A simple application of conditional probability

Question

I am practicing using conditional probability and have the following question

Suppose the probability of contracting a rare tropical disease Chickungunya on a holiday to India is 0.0001. Fred, on his return to the United Kingdom from a Jungle Safari in India, feels unwell and decides to see his doctor. Fred complains to his doctor of joint pain. The doctor had heard that there had been some cases of Chickungunya reported in India and also knew that 64% of people with Chickungunya complained of joint pain. There are other causes of Joint Pain however and the doctor estimates that the probability of joint pain in the absence of Chickungunya is 0.6. The doctor ordered a very reliable test used to diagnose Chickungunya that returned a positive result 99% of the time when the patient had Chickungunya and only returned a positive result 4% of the time when the patient tested was not suffering from the disease. The test result returned for Fred was positive.

Then I am asked three questions:

(a) Based on the evidence to date can the doctor conclude that Fred has Chickungunya?

(b) Realising that there are no other symptoms of the disease or any other more accurate tests available, the doctor decides to ask for the test to be repeated. Once again a positive test result is obtained. How, if at all, will this affect his belief in the diagnosis of Chickungunya?

(c) Can the strategy of repeating the same test over and over again improve his belief in a positive diagnosis of Chickengunya? If so, what is the minimum number of times that the test must be carried out, assuming a positive result each time, for the belief in Chickungunya to be greater than the belief in not Chickungunya?

Here is my attempted solution for part (a):

First I extracted all the information out of the given text using the notation $P(d),P(JP),P(+)$ as probability of disease, probability of joint pain and probability of getting a positive on the test respectively.

$$P(d)=0.0001\quad P(JP|d) = 0.64$$

$$P(JP|¬d) = 0.6 \quad P(+|d) = 0.99$$

$$P(+|¬d) = 0.04$$

$$P(d|(+, JP)) = \frac{P((+, JP)|d)P(d)}{P(+, JP)} = \frac{P(+|d)P(JP|d)P(d)}{P(+,JP)}$$

I have all the numbers above apart from $P(+, JP)$ but I figured you could calculate it by looking at disjoint cases of having the disease or not and get

$$P(+, JP) = P(d)P(+|d)P(JP|d) + P(¬d)P(+|¬d)P(JP|¬d)$$

Is this allowed?

Then the answer comes out as 0.0026.

For part (b) I know intuitively what is happening but I can't put it into symbols can anyone give me a suggestion?

I think if I can solve part (b) I should be able to work out (c).

Answer 1

The decomposition you used for part (a) is correct, since it is just an application of the Law of Total Probability.

$$P(+,JP) = P(+,JP|d)P(d) + P(+,JP|¬d)P(¬d)$$

$+$ and $JP$ can be assumed to be independent conditional on having or not having the disease, so your expression follows. (The test turning positive does not depend on having joint pain, given that the person has/does not have the disease)

For part (b) and (c), it can be assumed that any number of tests will be independent of each other (again conditional on having/not having the disease). Let $+_n$ denote the event that $n$ tests have been performed, and all have a positive result. Then using this idea of independence,

$$P(d|+_n,JP)=\frac{P(+_n,JP|d)P(d)}{P(+_n,JP)}=\frac{[P(+|d)]^nP(JP|d)P(d)}{P(+_n,JP)}$$

The denominator can be adjusted similarly to the numerator.