2

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

spiros
  • 59
  • 1
    1, You're not showing any effort; 2, You're either too lazy to press shift, or, just plain copy and paste; 3, more precisely this is a "probability" problem; 4, obviously you never learned alphabet: a$\to$b$\to$c$\to$c –  Jul 11 '13 at 02:36
  • i have solved so many probability problems today, for my test tomorrow but this seems to be giving trouble. I am sorry i didnt say anything. – spiros Jul 11 '13 at 02:39
  • heres what i thought – spiros Jul 11 '13 at 02:39
  • Please add your thoughts or attempts to the question by clicking "edit" –  Jul 11 '13 at 02:40
  • 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) – spiros Jul 11 '13 at 02:41
  • Alright, that's better, I'll give you an upvote instead of a downvote then ... –  Jul 11 '13 at 02:50

1 Answers1

1

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.

Brian M. Scott
  • 616,228
  • how do you know that the 5 of the 10 english majors were among the 20 who failed the bio? – spiros Jul 11 '13 at 02:59
  • so b) would be just 80/90? – spiros Jul 11 '13 at 03:03
  • @george: There are $20$ English majors who failed math, and there are $5$ English majors who failed both math and biology. Every one of the $5$ who failed both failed math, so all $5$ of them are among the $20$ who failed math. – Brian M. Scott Jul 11 '13 at 03:04
  • @george: For (b) you’re right that $90$ passed math. However, it’s not true that $80$ of these passed biology. There are $20$ English majors who failed biology, but $5$ of them also failed math, so there are $20-5=15$ who failed biology but passed math. There are another $5$ who failed both, and (as we saw earlier) $5$ who passed math and failed biology. That’s a total of $15+5+5=25$ who failed one or both of math and biology. How many does that leave who passed both math and biology? – Brian M. Scott Jul 11 '13 at 03:10
  • 25/90. for c) is it 15/20? – spiros Jul 11 '13 at 03:15
  • @george: Whoa! If $25$ of the $100$ failed at least one of math and biology, then $100-25=75$ passed both. Since $90$ passed math, the probability that an English major passed both given that he passed math is $\frac{75}{90}=\frac56$. Your answer for (c) is right, however. – Brian M. Scott Jul 11 '13 at 03:17