Missing Document [Probability Problem]

[SEstudent22]

Here is the problem:
" One programmer has to find a document, which is located on either disk 1 or disk 2 (but not on both). The document is on disk 1 with a probability of 2/5. The document is on disk 2 with a probability of 3/5. For this purpose the programmer has created a program to search for the document, which he can use on 4 computers. Each computer finds the document with a probability of 3/4, if it searches on the disk which contains the document (and it can't find the document if it's not on the disk which it's searching).

A) The programmer has arranged the computers so 1 computer searches disk 1 and the others search disk 2. Find the probability that the document will be found? "

Here is my work and thinking:

Okay so the document is on one of the disks, but not both at the same time. I define the following events and their respected probabilities:
A - the document is on disk 1
P(A) = 2/5 = 0.4
B - the document is on disk 2
P(B) = 3/5 = 0.6

Now 4 computers are searching for the document and the probability that a computer will find the document is the same, however that depends on if the document is actually on the disk the computer is searching on. So this is my thinking:

C1 - represents the computer searching on disk 1
C2 - represents the other computers searching on disk 2

1 computer is going to search disk 1. I need to find the probability that the document is actually on disk 1 and the machine will find it.
P(AC1) = P(A) * P(C1|A) = 0.4 * 3/4 = 0.3

3 computers are going to search disk 2. I need to find the probability that the document is actually on disk 2 and the machine will find it. This applies to all 3 computers
P(AC2) = P(B) * P(C2|B) = 0.6 * 3/4 = 0.45

I define the final event and it's probability is the answer I'm searching for:
D - the document is found.

I'm thinking that I need to add up the probability that computer 1 will find the document on disk 1 and the probability that the other 3 computers will find the document on disk 2. It's clear that the probability that computer 1 will find the document is 0.3, however I am having trouble figuring out the probability of the other 3 machines. I thought I would apply the same logic and use the Binomial Distribution to find the probability for the other 3 machines, although in the end I get a result that is larger than 1 so that can not be the correct probability. I also tried thinking about there being a Hypergeometric distribution, but it doesn't specify in the text of the problem that the experiment ends when we actually find the document once, so I don't think that's it.

Any ideas of where is the flaw in my logic?

 

[Cubist]

...I am having trouble figuring out the probability of the other 3 machines.

There are only 2 other machines, not 3. There's a total of 3 computers searching disk B (since one is assigned to searching disk A).

I think we need to assume that the 3 computers searching disk B will find the document with independent probabilities of 3/4. (Although this seems VERY strange since I'd expect this kind of search to be deterministic - ie. I'd expect all 3 to fail OR all 3 to succeed! But, let's roll with the "independent" assumption ? )

You could use the Binomial Distribution to find the answer, and I'm a little curious about how you managed to get a result greater than one. (Therefore please post your work.) BUT, there's a much simpler method. You just need to be aware that one computer could find the document, or two computers could find it, or even all three. It might be easier to switch your thinking, what's the chance of all three failing to find the document GIVEN that it's on disk B?

EDIT: When you've worked out the above, then you can introduce the probability of the document being on disk A or B

 

[SEstudent22]

what's the chance of all three failing to find the document GIVEN that it's on disk B?

I'm thinking: 1 - \binom{3}{0}*0.45^0* 0.55^3 . This gives me the probability that at least one computer will find the document.

You could use the Binomial Distribution to find the answer, and I'm a little curious about how you managed to get a result greater than one.

Well, I tried adding up the probability that 0 computers will find the document, 1 computer will find the document, 2 computers will find the document and all three. But I realize that always adds up to 1. On top of that I added the probability that the one computer searching disk A would find it (0.3) and that's how I got over 1, not very smart of me.

 

[Cubist]

I'm thinking: [imath]1 - \binom{3}{0}*0.45^0* 0.55^3[/imath]. This gives me the probability that at least one computer will find the document.

Almost, but you ought to be using the probability that a computer finds the document GIVEN that it's on disk 2. This is 3/4. Therefore, the probability that computer 2,3 OR 4 will find the document GIVEN that it's on disk 2 is...

P(C23or4|B) = [imath]1 - \binom{3}{0}\times \left(\frac{3}{4}\right)^0 \times \left(1-\frac{3}{4}\right)^3[/imath]

3 computers are going to search disk 2. I need to find the probability that the document is actually on disk 2 and the machine will find it. This applies to all 3 computers
P(AC2) = P(B) * P(C2|B) = 0.6 * 3/4 = 0.45

P(AC23or4) = P(B) * P(C23or4|B) = ?

--

I'll try to explain why it's wrong to use the 0.45 for the probability within the biniomial distribution. Consider these two Venn diagrams...



In the left diagram, the "disk 2" area is considered as the universal set during the binomial calculation.
In the right diagram, the "disk 1 & disk 2" area is considered as the universal set during the binomial calculation.

The gray areas represent the probability that ONE computer will find the document. The green represents probability that TWO find the document. You can probably work out what the white and red colours represent for yourself. Can you see that the overlaps would be quite different in the two scenarios? BTW: These are sketches, and they aren't drawn to scale. (If they were accurate, then I think the difference would be even more apparent!). We have to use the left version of the diagram because the 3 computers C2,C3 and C4 will just be operating on disk2.

( I hope that I'm correct in this description, there are other more experienced teachers on this site who will hopefully correct me if I'm wrong! )

... that's how I got over 1, not very smart of me.

Getting an answer wrong is how you learn. It DOESN'T mean that you're, "not smart". Just keep practicing and your knowledge will increase!

 

[SEstudent22]

Almost, but you ought to be using the probability that a computer finds the document GIVEN that it's on disk 2. This is 3/4. Therefore, the probability that computer 2,3 OR 4 will find the document GIVEN that it's on disk 2 is...

P(C23or4|B) = [imath]1 - \binom{3}{0}\times \left(\frac{3}{4}\right)^0 \times \left(1-\frac{3}{4}\right)^3[/imath]

P(AC23or4) = P(B) * P(C23or4|B) = ?

I knew that when I calculated that 0.45 probability it wasn't right! It was fishy to me when I decided to lump the 3 computers into one. Now I understand the problem and the solution. Thank you Cubist for taking the time to go into so much detail to explain it to me ?