Question

A scientist is working on a new test for cotton allergies. She knows that 2.5% of all the students at Annie's school are allergic to cotton. For those who are allergic, 80% test positive and 20% test negative. For those who are not allergic, the test shows 10% positive and 90% negative.Use Bayes's theorem to find the probability that a student at Annie's school is allergic to cotton, given that the test results are positive.

Bayes Theorem: P(A l B)= [P(B I A) x P(A)]/ P(B)

Answer 1

I am not very sure about this answer.. but worth a try.. lets someone else confirm it..

Bayes Theorem: P(A l B)= [P(B I A) x P(A)]/ P(B)

1. Test on the students who she knew are allergic..
p(A) = 0.025
p(B) = 0.2
p(B/A) = 0.8

so, p(A/B) = [0.8 *0.025]/0.2 = 0.1

2. similarly find for students who she thought are not allergic and test was run on them..
3. Add the two figures..

Answer 2

Yes, you could use Bayes theorem:
\(P(A|B)=\large \frac{P(B | A) \times P(A)}{P(B)}\)
Let events
A=allergic, A'=not alergic
B=tests positive, B'=tests negative
and use Bayes Theorem to calculate
P(A|B), or Probability that a student is allergic to cotton, given the test results are positive.
From the given data,
2.5% are allergic, so \(P(A)=0.025\)
Out of the allergic, 80% test positive, i.e. 80% test positive GIVEN that they are allergic.
so \(P(B|A)=0.80\).
The last quantity we need is P(B), which is not directly given.
We will need the law of total probability to find P(B).
\(P(B)=P(B\cap \Omega)=P(B\cap (A+A'))=P(B\cap A)+P(B\cap A')\)
where
\(P(B\cap A)=P(B|A)*P(A)=0.8*0.025\) test positive and alergic
\(P(B\cap A')=P(B|A')*P(A')=0.1*0.975\) test positive and not alergic
using \(P(X|Y)=P(X\cap Y)/P(Y)\)
Putting it altogether you'd get the required probability P(B).

Answer 3

An easier way is to draw a contingency diagram (at least to check your results) from which we deduce right away that P(A|B)=0.02/0.11175=0.1702.
However, the question requires you to use Baye's theorem, so you still have to go through the motions.