If a student goes through a multiple-choice exam with $20$ questions, each question having $5$ possible answers with only one of them being correct, what is the probability for the student to pass the exam (therefore getting a grade equal or greater than $10$) if each of her answers is totally random?
It seems rather simple as the law is $B(n,p)$ where $n=20$ and $p={1\over5}$.
My problem is that this law allows me to find $P(X=k)$ not $P(X\geq k)$. With usual small values of $n$ it's not an issue as I just need to sum all the probabilities.
If I want, for instance, to know the probability of a grade strictly smaller than $3$, I do $P(X=0)+ P(X=1) +P(X=2)$. And if I want the probability to get a grade larger or equal than $3$, I do $1 - [P(X=0)+ P(X=1) +P(X=2)]$. Fair enough.
But if I need to find the probability of a grade larger or equal to $10$, is my only option to do:
$$P(X\geq10) = P(X=10)+ P(X=11) +P(X=12)+P(X=13)+P(X=14)+P(X=15)+P(X=16)+P(X=17)+P(X=18)+P(X=19)+P(X=20)$$
Isn't there a quicker way I didn't think of (while only using binomial formulas)?
I searched the web but all the instances of that type (probability to pass a multiple-choice test) that I found are always limited to a very small amount of trials. So they always end up summing up one by one each probability. They don't say if there is a quicker method for larger amounts of trials.
PS: I tried to convert the binomial law to a normal one (using the expectation and standard deviation derived from the binomial law, and then converting it using the "T" formula) but it doesn't seem to help.
Thanks for your input.