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.

The correct answer is C.

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!

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?-
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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.

The correct answer is C.

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!

dzyubam says the answer is (235/250)^5
GMAT TIGER says the answer is (235c10 / 250c10)^5

Which is correct?

dzyubam says the answer is (235/250)^5
GMAT TIGER says the answer is (235c10 / 250c10)^5

Which is correct?

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.

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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.

Answer: C.

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.

Answer: C.

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