Binomial distribution to approximate 90 % successes A student is taking a multiple choice test where each question has four options for an answer. The student mastered mastered 70% of the material. Assume this means that the student has a 0.7 chance of knowing the correct answer to a random test question. On the other hand, if the student does not know the answer to the question, she randomly selects among the four answer choices. Finally, assume that this holds for each question independent of the others.
What is the probability that a specific question is answered correctly?
P(correct) = 0.7+0.25*0.3 = 0.775
Suppose that at least 90% of the choices has to be correct to pass the test.
If the test has 30 questions, approximate the probability that the student
will get at least 90% of the choices correct?
I tried using the binomial distribution to approximate this: 
P(90% correct)=$\binom{30}{27} 0.775^{27}  (1-0.775)^{30-27}$
$\binom{30}{27} = (30!)/((27!)((30-27)!)) = 4060 $
This gives me 0.047 as an result.
Is this the right way of approximating the probability that the student will get at least 90% of the choices correct? 4,7%

Set up our distribution:
"X has 30 trials, with the probability of success being 0.775".

$ X \sim B (30, 0.775) $

Now for our probability calculation (with 90% = 27 score):
"The probability of X being at least 27, is 1 subtract the probability of X being less than 27".
We do this because most conventional calculators use $ P(X<x) $ as the cumulative inequality, so we have to convert the inequality into this form. 

$ P(X \ge 27) = 1 - P(X < 27) = 1 - 0.9304 = 0.0696 $

Our final answer is 6.96% chance of getting at least 27 score, or 90%.
Your answer is correct for $ P(X=27) $, not $ P(X \ge 27) $.
