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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}&\color{white}=P(NNN)+P(NWNN)+P(NWWNN)\\&=\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)\\ &=\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?

Community
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  • For the first question, note that the sequences of not working versus working could have been any of the following: NNN, WNNN, NWNN, NNWN, WWNNN, WNWNN, WNNWN, NWWNN, ... NNWWN. It appears as though your thought process might have been to loop over each of these outcomes and calculate for that and then add them together (though... did you multiply these instead of add?). As alluded to in the answer below however, there is a much more efficient approach by considering what happens if we look at all seven regardless of if we had enough information and whether the last two work. – JMoravitz Feb 06 '20 at 15:18
  • I'm going to stick to my method but that makes sense I don't know why I didn't add so these other outcomes. How would you set up the remaining questions, meaning a lot of them are similar so if somebody can guide me through maybe one I can figure out the rest –  Feb 06 '20 at 16:05
  • If you stick with your current method, then you need to correct your current method. Add, not multiply. Then, I see the coefficient $3$ in the second term, that's good. That corresponds to the three ways in which we have one working computer in the first four such that the fourth is non-working, WNNN, NWNN, NNWN. Now, in the next term you used a $4$. This $4$ is incorrect. Think again as to what value should be used there and why. – JMoravitz Feb 06 '20 at 16:13
  • Yeah I understand that, that was a mistake on my part. Would it be 6 since the 5 combos has 6 different ways it can be rearranged –  Feb 06 '20 at 16:15
  • It would be $6$, yes. There appears to be another error in your attempt for the first problem... you had a $\frac{1}{4}$ appearing for some reason in the second term. Shouldn't this have been a $\frac{3}{7}$? – JMoravitz Feb 06 '20 at 16:27
  • You can write $5\times3\times2$ or $5\cdot3\cdot2.$ The use of an asterisk in that context is a workaround for occasions when you are limited to characters on the keyboard. – Michael Hardy Feb 06 '20 at 16:29

3 Answers3

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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.

Magma
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  • Okay that makes sense, for the other list of questions how can I set that up cause that was the main focus of the post for me –  Feb 06 '20 at 16:03
  • I just need one setup as a lot of them are similar in wording –  Feb 06 '20 at 16:05
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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 $$P(A|B)=\frac{P(A\cap B)}{P(B)}=\frac{\tfrac6{35}}{\tfrac57}=\frac6{25}$$

Alain Remillard
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  • That makes sense. So would the second to last question be $\frac{1}{4}$ since 2 computers are already in 3 orders 1,3,5 so there are 4 sports left 2,4,6,7 so there is one not working computer for 4 spots. Finally the last one would be $\frac{2}{5}$ since 5 orders are still available 1-5 and 2 computers that are not working are left –  Feb 06 '20 at 17:47
  • @user710744 on the second to last question, it will be easier to consider that computer 2 and 4 need to work in order to have a non working computer at 6 or 7. The last one is different since the last two computer could be non-working. I'll update my answer soon. – Alain Remillard Feb 06 '20 at 18:44
  • Your solution of the first question is incorrect. You don't necessarily need to find the three non-working computers in the first five tests, you just need to know which ones they are. So if you check the first five computers and four of them work, you still know that the remaining two computers do not work. In other words, your condition is satisfied if the two last computers are equal. So the correct probability is not just $\frac 47\times \frac 36 = \frac 27$, it's $\frac 47 \times \frac 36 + \frac 37 \times \frac 26 = \frac 37$. – Magma Feb 07 '20 at 10:52
  • Thanks @Magma, I missed that possibility. I'll update my answer accordingly. – Alain Remillard Feb 07 '20 at 15:15
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As alluded to elsewhere, for the first problem, it is much easier and cleaner to recognize that the three broken computers all being found within the first five computers checked is equivalent to the two computers which would have been checked sixth or seventh both being working computers.

The probability the seventh computer works is simply $\frac{4}{7}$. (If this is not immediately obvious to you why, read the answers to this question: If you draw two cards, what is the probability that the second card is a queen?). Now, given that the seventh computer works, the probability that the sixth computer also works is $\frac{3}{6}$. Multiplying these gives us the probability that the seventh and sixth computers both working as being $$\frac{4}{7}\times\frac{3}{6}=\frac{2}{7}$$ which is as mentioned equivalent to the question you were originally asked.

Now... for the next question... "Given that exactly one of the first three computers isn't working, what is the probability that both of the other two not working computers are found within the next three computers?" We could if we wanted go the long way and try to calculate each probability involved of the overall larger scenario... but it is much easier to just imagine ourselves in the position described and continue from there.

So... we have a non-working computer in the first three and two working computers in the first three. Great. We have just four computers left to check, two of which won't work and two of which will. We ask what the probability is that both of the not-working will be in the next three... in other words we ask what the probability is that the last computer is working.

Well... with four computers, each of which being equally likely to occur in the final place, two of which working, the probability that the last of these four will be one of the ones that does work will be $$\frac{2}{4}=\frac{1}{2}$$

For the next problem... we imagine ourselves in the situation that two of the not-working computers are found within the first three. So... we have four computers left, one of which doesn't work. We ask what the probability is that the remaining not-working computer is one of the next two. Well... the remaining not-working computer is equally likely to have been in any of the remaining four places... so the probability is $$\frac{2}{4}=\frac{1}{2}$$

The others continue similarly except for the last one which requires a bit more effort...

For the last, let us approach directly by definition... $Pr(A\mid B) = \dfrac{Pr(A\cap B)}{Pr(B)}$

The probability that the last computer that doesn't work was either the sixth or seventh computer tested... well.... this is just a rewording of the complementary event of the very first problem. Letting $B$ be the event that one of the sixth or seventh computers being broken gives us $Pr(B)=\dfrac{5}{7}$

Next, we ask the probability that not only is the final broken computer found to be either sixth or seventh, but also among the first three two are broken... This unfortunately, I don't see a clean clever way of rephrasing the problem so we'll do it the long way.

  • Pick which position within the first three computers is the working computer: $3$ choices

  • Pick which broken computer was the first of the broken computers in the first group of three: $3$ choices

  • Pick which other broken computer was in the first group of three: $2$ choices

  • Pick which working computer was in the first group of three: $4$ choices

  • Pick which working computer was in the fourth position: $3$ choices

  • Pick which working computer was in the fifth position: $2$ choices

  • Pick whether it was the sixth computer or the seventh computer that was the final broken computer: $2$ choices

This gives a total of $3\cdot 3\cdot 2\cdot 4\cdot 3\cdot 2\cdot 2=864$ outcomes out of the $7!$ possible orderings of the computers that yield this outcome.

Taking the ratio of the probabilities gives us the conditional probability:

$$\frac{864/7!}{5/7} = \frac{6}{25}$$

JMoravitz
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  • That makes sense. So would the second to last question be $\frac{1}{4}$ since 2 computers are already in 3 orders 1,3,5 so there are 4 sports left 2,4,6,7 so there is one not working computer for 4 spots. Finally the last one would be $\frac{2}{5}$ since 5 orders are still available 1-5 and 2 computers that are not working are left –  Feb 06 '20 at 17:47
  • Yes for the second to last question. The last question unfortunately now that I read it more closely is rather frustrating and might need to be approached by definition rather quick and loose like these others, since it is possible for the last two computers to both be broken, or merely one of the last two computers to be broken and the probabilities will change depending on that. – JMoravitz Feb 06 '20 at 17:53
  • So since there are 6 spots the computer can be tested, with two computers not working then it would be $\frac{1}{3}$ is this correct or no? –  Feb 06 '20 at 17:59
  • Your solution of the first question is incorrect. You don't necessarily need to find the three non-working computers in the first five tests, you just need to know which ones they are. So if you check the first five computers and four of them work, you still know that the remaining two computers do not work. In other words, your condition is satisfied if the two last computers are equal. So the correct probability is not just $\frac 47\times \frac 36 = \frac 27$, it's $\frac 47 \times \frac 36 + \frac 37 \times \frac 26 = \frac 37$. – Magma Feb 07 '20 at 10:52
  • @magma I can see how that interpretation is possible, however I disagree with the interpretation. In either case, I made it clear how I interpreted the problem and provided an answer corresponding to that interpretation. The interpretation I used very conveniently matches the hypothesis for the final conditional probability problem, lending further evidence that my interpretation was the intended one as it is common practice when writing a sequence of questions like this to reuse answers from earlier parts to reduce overall work. – JMoravitz Feb 07 '20 at 12:12