How to find probabilty? A new computer virus attacks a folder. Each file gets damaged with the probability 0.2 independently of other files.


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*What is the probability that fewer than 11 files get damaged if the folder consisted of 15 files?

*What is the probability that probability that the forth file attacked will be the third one to get damaged?


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*I solved part 1 using Binomial distribution PMF for all values of
from 0 to 10. Which seemed like a brute force approach. Is there a
better way to solve this?

*I am clueless about the 2nd part.
 A: There's an easier way to do the first part. The number of files damaged in the folder with $n=15$ files could be $k=0,1,\ldots,13,14$ or $15$. You can find probabilities for each one. The number damaged being fewer than $11$ is the complement of the number being at least $11$. Thus, you can find probabilities for $k=11,12,13,14$ and $15$, add those up, and then subtract from $1$.
(You can also do this using the normal approximation to the binomial distribution, but you'll get an approximate answer, not an exact one. I can say more about that, if you're interested.)
Now, for the second part: If the fourth file attacked in the third to get damaged, that means two things. 1) Among the first three files, exactly $2$ were damaged, and 2) The $4$th file was damaged. Those probabilities are easy enough to find. The first is a binomial formula calculation, and the second is simply $0.2$.

EDIT: A binomial distribution, $\operatorname{Binom}(n,p)$ is approximately equal to a normal distribution $\operatorname{Norm}(np,npq)$. (That's a variance of $npq$, meaning a standard deviation of $\sqrt{npq}$.) In this case, we're working with $\operatorname{Norm}(3,2.4)$, which has standard deviation $\sqrt{2.4}$.
To correct for the fact that the binomial distribution is discrete, while the normal distribution is continuous, you can adjust the exact question we're asking. Since we want fewer than $11$ files to be infected, then we look for area under the normal curve between $-\infty$ and $10.5$.
Thus, we turn $10.5$ into a $z$-score: $z_0=\frac{10.5-3}{\sqrt{2.4}}\approx 4.84$. So, what's the probability that $z<z_0$? Well, looking at a standard normal distribution, we see that it is very close to $1$: it's about $0.99999935$. Doing this problem with the binomial distribution, we get $0.99998754$, so it's not a perfect approximation. Then again, we only have $n=15$. As $n$ gets larger, the approximation is better.
