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A new disease has been discovered in the world, it is called Blue. Statistically 1:1000 people in the world has it and it makes your whole skin blue in a few weeks after getting the disease but not kill you somehow but you became like an blue alien. Moreover it is not known how it spreads.

Though if you can take antidote before becoming blue and while disease in your blood, you are most likely be saved from being bluefor the rest of your life. But if you take the antidote before getting disease, you will not be saved any more for sure since the disease becomes immune to the antidote after entering the body while antidote exists in your system.

There is a very very expensive lab test for that and the accuracy of that is 99%. In other words, 1% it will show you wrong result. You take the test and it seems you are positive.

If I take the antidote now,

What is the chance of becoming immune to the disease?

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    $\begingroup$ This is the strangest set-up for a probability question I've ever read. $\endgroup$ – hexomino May 14 at 9:35
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    $\begingroup$ Related: puzzling.stackexchange.com/questions/3792/… $\endgroup$ – GOTO 0 May 14 at 12:19
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    $\begingroup$ Why do I have the feeling that the real purpose of this question is to generate a chat page? $\endgroup$ – A. I. Breveleri May 14 at 13:56
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    $\begingroup$ This is almost a textbook stats problem. To make it the classic Pa given b, problem, we'd need to know the percent of the population that actually has the disease. $\endgroup$ – Chris Cudmore May 14 at 16:14
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    $\begingroup$ Uh... this one just feels really off. The setup is that your skin gets dark and that that's a bad thing? I would suggest changing it slightly since presumably the fact that people get darker is not important. Easily could change it to "Blue" etc. $\endgroup$ – Carley May 14 at 20:35
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I'd say about

one in eleven.

If you test

a thousand people, then, on average, one of them actually has the disease, and 10 will show false positives.

Since you tested positive, you are one of those people.

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You should

not take the antidote

because of

the base rate fallacy.

In numbers:

The chance that you have the disease and test positive is $\frac{99}{100}\cdot\frac{1}{1000}=\frac{99}{100000}$; the chance that you don't have it and test positive is $\frac{1}{100}\cdot\frac{999}{1000}=\frac{999}{100000}$. So the chance that the antidote works is $\frac{99}{99+999}$ which is only about 9%.

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    $\begingroup$ Is there another way that is more likely to know that you are infected? Because if not, a 9% chance vs 0% ... $\endgroup$ – Hermes May 14 at 9:51
  • $\begingroup$ Maybe if you have a second test whose results are independent of the first test... $\endgroup$ – Glorfindel May 14 at 9:59
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Just for the sake of it, the solution in terms of Bayes Theorem:

We are interested in the chance of beeing ill, given a positive test result: $P(dark | test_+)$

From the Question it is known:

$P(dark)=0.001$
$P(test_+|dark)=0.99$ $P(test_+)=P(dark)*P(test_+|dark)+P(\overline{dark})*P(test_+|\overline{dark}) = 0.010089$

Now, its just plug and play:

$P(dark | test_+)= \frac{P(test_+|dark)*P(dark)}{P(test_+)} = 0.0981$

So we can conclude:

With a chance of beeing ill at merly 9.81%, you should not take the antidote!

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    $\begingroup$ To clarify the notation (for those not familiar with it), P(x|y) means "the probability of x happening, given that y has happened) $\endgroup$ – Dancrumb May 14 at 15:43
  • $\begingroup$ Glorfindel's answer uses Bayes' Theorem. He just doesn't say that explicitly, nor does he explain which of his factors are conditional or marginal probabilities. $\endgroup$ – Bridgeburners May 14 at 19:46
  • $\begingroup$ @Bridgeburners, No, he calculated the conditional probability. Bottom line is the same, but Bayes theorem happens to be defined in terms of conditional or marginal probabilities $\endgroup$ – Lucas May 14 at 19:59
  • $\begingroup$ @Lucas He is clearly using it. It may not be immediately apparent because he cancelled the factor of 1/100000 in both the numerator and demonimator of the final computation. But, as you can plainly see, the numerator is the result of calculating P(pos | dis) * P(dis) and the denominator is P(pos) which he took by marginalizing over disease and non-disease in a previous step. $\endgroup$ – Bridgeburners May 15 at 13:52

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