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

- anonymous

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- anonymous

this is often very confusing
one good way to approach this is to use some actual numbers and see what happens

- anonymous

it makes "baye's theorem" far more undersantable

- anonymous

Okay but how do I choose my numbers?

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## More answers

- 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

okay

- 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

Umm...40? I'm not sure I did this right

- anonymous

yes that is right

- anonymous

oops sorry that is wrong

- anonymous

dang it

- anonymous

2% of 1,000 is \(.02\times 1000=20\)

- anonymous

that is why i picked a nice round number like 1000 so the it would be easy to compute the percents

- anonymous

That what I did but my calculator said i was 40...oh well what do I have to do next?

- anonymous

ok so 20 people have the allergy, how many do not?

- anonymous

By the way that already helps a lot

- anonymous

20 out of 1000 which is... 20/10000=0.2%?

- anonymous

hold on i think i have confused you

- 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

ooooh, sorry with the 10000 I got confused with another problem on my paper sorry.
Umm so 980 don't have the allergy?

- anonymous

right

- 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

of the 20 people that have it, the test is 98% accurate
what is 98% of 20?

- anonymous

btw i hope it is clear that you do \(.98\times 20\)

- anonymous

19.6?

- anonymous

right

- anonymous

yeah I'm good with the percentage....most of the time

- 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

now we know \(980\) so NOT have the allergy, lets see how many of those will also test positive

- 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

oops bad math

- anonymous

4% will test positive
what is 4% of 980

- anonymous

39.2?

- anonymous

yeah i get that too

- anonymous

ok so now how many people total (it is a decimal) will test positive?

- anonymous

0.04 people? I'm confused

- 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

oh, 58.8?

- anonymous

right, that is the total number that test positive

- anonymous

out of those, we know that 19.6 actually have the allergy

- anonymous

okay

- 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

or whatever you get when you write that as a decimal, i get \(0.337\) rounded, a little more than a third

- anonymous

okay I get 0.33333etc.

- anonymous

oh maybe i put the wrong numbers in
yeah you are right

- anonymous

one third in other words

- 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

yeah, it would help a lot, my teacher just throws a formula and exercises at us

- anonymous

ok do you have the formula you are supposed to use?
i can make a guess if you like

- anonymous

yeah I have it

- anonymous

go ahead and write it, we can walk through it slowly

- anonymous

It's \[P(A|B)=\frac{P(A|B)*P(A)}{ P(B)}\]

- anonymous

ok good

- anonymous

I have it right in front of me

- 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

oh crap your definition is wrong, but we will get to that in a second

- anonymous

It means probability of A given B

- anonymous

ok good

- anonymous

and the definition (not baye's theorem) is
\[P(A|B)=\frac{P(A\cap B)}{P(B)}\]

- anonymous

My teacher made us study about a formula like that, but I really didn't get a thing

- 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

ok lets see if we can get it

- anonymous

oh yeah my mistake

- 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

That's where I get extremely confused

- anonymous

lets use our example
you want the probability you have the allergy GIVEN you test positive

- anonymous

okay

- 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

let me know if i lost you, we need to go slow i am sure

- anonymous

okay so far I think I'm okay...I think so for now

- 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

so far so good

- anonymous

but we were told \(P(B)\) the probabily you have the disease, that was \(2\%=0.02\)

- anonymous

okay

- 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

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

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

so far so good?

- anonymous

Wait why did you change the 98% to a decimal. to make it easier to work with?

- 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

ready to find the denominator?

- anonymous

okay

- 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

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

so you divided it by 1000 right?

- anonymous

we also need the number of people who test positive who do NOT have the disease

- anonymous

well, yes, but we just found again computing \(.02\times .98=.00196\)

- anonymous

to compute the probability that you test positive if you do not have the disease we used \(.04\times .980=.0392\)

- anonymous

okay

- anonymous

I'm sorry I really want to learn but I have to go to bed

- anonymous

now add together to get the probability you test positive, that is \(.00196+.00392=.00588\)

- 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

yeah it is late
you can look at this later and see that we are really using the same numbers as before

- anonymous

okay

- anonymous

later

- anonymous

later!

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