Should these three add to unity because of their mutual exclusiveness and exhaustiveness? Please clarify my thoughts. 
Edited: Much thanks for every single answer and comment! In the mean time I came up with the tree diagram to visualize the solutions:

 A: As @JMoravitz stated in the comments, your answers to the second problem are correct.
In the first problem, sampling is done with replacement (which is not the best way to check for defective products).  Therefore, we can use the binomial distribution.
The probability of selecting $k$ good computers and $n - k$ bad computers when $n$ computers are selected with replacement is 
$$\binom{n}{k}p^k(1 - p)^{n - k}$$
where $\binom{n}{k}$ represents the number of orders in which exactly $k$ good computers can be selected in $n$ trials, $p$ is the probability that a good computer is selected, and $1 - p$ is the probability that a defective computer is selected. 
Two good computers are selected: Using the formula given above with $n = k = 2$ and $p = 195/200$ yields
$$\binom{2}{2}\left(\frac{195}{200}\right)^2\left(\frac{5}{200}\right)^0 = \left(\frac{39}{40}\right)^2$$
One good and one defective computer are selected:  Using the formula given above with $n = 2$, $k = 1$, and $p = 195/200$ yields

 $$\binom{2}{1}\left(\frac{195}{200}\right)^1\left(\frac{5}{200}\right)^1 = 2\left(\frac{39}{40}\right)\left(\frac{1}{40}\right)$$

Two defective computers are selected:  Using the formula given above with $n = 2$, $k = 0$, and $p = 195/200$ yields

 $$\binom{2}{0}\left(\frac{195}{200}\right)^0\left(\frac{5}{200}\right)^2 = \left(\frac{1}{40}\right)^2$$

Since the three events described above are mutually exclusive and exhaustive, their probabilities should add to $1$, which you should check.
