Probability of getting exam in specific course with 8 courses and two exams A person takes 8 courses and is randomly chosen to take an exam in 2 of those courses. All courses have the same probability of being chosen. 
What is the probability that the person is chosen for a specific course? 

Me and my friend can't agree which is the correct answer:
1) $\frac{1}{8} + \frac{1}{7}=\frac{15}{56}\approx 26.79\%$
Reasoning: Imagine all the courses are in a bowl. When the exams are drawn, the principal first picks one course from the bowl with 8. Then he picks another course, but now there are only 7 courses to pick from. 
2)  $\frac{1}{8}\times 2=\frac{1}{4}=25\%$
Reasoning: If there was just a single exam, the probability of drawing a specific course would be $\frac{1}{8}$. Now there are two exams, so one must simply multiply with the number of exams. 
3) Something else entirely 
We have tried Googleing for a very long time, but couldn't find the answer.
Edit: Replaced $\times$ with $+$ in first calculation (typo) 
 A: P(A) = 1-(P(not-A)). P(not-A) is usually denoted P(¬A).
The probability of drawing a specific course having an exam is the negation of that specific course not having an exam. https://en.wikipedia.org/wiki/Probability#Summary_of_probabilities Row 2.
What is the probability a specific course doesn't have an exam?
Well, if there's a hat with 8 pieces of paper in it, one with the name of each course, and we draw two papers from it, the probability we don't draw a given course (the 1st time) is 7/8. Since the drawn course isn't replaced, and the specific course is still in the hat, the probability of not drawing it the second time is (7-1)/(8-1) which is 6/7.
We multiply those probabilities together to find the probability of them both happening (the union): 7/8*6/7 = 6/8 = 3/4
Now we've found the probability of not getting an exam in a given course, the probability of getting an exam is the negation: 1 - 3/4 = 1/4.
TL;DR:
P(Course A has an exam) = 1-P(Course A doesn't have an exam)
P(Course A doesn't have an exam) = 7/8*6/7 = 6/8
P(Course A has an exam) = 1 - (6/8) = 2/8 = 1/4
It's (2).
It helps to draw it out with venn diagrams like the one on this page: https://en.wikipedia.org/wiki/Complement_(set_theory)#Absolute_complement
A: It's (2).
The error in the reasoning in (1) is as follows (by the way, I assume you meant $\frac18+\frac17$, not $\frac18\times\frac17$). The principle first picks one course from the bowl, and it's the specific one with probability $\frac18$. Then he picks another course from the bowl. If he didn't pick the specific course first -- an event which happens with probability $\frac78$ -- then he has a $\frac17$ chance of picking the specific one second. So the total probability is $\frac18+\frac78\times\frac17 = \frac18+\frac18 = \frac14$.
A: The second answer is correct.  The mistake in the first calculation is that you have to condition on the specific subject's not having been picked first.  The probability is $$
\frac{1}{8}+ \frac{7}{8}\cdot\frac{1}{7}=\frac{1}{4}$$
The factor of $\frac{7}{8}$ represents the probability that the given subject wasn't picked first.
A: Note: this answer provides a different way from the other two answers published here to solve the problem. However, this is not the simplest method to do it (or at least it is more convoluted than the other two answers here, even if it works fine). I published it just to have some different way to approach the problem and because this is a common way to solve such problems in probability.
You might ask how many ways are there to pick up two exams in that way, i.e., given the list of all possible exams, say $\{e_1,e_2,e_3,e_4,e_5,e_6,e_7,e_8\}$, how many sets with two exams you can possibly have. This number can be calculated with the Newton's binomial coefficient and it's equal to $$\binom{8}{2} = \frac{8!}{6!\cdot2!} = 28$$
Then we need to know how many of these sets contains the fixed exam $e_i$. And this is easy because you can just "fix $e_i$ in place" and then make the other "available seat inside the set" vary with the 7 remaining exams. So we have $7$ possible sets that contains the exam $e_i$.
Therefore the probability is $$\frac{7}{28} = \frac{1}{4}$$ 
A: Another way is to consider the 28 ways that 2 can be chosen from 8. 
$8!/(2!(8-2)!) = 28$. 
Of that 28, 7 of them will contain any one specific exam. That is, any one exam will pair with the other 7.
$7/28 = 1/4$
