anonymous
  • anonymous
I need someone to explain this to me, I really don't understand... Using Baye's Theorem: It is known that 2% of the population has a certain allergy. A test correctly identifies people who have the allergy 98% of the time. The test correctly identifies people who do not have the allergy 94% of the time. A doctor decides that anyone who tests positive for the allergy should begin taking anti-allergy medication. Do you think this is a good decision? Why or why not? I know it's pretty long and a little bit of reading but it's on paper so I had to copy it, I wouldn't have if I didn't need the
Mathematics
jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
this is often very confusing one good way to approach this is to use some actual numbers and see what happens
anonymous
  • anonymous
it makes "baye's theorem" far more undersantable
anonymous
  • anonymous
Okay but how do I choose my numbers?

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anonymous
  • anonymous
i would pick a large number for the population, so that we get all whole number answers (although once we get the answer, we can redo it without using that crutch) lets say the population is 1000 people ok?
anonymous
  • anonymous
okay
anonymous
  • anonymous
and if 2% of the population suffers from the allergies, how many people exactly will have it? i.e what is 2% of 1000 ?
anonymous
  • anonymous
Umm...40? I'm not sure I did this right
anonymous
  • anonymous
yes that is right
anonymous
  • anonymous
oops sorry that is wrong
anonymous
  • anonymous
dang it
anonymous
  • anonymous
2% of 1,000 is \(.02\times 1000=20\)
anonymous
  • anonymous
that is why i picked a nice round number like 1000 so the it would be easy to compute the percents
anonymous
  • anonymous
That what I did but my calculator said i was 40...oh well what do I have to do next?
anonymous
  • anonymous
ok so 20 people have the allergy, how many do not?
anonymous
  • anonymous
By the way that already helps a lot
anonymous
  • anonymous
20 out of 1000 which is... 20/10000=0.2%?
anonymous
  • anonymous
hold on i think i have confused you
anonymous
  • anonymous
we have a population of 1,000 2% have the allergy, so 20 have the allergy if out of 1,000 people, 20 have the allergy, how many (not what percent) do not have the allergy
anonymous
  • anonymous
ooooh, sorry with the 10000 I got confused with another problem on my paper sorry. Umm so 980 don't have the allergy?
anonymous
  • anonymous
right
anonymous
  • anonymous
so we know how many have it, and how many do not now lets see how many people will test positive for it
anonymous
  • anonymous
of the 20 people that have it, the test is 98% accurate what is 98% of 20?
anonymous
  • anonymous
btw i hope it is clear that you do \(.98\times 20\)
anonymous
  • anonymous
19.6?
anonymous
  • anonymous
right
anonymous
  • anonymous
yeah I'm good with the percentage....most of the time
anonymous
  • anonymous
i guess we have to work with the decimal, should have chosen a population of 10,000 then it would be whole numbers but too late now
anonymous
  • anonymous
now we know \(980\) so NOT have the allergy, lets see how many of those will also test positive
anonymous
  • anonymous
for them the test is 96% accurate so 96% of them will test negative, but that means 6% of them will test positive what is 6% of 980?
anonymous
  • anonymous
oops bad math
anonymous
  • anonymous
4% will test positive what is 4% of 980
anonymous
  • anonymous
39.2?
anonymous
  • anonymous
yeah i get that too
anonymous
  • anonymous
ok so now how many people total (it is a decimal) will test positive?
anonymous
  • anonymous
0.04 people? I'm confused
anonymous
  • anonymous
we computed two numbers of people that test positive right? the ones with the allergy, 19.6 and the ones without the allergy, 39.2 what is the total?
anonymous
  • anonymous
oh, 58.8?
anonymous
  • anonymous
right, that is the total number that test positive
anonymous
  • anonymous
out of those, we know that 19.6 actually have the allergy
anonymous
  • anonymous
okay
anonymous
  • anonymous
so the question is, IF you test positive, what is the probabilty you actually have the allergy that is the number of people who test positive and have the allergy, divided by the total number of people who test postive ie\[\frac{ 19.6}{58.8}\]
anonymous
  • anonymous
or whatever you get when you write that as a decimal, i get \(0.337\) rounded, a little more than a third
anonymous
  • anonymous
okay I get 0.33333etc.
anonymous
  • anonymous
oh maybe i put the wrong numbers in yeah you are right
anonymous
  • anonymous
one third in other words
anonymous
  • anonymous
would you like to redo this using "baye's formula" ? will will do pretty much the same arithmetic, and get the same answer
anonymous
  • anonymous
yeah, it would help a lot, my teacher just throws a formula and exercises at us
anonymous
  • anonymous
ok do you have the formula you are supposed to use? i can make a guess if you like
anonymous
  • anonymous
yeah I have it
anonymous
  • anonymous
go ahead and write it, we can walk through it slowly
anonymous
  • anonymous
It's \[P(A|B)=\frac{P(A|B)*P(A)}{ P(B)}\]
anonymous
  • anonymous
ok good
anonymous
  • anonymous
I have it right in front of me
anonymous
  • anonymous
first off, it this is going to make sense, you need to know what \(P(A|B)\) means do you know what it means?
anonymous
  • anonymous
oh crap your definition is wrong, but we will get to that in a second
anonymous
  • anonymous
It means probability of A given B
anonymous
  • anonymous
ok good
anonymous
  • anonymous
and the definition (not baye's theorem) is \[P(A|B)=\frac{P(A\cap B)}{P(B)}\]
anonymous
  • anonymous
My teacher made us study about a formula like that, but I really didn't get a thing
anonymous
  • anonymous
baye's theorem, at least one form, is not quite what you wrote, the condition is switched on the right it is \[P(A|B)=\frac{P(B|A)*P(A)}{ P(B)}\]
anonymous
  • anonymous
ok lets see if we can get it
anonymous
  • anonymous
oh yeah my mistake
anonymous
  • anonymous
ok now the numerator in your fraction \[P(B|A)P(A)\] is just another way to find \(P(A\cap B)\) the probability that both A and B occur
anonymous
  • anonymous
That's where I get extremely confused
anonymous
  • anonymous
lets use our example you want the probability you have the allergy GIVEN you test positive
anonymous
  • anonymous
okay
anonymous
  • anonymous
we can put A = you have the allergy, B = you test positive then we are trying to find \(P(A|B)\) the probability you have the allergy given that you test positive
anonymous
  • anonymous
let me know if i lost you, we need to go slow i am sure
anonymous
  • anonymous
okay so far I think I'm okay...I think so for now
anonymous
  • anonymous
ok to compute that we need the numerator \(P(A\cap B) \) i.e. the probability you have it AND test positive we were not told that number
anonymous
  • anonymous
so far so good
anonymous
  • anonymous
but we were told \(P(B)\) the probabily you have the disease, that was \(2\%=0.02\)
anonymous
  • anonymous
okay
anonymous
  • anonymous
and we were also told \(P(B|A)\) the probability you have the disease given that you test positive that was the \(98\%=0.98\)
anonymous
  • anonymous
to find the numerator, multiply those together, i.e. the probability you have the disease AND test positive is the probability you have the disease times the probability you test positive GIVEN that you have the disease, i.e. \[.02\times.98=.0196\]
anonymous
  • anonymous
you notice that is the same numerator we had before, just with the decimal moved over three places, because we did not start out with a population of 1,000 we just went straight for the number
anonymous
  • anonymous
so far so good?
anonymous
  • anonymous
Wait why did you change the 98% to a decimal. to make it easier to work with?
anonymous
  • anonymous
when doing any work in math, you use numbers, not percents \(98\%=0.98\) as a number percents are nice to look at, but think about what you did when you wanted to find 98% of 20 you multiplied 20 by .98 right, not by 98
anonymous
  • anonymous
ready to find the denominator?
anonymous
  • anonymous
okay
anonymous
  • anonymous
that is \(P(B)\) the probability you test positive remember how we found that before? we had to add two numbers, they were \(19.6\) and \(39.2\) they will be the same this time, but with the decimal in a different place
anonymous
  • anonymous
the fist one is the \(.00196\) we already found, that is the number of people who test positive given that they have the disease
anonymous
  • anonymous
so you divided it by 1000 right?
anonymous
  • anonymous
we also need the number of people who test positive who do NOT have the disease
anonymous
  • anonymous
well, yes, but we just found again computing \(.02\times .98=.00196\)
anonymous
  • anonymous
to compute the probability that you test positive if you do not have the disease we used \(.04\times .980=.0392\)
anonymous
  • anonymous
okay
anonymous
  • anonymous
I'm sorry I really want to learn but I have to go to bed
anonymous
  • anonymous
now add together to get the probability you test positive, that is \(.00196+.00392=.00588\)
anonymous
  • anonymous
you did help a lot already though I'll make sure I have a lot of time next time I ask a question though
anonymous
  • anonymous
yeah it is late you can look at this later and see that we are really using the same numbers as before
anonymous
  • anonymous
okay
anonymous
  • anonymous
later
anonymous
  • anonymous
later!

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