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M14-27

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M14-27  [#permalink]

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New post 15 Sep 2014, 23:54
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  35% (medium)

Question Stats:

67% (00:52) correct 33% (00:47) wrong based on 110 sessions

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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.

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Re M14-27  [#permalink]

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New post 15 Sep 2014, 23:54
Official Solution:


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. Given: \(0.04*total=scanned\). Not sufficient.

(1)+(2) Since \(0.04*total=10\), then \(total=250\). 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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M14-27  [#permalink]

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New post 30 Apr 2015, 02:09
Hi Bunuel,

I do understand that this is quite simple, but can you please check the below and inform if my understanding is correct for (1) + (2).

desired value = select 10 out of 235(do not have virus)
Total value = select 10 out of 250

probability = ( (235!/10!225!)/ (250!/10!240!) )^5

Thanks!
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Re: M14-27  [#permalink]

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New post 05 Jul 2016, 10:19
Hi Bunuel,

I think Statement 1 is sufficient, as we knew that during 5 days they will check totally 50 computers, out of which 15 are infected with virus and another 35 (50-15) are not.
So the answer is 35/50=0.7
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Re: M14-27  [#permalink]

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New post 05 Jul 2016, 11:16
shanahit wrote:
Hi Bunuel,

I think Statement 1 is sufficient, as we knew that during 5 days they will check totally 50 computers, out of which 15 are infected with virus and another 35 (50-15) are not.
So the answer is 35/50=0.7


10 computers are scanned every day for 5 days does not mean that there are 50 computers, there can be more.
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Re: M14-27  [#permalink]

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New post 07 Dec 2016, 08:49
If you go by the distinct formulation, actually none of the statements is sufficient.
Since the scan is arbitrarily, it could happen that the exact same computers are scanned throughout the 5 day process.

(1) Scanning the same 10 computers each day: You could get a virus rate of 100% to 0% - depending on how many computers are in the network
(2) Scanning the same 4% of computers each day: You could get a virus rate of 100% to 0% - depending on how many computers are in the network

-> E, Statements (1) and (2) TOGETHER are NOT sufficient.
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Re M14-27  [#permalink]

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New post 23 Jan 2017, 00:57
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I think this is a high-quality question and the explanation isn't clear enough, please elaborate. The solution to DS question is clear.
But was the probability calculated?
Considering that 10 computers were scanned and separated from the lot each day, wouldn't the probability be:
235/250 * 225/240* 215/230 *205/ 220 *195/ 210??
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Re: M14-27  [#permalink]

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New post 11 Sep 2018, 18:34
smkashyap wrote:
Considering that 10 computers were scanned and separated from the lot each day, wouldn't the probability be:
235/250 * 225/240* 215/230 *205/ 220 *195/ 210??


I also agree with smkashyap. Bunuel can you please confirm. Thanks
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Re: M14-27 &nbs [#permalink] 11 Sep 2018, 18:34
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