Find all School-related info fast with the new School-Specific MBA Forum

 It is currently 06 Oct 2015, 23:32

### GMAT Club Daily Prep

#### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History

# Events & Promotions

###### Events & Promotions in June
Open Detailed Calendar

# M14 #27

Author Message
Intern
Joined: 10 Sep 2008
Posts: 37
Followers: 1

Kudos [?]: 34 [1] , given: 0

M14 #27 [#permalink]  13 Nov 2008, 16:55
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

Source: GMAT Club Tests - hardest GMAT questions

[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!
 Kaplan Promo Code Knewton GMAT Discount Codes GMAT Pill GMAT Discount Codes
SVP
Joined: 29 Aug 2007
Posts: 2493
Followers: 62

Kudos [?]: 605 [1] , given: 19

Re: M14 #27 [#permalink]  13 Nov 2008, 21:22
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
_________________
CIO
Joined: 02 Oct 2007
Posts: 1217
Followers: 93

Kudos [?]: 799 [1] , given: 334

Re: M14 #27 [#permalink]  27 Nov 2008, 06:23
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!

_________________

Welcome to GMAT Club!

Want to solve GMAT questions on the go? GMAT Club iPhone app will help.
Result correlation between real GMAT and GMAT Club Tests
Are GMAT Club Test sets ordered in any way?

Take 15 free tests with questions from GMAT Club, Knewton, Manhattan GMAT, and Veritas.

GMAT Club Premium Membership - big benefits and savings

Intern
Joined: 16 Feb 2010
Posts: 29
Followers: 0

Kudos [?]: 4 [0], given: 0

Re: M14 #27 [#permalink]  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

Manager
Joined: 15 Apr 2010
Posts: 196
Followers: 3

Kudos [?]: 15 [0], given: 29

Re: M14 #27 [#permalink]  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!
Manager
Joined: 04 Dec 2009
Posts: 71
Location: INDIA
Followers: 2

Kudos [?]: 9 [0], given: 4

Re: M14 #27 [#permalink]  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
_________________

MBA (Mind , Body and Attitude )

Manager
Joined: 27 Feb 2010
Posts: 105
Location: Denver
Followers: 1

Kudos [?]: 201 [0], given: 14

Re: M14 #27 [#permalink]  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.
Intern
Joined: 04 Jan 2010
Posts: 14
Location: Detroit
Schools: MIT LGO, Tepper
WE 1: Product Development Engineer
Followers: 1

Kudos [?]: 0 [0], given: 3

Re: M14 #27 [#permalink]  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?
CIO
Joined: 02 Oct 2007
Posts: 1217
Followers: 93

Kudos [?]: 799 [1] , given: 334

Re: M14 #27 [#permalink]  17 May 2010, 07:04
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?

_________________

Welcome to GMAT Club!

Want to solve GMAT questions on the go? GMAT Club iPhone app will help.
Result correlation between real GMAT and GMAT Club Tests
Are GMAT Club Test sets ordered in any way?

Take 15 free tests with questions from GMAT Club, Knewton, Manhattan GMAT, and Veritas.

GMAT Club Premium Membership - big benefits and savings

Intern
Joined: 02 Sep 2010
Posts: 21
Followers: 3

Kudos [?]: 5 [0], given: 1

Re: M14 #27 [#permalink]  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.

SVP
Joined: 16 Nov 2010
Posts: 1676
Location: United States (IN)
Concentration: Strategy, Technology
Followers: 31

Kudos [?]: 382 [0], given: 36

Re: M14 #27 [#permalink]  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

_________________

Formula of Life -> Achievement/Potential = k * Happiness (where k is a constant)

GMAT Club Premium Membership - big benefits and savings

Intern
Joined: 17 Dec 2010
Posts: 12
Followers: 0

Kudos [?]: 0 [0], given: 0

Re: M14 #27 [#permalink]  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

Intern
Joined: 19 May 2011
Posts: 3
Followers: 0

Kudos [?]: 0 [0], given: 9

Re: M14 #27 [#permalink]  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
Manager
Joined: 01 May 2012
Posts: 57
GPA: 1
Followers: 2

Kudos [?]: 6 [1] , given: 0

M14-27 [#permalink]  08 Jun 2012, 06:59
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: 29750
Followers: 4894

Kudos [?]: 53400 [1] , given: 8160

Re: M14-27 [#permalink]  08 Jun 2012, 07:06
1
KUDOS
Expert's post
1
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.

_________________
Manager
Status: I will not stop until i realise my goal which is my dream too
Joined: 25 Feb 2010
Posts: 235
Schools: Johnson '15
Followers: 2

Kudos [?]: 36 [0], given: 16

Re: M14-27 [#permalink]  31 Jul 2012, 07:29
Bunuel, can you please explain your answer more clearly than now...i am not understanding the solution put here...
_________________

Regards,
Harsha

Note: Give me kudos if my approach is right , else help me understand where i am missing.. I want to bell the GMAT Cat

Satyameva Jayate - Truth alone triumphs

Manager
Joined: 12 Feb 2012
Posts: 108
Followers: 1

Kudos [?]: 22 [0], given: 28

Re: M14-27 [#permalink]  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.

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
Followers: 1

Kudos [?]: 52 [0], given: 8

Re: M14 #27 [#permalink]  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.
Senior Manager
Joined: 13 Jan 2012
Posts: 307
Weight: 170lbs
GMAT 1: 730 Q48 V42
GMAT 2: 740 Q48 V42
WE: Analyst (Other)
Followers: 11

Kudos [?]: 108 [0], given: 38

Re: M14-27 [#permalink]  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.

What is the probability calculation? Can you please walk through that?
Intern
Joined: 05 Jun 2013
Posts: 1
Followers: 0

Kudos [?]: 0 [0], given: 0

Re: M14 #27 [#permalink]  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".

((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
Similar topics Replies Last post
Similar
Topics:
9 m14#37 19 18 May 2009, 15:32
19 M14#10 19 18 Mar 2009, 12:41
19 M14 #19 19 02 Feb 2009, 21:06
4 M14 #18 19 02 Feb 2009, 20:55
17 M14 #13 25 11 Nov 2008, 20:40
Display posts from previous: Sort by

# M14 #27

Moderators: WoundedTiger, Bunuel

 Powered by phpBB © phpBB Group and phpBB SEO Kindly note that the GMAT® test is a registered trademark of the Graduate Management Admission Council®, and this site has neither been reviewed nor endorsed by GMAC®.