Could someone double check my answers for me? They differ substantially from the text and I'm not sure why.

An elementary school is offering 3 language classes: one in Spanish, one in French, and one in German. The classes are open to any of the 100 students in the school. There are 28 students in the Spanish class, 26 in the French class, and 16 in the German class. There are 12 students that are in both Spanish and French, 4 that are in both Spanish and German, and 6 that are in both French and German. In addition, there are 2 students taking all 3 classes.

(a) If a student is chosen randomly, what is the probability that he or she is not in any of the language classes?

the probability that someone is in a language class should just be the total number of students in each class over the total number of students,

$\displaystyle \frac{28+26+16}{100}=\frac{70}{100}$

The question wants the compliment of this, so there's a 30% chance of selecting a non-language student from the school.

The book says .5

(b) If a student is chosen randomly, what is the probability that he or she is taking exactly one language class?

This answer should just be the quotient, of the sum of each class, minus duplicates, over the total. So take for example the Spanish class, of the 28 total students, 12 are taking French too, 4 are taking German as well, and 2 are in all three classes; so there are only 10 students taking only Spanish.

$\displaystyle \frac{(28-12-4-2)+(26-12-6-2)+(16-6-4-2)}{100}=\frac{10+6+4}{100}=\frac{20}{100}$

The given answer is .32

(c) If 2 students are chosen randomly, what is the probability that at least 1 is taking a language class?

So if you randomly select 2 students there are 4 outcomes only one of which involves both students not in a language class. I should just be able to subtract the probability of both students not being in a language class, in decimal forum, from 1 and have my answer?

$\displaystyle 1-0.2^2=0.96$

Here the text gives a solution of 149/198