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M14 #27

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M14 #27 [#permalink] New post 13 Nov 2008, 16:55
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15 computers in the corporate network are infected with virus. If every day several computers are arbitrarily selected for scanning, what is the probability that no computer containing virus will be scanned in the course of the next five days?

1. 10 computers are scanned every day
2. 4% of all computers in the corporate network are scanned every day

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S2 is not sufficient because we do not know what percentage of computers in the network are infected with virus.

S1 + S2 elucidates the situation completely: the network has 250 computers, 15 are infected, 10 are scanned every day. With this information we can answer the question.

The correct answer is C.

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Hello. I tried to search for this question but I don't see it anywhere. Just curious..I am clicking on search and putting in only M14 and it says nothing found..I know that's not true. Is there something I'm doing wrong?-
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QUESTION: I am asking for someone to tell me what the probability is of not finding an infected computer within the 5 days? Without the five day restriction and randomly picking a computer among all computers, the probability would be 235/250 of not picking one that's infected, right? Then, how would you adjust that to fit a five day period? From S1, I know that within 5 days 50 computers will be scanned. So, do we take the general probability of 235/250 and divide that by 50 to receive the 5 day probability of not picking a computer with a virus? Thank you!
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Re: M14 #27 [#permalink] New post 13 Nov 2008, 21:22
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dczuchta wrote:
15 computers in the corporate network are infected with virus. If every day several computers are arbitrarily selected for scanning, what is the probability that no computer containing virus will be scanned in the course of the next five days?

1) 10 computers are scanned every day
2) 4% of all computers in the corporate network are scanned every day

S1 is not sufficient because we do not know how many computers the corporate network has.
S2 is not sufficient because we do not know what percentage of computers in the network are infected with virus.
S1 + S2 elucidates the situation completely: the network has 250 computers, 15 are infected, 10 are scanned every day. With this information we can answer the question.

The correct answer is C.

QUESTION: I am asking for someone to tell me what the probability is of not finding an infected computer within the 5 days? Without the five day restriction and randomly picking a computer among all computers, the probability would be 235/250 of not picking one that's infected, right? Then, how would you adjust that to fit a five day period? From S1, I know that within 5 days 50 computers will be scanned. So, do we take the general probability of 235/250 and divide that by 50 to receive the 5 day probability of not picking a computer with a virus? Thank you!


no computer selecting day 1 = 235c10/250c10
no computer selecting day 2 = 235c10/250c10
no computer selecting day 3 = 235c10/250c10
no computer selecting day 4 = 235c10/250c10
no computer selecting day 5 = 235c10/250c10

no computer is selecting for all 5 days = (235c10/250c10)^5
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Re: M14 #27 [#permalink] New post 27 Nov 2008, 06:23
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The trick here is that we choose computers that will be scanned arbitrarily, which means those computers scanned one day could be chosen to be scanned the next day as well. That is why the probability of choosing a computer that is not infected within the 5-day period is \left(\frac{235}{250}\right)^5. Hope this helps.

As to the search, we are having some problems with indexing short strings such as m07 or m7. We're working on fixing this problem.

dczuchta wrote:
Hello. I tried to search for this question but I don't see it anywhere. Just curious..I am clicking on search and putting in only M14 and it says nothing found..I know that's not true. Is there something I'm doing wrong?-
_________________________________________________________________

15 computers in the corporate network are infected with virus. If every day several computers are arbitrarily selected for scanning, what is the probability that no computer containing virus will be scanned in the course of the next five days?

1) 10 computers are scanned every day

2) 4% of all computers in the corporate network are scanned every day

Statement (1) ALONE is sufficient, but Statement (2) ALONE is not sufficient
Statement (2) ALONE is sufficient, but Statement (1) ALONE is not sufficient
BOTH statements TOGETHER are sufficient, but NEITHER statement ALONE is sufficient
EACH statement ALONE is sufficient
Statements (1) and (2) TOGETHER are NOT sufficient
S1 is not sufficient because we do not know how many computers the corporate network has.

S2 is not sufficient because we do not know what percentage of computers in the network are infected with virus.

S1 + S2 elucidates the situation completely: the network has 250 computers, 15 are infected, 10 are scanned every day. With this information we can answer the question.

The correct answer is C.

QUESTION: I am asking for someone to tell me what the probability is of not finding an infected computer within the 5 days? Without the five day restriction and randomly picking a computer among all computers, the probability would be 235/250 of not picking one that's infected, right? Then, how would you adjust that to fit a five day period? From S1, I know that within 5 days 50 computers will be scanned. So, do we take the general probability of 235/250 and divide that by 50 to receive the 5 day probability of not picking a computer with a virus? Thank you!

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Re: M14 #27 [#permalink] New post 23 Apr 2010, 10:28
S1 is insufficient as till now we do not have total number of computers

S2 is insufficient 4% are scanned daily but out of how many is not given

combining S1 + S2 we get the network has 250 computers and given 15 are infected also given 10 are scanned every day.

with the given info probability that no computer containing virus will be scanned in the course of the next five days can be calculated

The answer is C.
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Re: M14 #27 [#permalink] New post 23 Apr 2010, 11:06
thanks for the detailed answers -- I was confused on the prob part as well but it is clear now. Note- we luckily dont have to give the final answer, just a conclusion that it can be solved!
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Re: M14 #27 [#permalink] New post 23 Apr 2010, 19:23
S1 not sufficient , not give any detail about total no of comp.
s2 not sufficient , same as s1

s1+s2 sufficient for ans.

ans:c
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Re: M14 #27 [#permalink] New post 24 Apr 2010, 07:39
I think, we if read the question and choice A and B's carefully, we should be able to guess C as the answer. The 5 day part confuses little bit but GmatTiger explained it clearly.
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Re: M14 #27 [#permalink] New post 15 May 2010, 10:24
dzyubam says the answer is (235/250)^5
GMAT TIGER says the answer is (235c10 / 250c10)^5

Which is correct?
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Re: M14 #27 [#permalink] New post 17 May 2010, 07:04
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It's so good we don't have to give exact answers in DS questions :). GMAT TIGER is more experienced than I am, so I think he's right.
jnelson5446 wrote:
dzyubam says the answer is (235/250)^5
GMAT TIGER says the answer is (235c10 / 250c10)^5

Which is correct?

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Re: M14 #27 [#permalink] New post 27 Apr 2011, 05:17
This might be a silly question, but where do you get the 250 computers from?

I said E, as I did not know the total amount of computers in the corporate network. :)

Thanks in advance...
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Re: M14 #27 [#permalink] New post 27 Apr 2011, 06:10
Probability = 1 - (N-15Cn)/NCn for 1 day and for 5 days this can be multiplied to itself 5 times, Where N is the total number of computers and n = no. of computers scanned everyday.

(1) and (2) are insufficient.

but 10 = 0.04N

=> N = 250

and n = 10 from (1)

So the rest can be found

Answer - C
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Re: M14 #27 [#permalink] New post 27 Apr 2011, 06:18
S1 is insufficient as till now we do not have total number of computers

S2 is insufficient 4% are scanned daily but out of how many is not given

combining S1 + S2 we get the network has 250 computers and given 15 are infected also given 10 are scanned every day.

In our ability to determine total number of computers we are able to calculate the probability

The answer is C.
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Re: M14 #27 [#permalink] New post 07 May 2012, 20:55
jnelson5446 wrote:
dzyubam says the answer is (235/250)^5
GMAT TIGER says the answer is (235c10 / 250c10)^5

Which is correct?


i think ..

(235/250)^5----this is the probability to select any non infected computer consecutively 5 time

(235c10 / 250c10)^5----this is the probability to select 10 non infected computer consecutively 5 time
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M14-27 [#permalink] New post 08 Jun 2012, 06:59
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15 computers in a corporate network are infected with a virus. If every day several computers are arbitrarily selected for scanning, what is the probability that no computer containing the virus will be scanned in the course of the next five days?

(1) 10 computers are scanned every day.

(2) 4% of all computers in the corporate network are scanned every day.

Notice that in order to find the probability we need to know how many computers are there in the network and how many are scanned every day.

(1) 10 computers are scanned every day. We still don't know how many computers are there in the network.

(2) 4% of all computers in the corporate network are scanned every day → 0.04∗total=scanned. Not sufficient.

(1)+(2) 0.04∗total=10 → total=250. We have all information needed: total=250 and scanned=10, so P=←(C10235C10250→)5. Sufficient.

---------------

I totally do not understand the explanation of this question. Why can we determine the probability of computer containing no virus being scanned, just by knowing that the total number of computers is 250?
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Re: M14-27 [#permalink] New post 08 Jun 2012, 07:06
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heintzst wrote:
15 computers in a corporate network are infected with a virus. If every day several computers are arbitrarily selected for scanning, what is the probability that no computer containing the virus will be scanned in the course of the next five days?

(1) 10 computers are scanned every day.

(2) 4% of all computers in the corporate network are scanned every day.

Notice that in order to find the probability we need to know how many computers are there in the network and how many are scanned every day.

(1) 10 computers are scanned every day. We still don't know how many computers are there in the network.

(2) 4% of all computers in the corporate network are scanned every day → 0.04∗total=scanned. Not sufficient.

(1)+(2) 0.04∗total=10 → total=250. We have all information needed: total=250 and scanned=10, so P=←(C10235C10250→)5. Sufficient.

---------------

I totally do not understand the explanation of this question. Why can we determine the probability of computer containing no virus being scanned, just by knowing that the total number of computers is 250?


That's not the only thing we know. We also know that 10 computers are scanned every day. Combining these two we can get the probability needed. Below is a solution of this problem with proper formatting.

15 computers in a corporate network are infected with a virus. If every day several computers are arbitrarily selected for scanning, what is the probability that no computer containing the virus will be scanned in the course of the next five days?

Notice that in order to find the probability we need to know how many computers are there in the network and how many are scanned every day.

(1) 10 computers are scanned every day. We still don't know how many computers are there in the network.
(2) 4% of all computers in the corporate network are scanned every day --> 0.04*total=scanned. Not sufficient.

(1)+(2) 0.04*total=10 --> total=250. We have all information needed: total=250 and scanned=10, so P=(\frac{C^{10}_{235}}{C^{10}_{250}})^5. Sufficient.

Answer: C.
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Re: M14-27 [#permalink] New post 31 Jul 2012, 07:29
Bunuel, can you please explain your answer more clearly than now...i am not understanding the solution put here...
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Re: M14-27 [#permalink] New post 02 Sep 2012, 19:53
Bunuel wrote:
heintzst wrote:
15 computers in a corporate network are infected with a virus. If every day several computers are arbitrarily selected for scanning, what is the probability that no computer containing the virus will be scanned in the course of the next five days?

(1) 10 computers are scanned every day.

(2) 4% of all computers in the corporate network are scanned every day.

Notice that in order to find the probability we need to know how many computers are there in the network and how many are scanned every day.

(1) 10 computers are scanned every day. We still don't know how many computers are there in the network.

(2) 4% of all computers in the corporate network are scanned every day → 0.04∗total=scanned. Not sufficient.

(1)+(2) 0.04∗total=10 → total=250. We have all information needed: total=250 and scanned=10, so P=←(C10235C10250→)5. Sufficient.

---------------

I totally do not understand the explanation of this question. Why can we determine the probability of computer containing no virus being scanned, just by knowing that the total number of computers is 250?


That's not the only thing we know. We also know that 10 computers are scanned every day. Combining these two we can get the probability needed. Below is a solution of this problem with proper formatting.

15 computers in a corporate network are infected with a virus. If every day several computers are arbitrarily selected for scanning, what is the probability that no computer containing the virus will be scanned in the course of the next five days?

Notice that in order to find the probability we need to know how many computers are there in the network and how many are scanned every day.

(1) 10 computers are scanned every day. We still don't know how many computers are there in the network.
(2) 4% of all computers in the corporate network are scanned every day --> 0.04*total=scanned. Not sufficient.

(1)+(2) 0.04*total=10 --> total=250. We have all information needed: total=250 and scanned=10, so P=(\frac{C^{10}_{235}}{C^{10}_{250}})^5. Sufficient.

Answer: C.


Hey Bunuel,

Good thing we actually dont need to compute it but shouldnt the answer be:

P=(\frac{{C^{0}_{15}C^{10}_{235}}}{C^{10}_{250}}*\frac{{C^{0}_{15}C^{10}_{225}}}{C^{10}_{240}}*\frac{{C^{0}_{15}C^{10}_{215}}}{C^{10}_{230}}*\frac{{C^{0}_{15}C^{10}_{205}}}{C^{10}_{220}}*\frac{{C^{0}_{15}C^{10}_{195}}}{C^{10}_{210}}).

Guess, it really depends if they reinspect the same computers.
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Re: M14 #27 [#permalink] New post 02 May 2013, 04:52
What is the number of computers that do not have virus?
What is total number of computers?

S1: This gives number of computers scanned each day = 10
But S1 does not answer the above two questions.
So, S1 is not sufficient. Eliminate A and D.

S2: This gives percentage of all computers that is scanned each day.
But S2 does not tell about the number of all computers.
So, S2 is not sufficient. Eliminate B.

Both: (4/100) * All Computer = 10
This gives the answer to total number of computers, which is 250.
Thus, we can get to number of computers with no virus and the probability for the same.

So both are sufficient. C is the correct answer.
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Re: M14-27 [#permalink] New post 02 May 2013, 07:32
Bunuel wrote:
heintzst wrote:
15 computers in a corporate network are infected with a virus. If every day several computers are arbitrarily selected for scanning, what is the probability that no computer containing the virus will be scanned in the course of the next five days?

(1) 10 computers are scanned every day.

(2) 4% of all computers in the corporate network are scanned every day.

Notice that in order to find the probability we need to know how many computers are there in the network and how many are scanned every day.

(1) 10 computers are scanned every day. We still don't know how many computers are there in the network.

(2) 4% of all computers in the corporate network are scanned every day → 0.04∗total=scanned. Not sufficient.

(1)+(2) 0.04∗total=10 → total=250. We have all information needed: total=250 and scanned=10, so P=←(C10235C10250→)5. Sufficient.

---------------

I totally do not understand the explanation of this question. Why can we determine the probability of computer containing no virus being scanned, just by knowing that the total number of computers is 250?


That's not the only thing we know. We also know that 10 computers are scanned every day. Combining these two we can get the probability needed. Below is a solution of this problem with proper formatting.

15 computers in a corporate network are infected with a virus. If every day several computers are arbitrarily selected for scanning, what is the probability that no computer containing the virus will be scanned in the course of the next five days?

Notice that in order to find the probability we need to know how many computers are there in the network and how many are scanned every day.

(1) 10 computers are scanned every day. We still don't know how many computers are there in the network.
(2) 4% of all computers in the corporate network are scanned every day --> 0.04*total=scanned. Not sufficient.

(1)+(2) 0.04*total=10 --> total=250. We have all information needed: total=250 and scanned=10, so P=(\frac{C^{10}_{235}}{C^{10}_{250}})^5. Sufficient.

Answer: C.


What is the probability calculation? Can you please walk through that?
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Re: M14 #27 [#permalink] New post 06 Jun 2013, 19:33
SUDHANSHUKNIT99 wrote:
jnelson5446 wrote:
dzyubam says the answer is (235/250)^5
GMAT TIGER says the answer is (235c10 / 250c10)^5

Which is correct?


i think ..

(235/250)^5----this is the probability to select any non infected computer consecutively 5 time

(235c10 / 250c10)^5----this is the probability to select 10 non infected computer consecutively 5 time


Based on the above, we see that:

P(Picking first uninfected computer) = 235/250
P(Picking second uninfected computer) = (235 - 1)/(250 - 1) = 234/249
P(Picking third uninfected computer) = (234 - 1)/(249 - 1) = 233/248
...
P(Picking 10th uninfected computer) = (227 - 1)/(242 - 1) = 226/241

In order to figure out the probability for a day's worth of scanning and not picking/finding an infected machine, we multiply the series of probabilities listed above = P(scanning 10 and finding nothing)

In order to repeat the above process for 5 days = P(scanning 10 and finding nothing)^5.

This seems like a very complex calculation, but with a bit of computational magic, we can verify that the this solution matches "(235c10 / 250c10)^5". 8-)

((235/250) (234/249) (233/248) (232/247) (231/246) (230/245) (229/244) (228/243) (227/242) (226/241))^5 - (Binomial[235, 10]/Binomial[250, 10])^5

The above equation nets out to zero! I hope this helps.
Re: M14 #27   [#permalink] 06 Jun 2013, 19:33
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