SOLUTION: Suppose that of 100 computers checked for viruses 30 where infected and 70 where not. If we select 10 computers at random what is the probability that at most 4 of them are infecte

Question 995872: Suppose that of 100 computers checked for viruses 30 where infected and 70 where not. If we select 10 computers at random what is the probability that at most 4 of them are infected
Answer by Theo(13342) (Show Source): You can put this solution on YOUR website!
the probability that at most 4 of them are infected is equal to the probability that 0 of them are infected plus 1 of them are infected plus 2 of them are infected + 3 of them are infected + 4 of them are infected.

this uses the binomial probability formula which is:

p(x) = nCx * p^x * q^(n-x)

p is the probability of infected which is 30 / 100 = .3
q is the probability of not infected which is 1 - .3 = .7

for example:

the probability that exactly 1 out of the 10 will fail is 10C1 * .3^1 * .7^9.

p(1) is therefore equal to 10 * .3 * .040353607 = .121060821.

the sum of all the probabilities must be equal to 1 or you did something wrong.

here's the probabilities for all the possibilities.

in this spreadsheet p*(3) whould really have been just p(3).
the * is a typo.
my keyboard likes to add extra characters sometimes and i don't always catch it before publishing.

$$$

here's a reference on how to use the binomial probability formula.

http://regentsprep.org/regents/math/algtrig/ATS7/BLesson.htm

the combination formula can be referenced as nCx, C(n,x) or the form that is shown in the reference.

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