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# M14 #27

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Intern
Joined: 10 Sep 2008
Posts: 36

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13 Nov 2008, 17:55
1
KUDOS
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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[Reveal] Spoiler: OE
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.

_________________________________________________________________
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?-
_________________________________________________________________

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!
SVP
Joined: 29 Aug 2007
Posts: 2473

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13 Nov 2008, 22:22
1
KUDOS
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.

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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Joined: 02 Oct 2007
Posts: 1218

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27 Nov 2008, 07:23
1
KUDOS
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.

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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Intern
Joined: 16 Feb 2010
Posts: 28

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23 Apr 2010, 11:28
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

Manager
Joined: 15 Apr 2010
Posts: 192

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23 Apr 2010, 12:06
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!
Manager
Joined: 04 Dec 2009
Posts: 70
Location: INDIA

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23 Apr 2010, 20:23
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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Manager
Joined: 27 Feb 2010
Posts: 105
Location: Denver

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24 Apr 2010, 08:39
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.
Intern
Joined: 04 Jan 2010
Posts: 14
Location: Detroit
Schools: MIT LGO, Tepper
WE 1: Product Development Engineer

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15 May 2010, 11:24
dzyubam says the answer is (235/250)^5
GMAT TIGER says the answer is (235c10 / 250c10)^5

Which is correct?
CIO
Joined: 02 Oct 2007
Posts: 1218

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17 May 2010, 08:04
1
KUDOS
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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Intern
Joined: 02 Sep 2010
Posts: 20

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27 Apr 2011, 06:17
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.

SVP
Joined: 16 Nov 2010
Posts: 1663
Location: United States (IN)
Concentration: Strategy, Technology

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27 Apr 2011, 07:10
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

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Intern
Joined: 17 Dec 2010
Posts: 12

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27 Apr 2011, 07:18
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

Intern
Joined: 19 May 2011
Posts: 3

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07 May 2012, 21:55
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
Manager
Joined: 01 May 2012
Posts: 57
GPA: 1

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08 Jun 2012, 07:59
1
KUDOS
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?
Math Expert
Joined: 02 Sep 2009
Posts: 39755

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08 Jun 2012, 08:06
1
KUDOS
Expert's post
2
This post was
BOOKMARKED
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.

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31 Jul 2012, 08:29
Bunuel, can you please explain your answer more clearly than now...i am not understanding the solution put here...
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Manager
Joined: 12 Feb 2012
Posts: 135

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02 Sep 2012, 20:53
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.

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.
Intern
Joined: 11 Jan 2010
Posts: 38

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02 May 2013, 05:52
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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Posts: 307
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GMAT 2: 760 Q50 V42
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02 May 2013, 08:32
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.

What is the probability calculation? Can you please walk through that?
Intern
Joined: 05 Jun 2013
Posts: 1

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06 Jun 2013, 20:33
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".

((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, 20:33
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# M14 #27

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