if 1:1000 of people is sick. the probability to be false positive is 0.07. if a person is sick there is not chance the test for the disease is wrong. If someone random is got a positive result, what are the chances he's actually sick?
I got 1.5% and wanted to check because I feel it should be more since the diagnose is never wrong. I took 1/1000 and divided it by 0.07+1/1000 and multiplied by 100 to get the percent. (by the formula for conditioned probability).
On
Probabilities have counter intuitive results all of the time. Its one reason why human beings are very weak at probability, and even sometimes the best statisticians get tripped up. Many of us mathematicians find it to be a very difficult subject to wrap your mind around because you always think your answers are wrong and end up second guessing yourself. This problem and problems like it are famous for their weird results.
If I understood you correctly... The probability of being sick is $P(S) = 0.001$. (by the way, dont write it 1:1000; this is how you write odds and odds are different than probabilities)
This implies that the probability of not being sick is $P(\neg S) = 0.999$
The rate of false positives is $P(+|\neg S)=0.07$.
Implying the rate of true negatives is $P(-|\neg S)=0.93$
And the rate of false negatives is $P(-|S)=0$
Implying the rate of true positives is $P(+|S) = 1$.
Thus the probability of being both sick and testing one way or the other: $P(-\wedge S) = 0 $ and $P(+\wedge S) = 0.001$
While the probability of being well but testing either way: $P(-\wedge \neg S) = 0.92907 $ and $P(+\wedge \neg S) = 0.06993$
Thus the probability of a random person testing positive is $P(+)=0.07093$
So $P(S|+) = 0.01409840688$.
Yes, so you are correct. If you test positive on such a high quality test you still stand a very little chance of actually being ill. Weird, huh?
This is why they (should) perform several confirmation tests afterwards.
Your calculations are not quite correct. $$ P(\text{sick} \, | \, \text{positive test}) = \frac{P(\text{positive test} \, | \, \text{sick}) P(\text{sick}) }{P(\text{positive test})} $$ We know $P(\text{positive test} \, | \, \text{sick}) =1$ and $P(\text{sick}) = 1/1000$. As for the denominator, $$ P(\text{positive test}) = P(\text{positive test} \, | \, \text{sick}) P(\text{sick}) + P(\text{positive test} \, | \, \text{not sick}) P(\text{not sick}) $$ We know $P(\text{positive test} \, | \, \text{sick}) P(\text{sick}) = P(\text{sick}) = 1/1000$ and $P(\text{positive test} \, | \, \text{not sick}) P(\text{not sick}) = (7/100)\cdot(999/1000)$. So it should be $$ P(\text{sick} \, | \, \text{positive test}) = \frac{0.001}{0.001 + 0.07\cdot0.999} \approx 0.0141 $$ or $1.41\%$.
But your question remains: this is surprisingly low, right?! The reason is that very few people have the disease! If $1000$ people get the test, on average only one will have the disease, but about $70 \approx .07\cdot999$ healthy people will test positive.