Question 1160252: A lab network consisting of 20 computers was attacked by a computer virus. This virus enters each computer with probability 0.4, independently of other computers. Find the probability that it entered at least one computer. Ans: 0.99996
Found 2 solutions by solver91311, Theo:
Answer by solver91311(24713)   (Show Source):
You can put this solution on YOUR website!
The probability of at least one is the probability of exactly one, P(1), plus the probability of exactly two, P(2), plus P(3), plus P(4), ..., plus P(20).
Each of the listed probabilities is calculated with the formula:
\ =\ {{n}\choose{k}}\,p^k\(1\,-\,p\)^{n-k})
Where  is the desired number of successes,  is the total number of trials, and  is the probability of success on any given trial.
So the entire sum is:
\ =\ \sum_{r=k}^n\,{{n}\choose{r}}\,p^r\(1\,-\,p\)^{n-r})
and for your problem  ,  , and 
As should be readily apparent, this calculation would require a tremendous amount of tedious arithmetic. However, note that the probability of at least one success plus the probability of zero successes is equal to 1. Hence, the probability of at least one is equal to 1 minus the probability of zero successes, to wit:
\ =\ 1\ -\ P(0,20,0.4)\ = 1\ - \ {{20}\choose{0}}\,(0.4)^0\(0.6\)^{20})
Which reduces to:
(1)(0.6)^{20})
And I leave you with that little bit of calculator work.
John
My calculator said it, I believe it, that settles it
Answer by Theo(13342)  (Show Source):
You can put this solution on YOUR website! p = .4
q = 1 - p = .6
n = 20
p(x) = p^x * q^(n-x) * c(n,x)
this is the binomial probability formula you would use to solve this problem.
the probability that it entered at least 1 computer is equal to 1 minus the probability that it entered 0 computers.
p(0) = .4^0 * .6^20 * c(20,0) = 3.65615844 * 10^-5.
1 minus that = .9999634384.
that's your answer.
you can use your calculator to confirm.
all the probabilities, given n = 10, p = .4 and q = .6 are shown below.
the sum of all probabilities is equal to 1, as it should be.
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