M14-27

Question

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.

Bunuel · Answer

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

smkashyap · Answer

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??

JIAA · Answer

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!

XyLan · Answer

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:  
 
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

carma19 · Answer

I think this is a  high-quality question  and I agree with explanation.

Basshead · Answer

The computers can be scanned again the following day. If the computer is scanned on day 1, it can still be scanned again on day 2.

KarishmaB · Answer

The question says that "every day several computers are arbitrarily selected for scanning".

It does not say that the ones scanned today are set aside. Every day 10 are arbitrarily selected (say by generating random numbers) and then scanned. Next day, again 10 are arbitrarily selected and scanned. It is possible that one computer gets scanned on two consecutive days.

So if we have total 250 computers, 235 of them clean and 15 infected and we select and scan 10 every day, the probability that only clean computers are scanned on any given day is
= 235C10 / 250C10
This probability stays the same every day.

Bunuel · Answer

I have edited the question and the solution by adding more details to enhance its clarity. I hope it is now easier to understand.

alamin8 · Answer

I like the solution - it’s helpful.

Hbsvik · Answer

Sample reset each day so 235 will not decrease to 225 215

M14-27

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