Probability of Broken Computers A school orders  7 new computers for a classroom but they are told 3 will not work properly when they recieve them.  The school begins to turn each computer on, one by one, to figure out which computers do not work.
There are quite a few questions but most of them I am just having trouble setting up and I feel mostly are similar in set up.

What is the probability that no more than five computers need to be turned on to find the three computers that don't work?

For this question N is the event the computer does not work and W means it works:
\begin{align}&amp;\color{white}=P(NNN)+P(NWNN)+P(NWWNN)\\&amp;=\left(\frac{3}{7}\cdot\frac{2}{6}\cdot\frac{1}{5}\right)\left(3\cdot\frac{1}{4}\cdot\frac{2}{6}\cdot\frac{4}{5}\cdot\frac{1}{4}\right)\left(4\cdot\frac{3}{7}\cdot\frac{2}{6}\cdot\frac{4}{5}\cdot\frac{3}{4}\cdot\frac{1}{3}\right)\\
&amp;=\frac{12}{42875}\end{align}
The next couple questions are very similar I just don't know how to set up.  These all don't have specific order of which computers are picked but are told the computers are turned on in a couple possible spots.  These are the remaining questions if someondy can explain how these can be set up that be appreciated.

Given that exactly one of the computers not working was found within the first three computers, what is the probability that the other two computers that aren't wokring are found within the next three computers turned on?
Given that exactly two of the computers that don't work were found within the first three computers, what is the probability the last computer that doesn't work is found within the next two computers turning on?
Given that exactly two of the computers that don't work were found within the computers 1, 3, 5, what is the probability the other computer not working was found on tests 6 or 7?
Given the last computer doesn't work was found within the last two tests, what is the probability that the first two computers that don't work were found within the first three computers?
For the first question, it helps to consider the opposite event. What is the probability that more than five computers need to be turned on to find the three that do not work? In other words, what is the probability that after knowing the state of five computers, you still don't know the state of the two remaining computers?

Note that it really doesn't matter in which order the first five computers work or do not work.
For the first, we could use the fact that if the three computer that doesn't have been found it the give first attempts, the the last two are working. Your problem is equivalent to "find the probability that the last two computers are working." The order in which the non-working computers are found is irrelevent.  This probability is 
$$\frac47\times \frac36=\frac27$$

EDIT

Has pointed out in the comments, There is also the possibility that the last two computer are non-working. Since it is known there is three non-working computers, if we found only one in the first five attempts, the last two are non-working. The probability that the last two are non-working is
$$\frac37\times\frac26=\frac17$$
Then the probability that we need only five attempts to find the non-working computers is
$$\frac27+\frac17=\frac37$$



For the other problems, we could do it with conditiinal probability or consider what is left and do it like a new problem.


  Given that exactly one of the computers not working was found within the first three computers, what is the probability that the other two computers that aren't wokring are found within the next three computers turned on?


After three attempts, we found two working and one non-working. We are left with four computers where two are not working. To found the two non-working within the next three attempts, the last remaining computer must be working. There are two working computers in the four we have left. The probability is
$$\frac24=\frac12$$


  Given that exactly two of the computers that don't work were found within the first three computers, what is the probability the last computer that doesn't work is found within the next two computers turning on?


This case is similar to the previous one. Consider that the out of the last four, the last two need to be working.
$$\frac34\times\frac23=\frac12$$


  Given that exactly two of the computers that don't work were found within the computers 1, 3, 5, what is the probability the other computer not working was found on tests 6 or 7?


It left computers tested on 2, 4, 6 and 7. Three of them work, and one doesn't. The one that doesn't work need to be tested on 6 or 7. It is the same as saying 2 and 4 need to work.
$$\frac34\times\frac23=\frac12$$


  Given the last computer doesn't work was found within the last two tests, what is the probability that the first two computers that don't work were found within the first three computers?


This one is a bit trickier since we don't know if there is exactly one non-working computer in the last two tests. We will have to do it with conditional probability. Let
$$A=\text{two non-working on first three attempts}$$
$$B=\text{at least one non-working in last two attempts}$$
$$P(A|B)=\frac{P(A\cap B)}{P(B)}$$
Event $B$ will be true unless the last two are working so
$$P(B)=1-\frac47\times\frac36=\frac57$$
It left to evaluate the events $A$ and $B$ will be true together. We have to do it the hard way.
$$P(A\cap B)=3\times\frac37\times\frac26\times\frac45\times2\times\frac14\times\frac33=\frac6{35}$$
Finally, the answer is 
$$P(A|B)=\frac{P(A\cap B)}{P(B)}=\frac{\tfrac6{35}}{\tfrac57}=\frac6{25}$$