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There is a method to detect certain disease with 99% sensitivity and 90% specificity. And also it is known that 5% of the population has this disease.

What is the probability a random person selected from the population actually has this disease given that the method detects so?

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See Bayes theorem. – guy Feb 4 '12 at 17:00
Your header is Bayesian probability question, so you must know Bayes' rule, what exactly is your specific difficulty ? – true blue anil Feb 4 '12 at 17:31
@trueblueanil: I am not sure how to understand the reasoning behind sensitivity and specificity in this case. So can anyone please provide a complete solution? – littleEinstein Feb 4 '12 at 17:48
$\dfrac{\Pr(\text{ill})}{\Pr(\text{well})} \cdot \dfrac{\Pr(\text{positive} \mid \text{ill})}{\Pr(\text{positive} \mid \text{well})} = \dfrac{\Pr(\text{ill} \mid \text{positive})}{\Pr(\text{well}\mid\text{positive})}$. – Michael Hardy Feb 4 '12 at 19:07
up vote 2 down vote accepted

Sensitivity is the probability that someone that has the disease tests positive for it, and specificity is the probability that someone who does not have the disease tests negative. See here for a more detailed explanation. So if we let $X$ represent a positive test and $D$ represent the patient having the disease then $P(X|D)=0.99$ and $P(X|\neg D)=0.1$. Apply Bayes rule and take two asprin if symptoms persist.

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The binary nature of the "parameter space" $\{\text{ill},\text{well}\}$ makes this simple formuation possible:

$$\frac{\Pr(\text{ill})}{\Pr(\text{well})} \cdot \frac{\Pr(\text{positive} \mid \text{ill})}{\Pr(\text{positive} \mid \text{well})} = \frac{\Pr(\text{ill} \mid \text{positive})}{\Pr(\text{well}\mid\text{positive})}.$$

The numerator and denominator of the first fraction above are "prior probabilities"; those of the last are "posterior probabilities".

One may write it thus: $$ \operatorname{logit} \Pr(\text{ill}) + \log\frac{\Pr(\text{positive} \mid \text{ill})}{\Pr(\text{positive} \mid \text{well})} = \operatorname{logit} \Pr(\text{ill} \mid \text{positive}). $$

where $$ \operatorname{logit} p = \log \frac{p}{1-p}. $$ (Google the word "logit" and you'll see it.)

So $\dfrac{\Pr(\text{ill})}{\Pr(\text{well})}= \dfrac{5}{95}=\dfrac{1}{19}$ and $\dfrac{\Pr(\text{positive} \mid \text{ill})}{\Pr(\text{positive} \mid \text{well})} = \dfrac{99}{10}$. Plug in the numbers.

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deinst has explained sensitivity & specificity.

Here's a less technical way of solving.

P[true positive] = 99% of 5% = 4.95%

P[false positive] = 10% of 95% = 9.5%

P[true positive | positive] = 4.95/(4.95+9.5)

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