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I'm not sure how to best phrase this, but I'm not sure how to combine two probabilities (or if that even makes sense).

Example:

Assume that 5% percent of the population has a particular gene is 20 times more likely to become a professional athlete than those without. Considering all athletes, what percentage of them has this gene?

It seems like the percentage of athletes would be greater than 5%, but I'm not sure how to figure this out. Is this something that can be determined?


Note: This seems super simple, so it might have already been asked. I searched a lot, but I don't have the right language for this field yet.

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    $\begingroup$ bayes' theorem and the law of total probability together is sufficient to solve this $\endgroup$ – shimao May 17 '18 at 20:35
  • $\begingroup$ Thanks @shimao, I wrote up an answer using those two. I think it makes sense. Can you look it over for correctness? $\endgroup$ – tylerware May 17 '18 at 22:00
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Per suggestion of shimao, the answer lies in the Bayes theorem and the Law of Total Probability.

Let $G$ be the event that someone has the gene and let $A$ be the event that someone becomes a professional athlete. From the example we know:

$$ P(G) = 0.05 \\ P(G^\prime) = 0.95 \\ P(A|G)/P(A|G^\prime) = 20 $$

What we want to know is what $P(G|A)$. Enter Bayes Theorem

$$ P(G|A) = \frac{P(A|G)P(G)}{P(A)} $$

Which has two unknowns $P(A|G)$ and $P(A)$, using the Law of Total probability we can find $P(A)$.

$$ P(A) = P(G)P(A|G) + P(A|G^\prime)P(G^\prime) $$

From $P(A|G)/P(A|G^\prime) = 20$ we can get $P(A|G^\prime) = P(A|G)/20$ and can substitute in to get:

$$ P(A) = P(G)P(A|G) + \frac{P(A|G)P(G^\prime)}{20} $$

Returning to our application of Bayes Theorem and substituting in this expression of $P(A)$ we get:

$$ P(G|A) = \frac{20\cdot P(A|G)P(G)}{P(A|G)(20 \cdot P(G) + P(G^\prime))} $$

Which simplifies to $$ P(G|A) = \frac{20 \cdot P(G)}{20 \cdot P(G) + P(G^\prime)} $$

all are known, so $$ P(G|A) = \frac{20 \cdot 0.05}{20 \cdot 0.05 + 0.95} \approx 0.51 $$

So the percentage of professional athletes that have this gene is 51%.

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  • $\begingroup$ You're missing an extra 20 in your denominator after combining the two. $$P(G|A) = \frac{20P(A|G)P(G)}{P(A|G)(20P(G) + P(G^\prime))}$$ $\endgroup$ – dankernler May 17 '18 at 23:31
  • $\begingroup$ @dankernler Fixed, thank you for pointing this out :) $\endgroup$ – tylerware May 18 '18 at 16:43
  • $\begingroup$ No problem. I haven't done one like this in a while, but the answer of 1 was a red flag! $\endgroup$ – dankernler May 18 '18 at 17:30

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