A dangerous computer virus attacks a folder consisting of 250 files. Files are affected by the virus independently of one another. Each file is affected with the probability 0.032. What is the probability that more than 7 files are affected by this virus?
I was looking at this problem and wondering if the answer to it is simply:
$P = (1 - 0.032^7)$
On
this is a classic binomial probability distribution question. You're asking to choose r from n with success & failure. Hence:
n=250 (number of trials)
r=7 (number of success or more)
p=0.032 (will be effected; probability of success in a signal trial)
q=1-.032=0.968 (will not be effected)
since you're looking for
more than 7
we're looking for the cumulative probability of P(x>7) so if we use the nice equation of Math Machine you'll get 0.549313779150176
When to use the binomial distribution (Yale University):
1: The number of observations n is fixed.
2: Each observation is independent.
3: Each observation represents one of two outcomes ("success" or "failure").
4: The probability of "success" p is the same for each outcome.
If 1 file is affected, there is a 250*.032*(1-.032)^249 probability, since there are 250 different ways that could happen 2 files would be 250*249*.032^2*(1-.032)^248 /2 probability, since there are 250 nCr 2 ways we could do this.
The Odds of any r amount getting the virus would be $$ \binom{250}{r} * .032^{r} *(1-.032)^{(250-r)} $$ Assuming we're using the nCr Function
Therefore, the odds that at least 7 would get it would be
$$1- \Sigma_{r=0}^7 \binom{250}{r} .032^{r}(1-.032)^{250-r} $$
Which yields us an answer of about 54.931377915%