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If

  • People X are N times more likely to have attribute A than non-X

  • People X are P percent of the population

then

  • What percentage of A's are X?

Example (the numbers are just for illustration).

If

  1. A redhead is 3 times more likely to get skin cancer than a non-redhead.
  2. Redheads are 7% of the population.

Then

What percentage of skin cancer victims are redheads?

X = Is a Redhead, A = Suffer Skin cancer, N = 3, P = 7

Zhubarb
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Stephen Hosking
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1 Answers1

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I will go through the numerical example. We are given: $\frac{P(C|R)}{P(C|\lnot R)}=3$ and $P(R)=0.07$. We need to find $P(R|C)$.

Bayes Theorem:

$P(C|R)= \frac{P(R|C)P(C)}{P(R)} \quad P(C|\lnot R)= \frac{P(\lnot R|C)P(C)}{P(\lnot R)} $

Dividing these terms will yield:

$ \frac{P(C|R)}{P(C|\lnot R)}= 3 =\frac{P(R|C)}{P(\lnot R|C)}\frac{P(\lnot R)}{P(R)}=\frac{P(R|C)}{1-P(R|C)}\frac{1-0.07}{0.07}$

Then, you can solve for $P(R|C)$.

Monotros
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  • Many thanks for a well worked answer, @Ozan! I've upvoted it, but will have to work through it to confirm before awarding the bounty. I'm broadly familiar with the mathematics you've presented here (which I trust is correct), but rusty. I'll try to finish this within 24 hours and award the bounty. It looks lovely! – Stephen Hosking May 01 '19 at 01:48
  • @StephenHosking I hope it helps :) – Monotros May 01 '19 at 18:03
  • Once again, thanks for the solution, for invoking Bayes Theorem (I had thought it was a purely algebraic problem), and for presenting it so elegantly. Bonus awarded! – Stephen Hosking May 02 '19 at 12:03
  • Thank you. I am glad that it was useful for you. – Monotros May 02 '19 at 20:35
  • My algebraic manipulation solving for P(R|C) comes up with P(R|C) = A / 1 - A, where A = 1-0.07 / 0.07. Which is 0.18. I'll have to check it a couple of times, and try on other examples fr. – Stephen Hosking May 03 '19 at 06:23