Exam Probability A student has discovered that the probability of failing any exam is $p = .8$ if she studies less than 2 hours. Suppose she has three exams one day and studies only one hour for each. What is the probability that she fails two or more exams?
I have the feeling that I need to use the Binomial Distribution with $P (X \geq 2) = 1 - P(X = 0) - P(X = 1)$, using $n = 3$ for each, $k = 0, 1$ respectively for the two probabilities, and $p = 0.8$. Is this on the right track?
 A: Assuming the events are independent (which I would assume because they are exams on the same day) you and hunter were on the right track. There are four total events that can happen, but 8 different outcomes:
0. She can pass all the exams(there is one way this can happen)
1. She can fail one exam and pass two (there are 3 ways this can happen)
2. She can fail two exams and pass one (there are 3 ways this can happen)  
3. She can fail all 3 (there is one way this can happen)

There are two ways to solve this problem. 
You can add the probability of event 0 and 1 together and subtract the sum from 1:
P(e) = 1 - (P(0) + P(1))
P(0) = 0.2 * 0.2 * 0.2 * 1
P(1) = 0.2 * 0.2 * 0.8 * 3
P(e) = 1 - ((0.008) + (0.096))
P(e) = 0.896

Or you can add the probability of event 2 and 3 together:
P(e) = P(2) + P(3)
P(2) = 0.8 * 0.8 * 0.2 * 3
P(3) = 0.8 * 0.8 * 0.8 * 1
P(e) = (0.384) + (0.512)
P(e) = 0.896

They are the same and required almost the same amount of work. And just to double check if you add P(0), P(1), P(2), and P(3) together you get 1.
P(t) = (0.008) + (0.096) + (0.384) + (0.512) = 1

The big key here is knowing that even there are a total of 4 events happening, there are 8 different possibilities.
