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