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What is the probability that a student takes exactly one:

$P(S)+P(F)+P(G)$

originally i started out as $P(S)=P(S)-P(SF)-P(SG) \to P(S)=.28-.12-.04=.12$

$P(F)=P(F)-P(FS)-P(FG) \to P(F)=.26-.12-.06=.08$

Anyways following the same pattern gave me .26 but the answer is .32. The only conclusion that I could arrive at is I should have added $P(SFG)=.02$ at the end of all three. But why is that the case? Why when trying to find out exactly how many take one why would I would add the number that take all three? To me that just does not follow.

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It would have helped if you had been more complete in your answer so we could be sure what "FS", "FG" mean! I guess that "FS" means that student is taking both French and Spanishs and that "FG" means that student is taking both French and German. And you say that "28- S", "26- F", etc. I interpret that to mean that there are 28 students taking Spanish, 26 students taking French, etc. But how did you then get "P(S)= 0.28"? Are you told that there are a total of 100 students? You don't say that in your post.

Anyway, you subtract P(FS) and P(SG) to begin with because "F" includes not only those students who are taking only French but also those who are taking French and Spanish and those who are taking French and German. However, FS includes those students who are also taking German and FG includes those students who are also taking Spanish. By subtracting both P(FS) and P(FG) you have subtracted those student twice. You need to add P(FGS) to add them back in.

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    Because P(S) clearly not the sample space but Spanish so P(S) is the probability of someone taking Spanish which I specify as S – 2017-02-03