Answer does not make sense using conditional probability As of now, there are 64.1 million people residing in the UK. 5.4 million of them are thought to be asthmatic. A new test for asthma has recently been introduced and medical trials indicate that


*

*in patients with asthma, the test correctly returns positive 68% of the time

*in patients that does not suffer from asthma, the test correctly returns 82% of the time.


Assuming a patient undergo medical inspection. What is the probability that the patient has asthma,


*

*if the test comes back positive?

*if the test comes back negative?


For Part 1, I used this method:
Step 1:
$$
P(A|+)=\frac{P(+|A)P(A)}{P(+)}
$$
Step 2:
$$
P(+)=P(+│A)P(A)+P(+│N)P(N)=(0.68)\left(\frac{54}{641}\right)+(0.18)\left(\frac{587}{641}\right)=\frac{7119}{32050}
$$
Step 3:
$$
P(A│+)=\frac{P(+|A)P(A)}{P(+)}=\frac{0.68\times\frac{54}{641}}{\frac{7119}{32050}}=\frac{204}{791}=25.79\%
$$
For Part 2, I used the same method as Part 1 but I got a very small answer. Assuming if I insert all the values in the equations correctly, would it be sensible if my answer for Part 2 is 3.47%?
However, personally, it does not make logical sense if the probability of test returning negative is 3.47% because that would mean almost everyone in the nation would be asthmatic.
Or I could just write it as 
$$
100-25.79 = 74.21%
$$
but I'm afraid this isn't the answer given the complexity of the question.
 A: One instructive way to do these kinds of $2 \times 2$ problems, since there are only four distinct populations, is to enumerate them and calculate the marginal probabilities by inspection.
$$
\begin{array}{|c|c|c|c|} \hline
& \text{asthmatic} & \text{not asthmatic} & \text{TOTALS} \\ \hline
\text{positive} & 3.672 & 10.566 & 14.238 \\ \hline
\text{negative} & 1.728 & 48.134 & 49.862 \\ \hline
\text{TOTALS} & 5.400 & 58.700 & 64.100 \\ \hline
\end{array}
$$
From this, one can read off
$$
P(\text{asthmatic} \mid \text{positive}) = 3.672/14.238 \doteq 0.25790
$$
and
$$
P(\text{asthmatic} \mid \text{negative}) = 1.728/49.862 \doteq 0.34656
$$
ETA: Two of the cells—$3.672 = (0.68)(5.400)$ and $48.134 = (0.82)(58.7)$—are filled directly by the given parameters.  Everything else is bookkeeping.
The abysmal false positive rate is due to a combination of the low specificity of the test ($82$ percent) and the relatively low prevalence of asthma ($5.4/64.1 \doteq 8.4$ percent), so that although only a minority (sizable, but still a minority) of non-asthmatics test positive, they still dwarf the asthmatics who test positive, because there are so few asthmatics available to test positive in the first place.  Even if the test were $100$ percent sensitive, the false positive rate would still have been $10.566/(10.566+5.400) \doteq 66$ percent.
