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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 M1427
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16 Sep 2014, 00:54
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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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
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Re: M1427
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05 Jul 2016, 11:19
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 (5015) are not. So the answer is 35/50=0.7



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Re: M1427
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05 Jul 2016, 12:16
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 (5015) 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: M1427
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07 Dec 2016, 09:49
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 M1427
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23 Jan 2017, 01:57
I think this is a highquality 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: M1427
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11 Sep 2018, 19:34
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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Bunuel wrote: 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 Hi Bunuel, I would really appreciate if you could elaborate the last step? Q is asking that what is the probability that no computer containing the virus will be scanned in the course of the next five days? So, out of total 250, we know infected = 15 & not infected = 235 and 10 are scanned everyday. Why are we using Combinatorics in the final step of calculating Probability ? Can you please explain? THANKS IN ADVANCE!



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Re: M1427
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25 Mar 2019, 01:12
JIAA wrote: Bunuel wrote: 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 Hi Bunuel, I would really appreciate if you could elaborate the last step? Q is asking that what is the probability that no computer containing the virus will be scanned in the course of the next five days? So, out of total 250, we know infected = 15 & not infected = 235 and 10 are scanned everyday. Why are we using Combinatorics in the final step of calculating Probability ? Can you please explain? THANKS IN ADVANCE! JIAA Let's expand on the premises which you have got correctly! Qstatement:For the 5 days: No virus should be scanned. NV: No virus V: Virus NV NV NV NV NV V Probability = Desired value/ Total value So, out of total 250, we know infected = 15 & not infected = 235 and 10 are scanned every day. The Total value: We need to select and scan 10 out the 250 computers irrespective of whether they have a virus: Thus, the selection of 10 out of the 250. \({C^{10}_{250}}\) The Desired value: We need no virus detection for each of the scans for the next 5 days. Hence, subtracting the infected ones from the total: \(25015 = 235\) Now, we need to select and scan 10 out of the remaining 235 (noninfected ones) such that no scan results in a virus. Thus, the selection of 10 out of the 235. \({C^{10}_{235}}\) To answer this:Quote: Why are we using Combinatorics in the final step of calculating Probability We are using combinations to calculate: 1) All the possible iterations to fetch the total value. 2) All the desired values to fetch the possible favorable iterations.
and then to substitute in the formulae:
Probability = Desired value/ Total value



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Xylan wrote: JIAA wrote: Bunuel wrote: 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 Hi Bunuel, I would really appreciate if you could elaborate the last step? Q is asking that what is the probability that no computer containing the virus will be scanned in the course of the next five days? So, out of total 250, we know infected = 15 & not infected = 235 and 10 are scanned everyday. Why are we using Combinatorics in the final step of calculating Probability ? Can you please explain? THANKS IN ADVANCE! JIAA Let's expand on the premises which you have got correctly! Qstatement:For the 5 days: No virus should be scanned. NV: No virus V: Virus NV NV NV NV NV V Probability = Desired value/ Total value So, out of total 250, we know infected = 15 & not infected = 235 and 10 are scanned every day. The Total value: We need to select and scan 10 out the 250 computers irrespective of whether they have a virus: Thus, the selection of 10 out of the 250. \({C^{10}_{250}}\) The Desired value: We need no virus detection for each of the scans for the next 5 days. Hence, subtracting the infected ones from the total: \(25015 = 235\) Now, we need to select and scan 10 out of the remaining 235 (noninfected ones) such that no scan results in a virus. Thus, the selection of 10 out of the 235. \({C^{10}_{235}}\) To answer this:Quote: Why are we using Combinatorics in the final step of calculating Probability We are using combinations to calculate: 1) All the possible iterations to fetch the total value. 2) All the desired values to fetch the possible favorable iterations.
and then to substitute in the formulae:
Probability = Desired value/ Total value Xylan THANKS for the detailed explanation! Going forward i'll tag you in Qs i'm not clear in understanding! Maybe we can help out each other with the prep ! Thanks again!



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Re: M1427
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25 Nov 2019, 04:18
harsh8686 wrote: 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 Bunnel, I am also struck here.Can you please explain!!



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Re: M1427
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19 Dec 2019, 12:16
prabsahi06 wrote: harsh8686 wrote: 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 Bunnel, I am also struck here.Can you please explain!! Hi, I am not expert but will try my best. Nowhere, It is mentioned that the computers those are scanned separated from the lot that's why we will not use this "235/250 * 225/240* 215/230 *205/ 220 *195/ 210". In above one we are decreasing the no of computers but it is not mentioned anywhere that computers are separated from lot after scanning. We will have to solve in the same way like Bunuel did as every day is independent of previous day hence we will have to multiply which is done through power i.e. 5. Hope, I will be able to clear your doubt. Thanks










