# 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,

1. if the test comes back positive?
2. 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.

• 3.47% is the right answer here... – Clement C. Apr 29 '15 at 13:33
• You're computing the probability of a false negative not the probability someone doesn't have asthma. False negatives are rare and should be rare intuitively. Otherwise, the test would be a piece of crap. So, you're right. – Jamie Lannister Apr 29 '15 at 13:34
• I don't understand what you mean. If 3.47% is the correct answer, this is quite reasonable. It says: "If the test says you don't have asthma, then it's very likely you really don't have asthma (i.e. the test is good)." The first answer is actually more surprising. It says: "If the test says you have asthma, it's only 25% likely you actually have it!" But that isn't that surprising given the low figure of 68% in the hypotheses. – Frank Apr 29 '15 at 13:34
• @iterence: that would not give the probability $\Pr[A\mid -]$, but instead the quantity $\Pr[A^c\mid +] = 1 - \Pr[A\mid +]$. That is, instead of "the probability to have asthma knowing the test says you don't," you would compute "the probability not to have asthma, knowing the test says you do." – Clement C. Apr 29 '15 at 13:38
• In part 2, "if the test comes back negative", it is assumed we know that the result was negative, i.e., the probability of the test returning negative is $100$%, not $3.47$%. The $3.47$% figure arises from the cases where people do have asthma but it is not detected by the test. – David K Apr 29 '15 at 13:41

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.