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!