M14-27

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