Statistics percentage problem! the english major at a university revealed the following: 
10% failed math, 20% failed biology, 5% failed both math and biology.
find prob:
a) failed math, given he passed bio
b) passed bio, given passed math
c)passed math, known that he failed bio
c)pass both courses\
p(a|b)= p(aU(upsidedown)b) / p(b) i would use that but i dont know the prob of a(failed math) and b (passing bio) it only gives the failures but even then i could use the a^c but still i wouldnt know the prob of thoe event together
 A: I’ll get you started by working (a). Imagine that there are $100$ English majors altogether. Then $10$ of them failed math, $20$ of them failed biology, and $5$ of them failed both math and biology. For (a) we want the probability that a randomly chosen English major who passed biology failed math. There are $100-20=80$ English majors who passed biology; in effect we’ve chosen one of these $80$ people at random. $10$ English majors failed math, but $5$ of those were among the $20$ who failed biology; that leaves $10-5=5$ who passed biology but failed math. Thus, among the $80$ English majors who passed biology are $5$ who failed math. The probability of getting one of them when we pick one of these $80$ people at random is $\frac5{80}=\frac1{16}=0.0625$.
You should wonder whether it’s legitimate to pick a particular size for the group of English majors. It is, because we’re actually working with fractions (or percentages) of the group of students. You could replace my $100$ English majors by $n$, my $20$ who failed biology by $0.2n$, and so on. If you did, and if you then carried out the analogous computations, you’d find that in the end all of the $n$’s cancelled out. In problems of this kind it’s always permissible to work with a universe of a specific size, which you can choose to make the arithmetic easy.
