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16 Sep 2014, 00:54
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C
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
35%
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Question Stats:
71% (01:22) correct
29%
(01:27)
wrong
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15 computers in a corporate network are infected with a virus. If several computers are chosen at random for scanning each day, what is the probability that none of the virus-infected computers will be scanned within the next five days?
(1) 10 computers are scanned daily.
(2) Each day, 4% of the total number of computers in the corporate network are scanned.
(1) 10 computers are scanned daily.
(2) Each day, 4% of the total number of computers in the corporate network are scanned.
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25 Mar 2019, 01:12
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JIAA
JIAA Let's expand on the premises which you have got correctly!
Q-statement:
- 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: \(250-15 = 235\)
Now, we need to select and scan 10 out of the remaining 235 (non-infected 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:
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
General Discussion
16 Sep 2014, 00:54
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Official Solution:
15 computers in a corporate network are infected with a virus. If several computers are chosen at random for scanning each day, what is the probability that none of the virus-infected computers will be scanned within the next five days?
In order to determine the probability, we need to know the number of computers in the network and the number of computers scanned every day. It is important to note that the question does not specify that a different set of computers are scanned each day, so it is possible that the same computers could be scanned on multiple days.
(1) 10 computers are scanned daily.
We are given the number of computers scanned every day, but the number of computers in the network is unknown. Not sufficient.
(2) Each day, 4% of the total number of computers in the corporate network are scanned.
The above implies that \(0.04 * total = scanned\). This information alone is not sufficient.
(1) + (2) Given that 10 computers are scanned every day and that \(0.04 * total = 10\), we can find the total number of computers to be 250. Knowing both the total number of computers (250) and the number of computers scanned daily (10), we can now calculate the probability. Each day, the probability that none of the virus-infected computers will be scanned is the same and can be calculated as \(\frac{C^{10}_{235}}{C^{10}_{250}}\). The probability that none of the virus-infected computers will be scanned in the next five days will then be \(P = ( \frac{C^{10}_{235}}{C^{10}_{250}} )^5\). Sufficient.
Answer: C
15 computers in a corporate network are infected with a virus. If several computers are chosen at random for scanning each day, what is the probability that none of the virus-infected computers will be scanned within the next five days?
In order to determine the probability, we need to know the number of computers in the network and the number of computers scanned every day. It is important to note that the question does not specify that a different set of computers are scanned each day, so it is possible that the same computers could be scanned on multiple days.
(1) 10 computers are scanned daily.
We are given the number of computers scanned every day, but the number of computers in the network is unknown. Not sufficient.
(2) Each day, 4% of the total number of computers in the corporate network are scanned.
The above implies that \(0.04 * total = scanned\). This information alone is not sufficient.
(1) + (2) Given that 10 computers are scanned every day and that \(0.04 * total = 10\), we can find the total number of computers to be 250. Knowing both the total number of computers (250) and the number of computers scanned daily (10), we can now calculate the probability. Each day, the probability that none of the virus-infected computers will be scanned is the same and can be calculated as \(\frac{C^{10}_{235}}{C^{10}_{250}}\). The probability that none of the virus-infected computers will be scanned in the next five days will then be \(P = ( \frac{C^{10}_{235}}{C^{10}_{250}} )^5\). Sufficient.
Answer: C
