A particular disease is known to infect approximately 1 in every 10 000 people. A test for this disease is 99.9% accurate, meaning the outcome of the test matches the true result 99.9% of the time. Find the probability that a person has the disease given that the person is tested positive. Researchers looking into whether wearing masks helps reduce the spread of a particular disease. A representative sample of 50 people is taken. In that sample, 7 people who wore masks got sick, and 6 people who did not wear masks got sick. A total of 31 people in the sample wore masks. The events M and C are defined as follows: M: The event that a randomly chosen person wore a mask.
C: The event that a randomly chosen person caught the virus. Determine if M and C are independent. After looking at the result of the sample collected, the researcher comments that: "Wearing a mask is pointless, since more people who got sick wore masks than didn't." By comparing suitable probabilities, justify if you agree with the comment.

1. Determine the probabilities of having the disease and the respective test outcomes.

Let $D$ be the event of having the disease, and $P$ be the event of testing positive.
$\mathrm{P}(D) = \frac{1}{10000} = 0.0001$
$\mathrm{P}(D') = 1 - 0.0001 = 0.9999$
Given the 99.9% accuracy of the test:
$\mathrm{P}(P \mid D) = 0.999 \text{ (True positive)}$
$\mathrm{P}(P \mid D') = 1 - 0.999 = 0.001 \text{ (False positive)}$

2. Calculate the total probability of a positive test.

Using the addition and multiplication rules of probability (often visualised with a tree diagram):
$\mathrm{P}(P) = \mathrm{P}(D \cap P) + \mathrm{P}(D' \cap P)$
$\mathrm{P}(P) = \mathrm{P}(D) \times \mathrm{P}(P \mid D) + \mathrm{P}(D') \times \mathrm{P}(P \mid D')$
$\mathrm{P}(P) = (0.0001 \times 0.999) + (0.9999 \times 0.001)$
$\mathrm{P}(P) = 0.0000999 + 0.0009999 = 0.0010998$

3. Find the conditional probability of having the disease given a positive test.

$\mathrm{P}(D \mid P) = \frac{\mathrm{P}(D \cap P)}{\mathrm{P}(P)}$
$\mathrm{P}(D \mid P) = \frac{0.0000999}{0.0010998}$
$\mathrm{P}(D \mid P) = \frac{999}{10998} = \frac{111}{1222} \approx 0.0908 \text{ (to 3 s.f.)}$

1. Determine the relevant probabilities from the given data.

$P(M) = \frac{31}{50}$
Total number of people who caught the virus = $7 + 6 = 13$
$P(C) = \frac{13}{50}$
$P(M \cap C) = \frac{7}{50} = 0.14$

2. Test the condition for independence and state the conclusion.

$P(M) \times P(C) = \frac{31}{50} \times \frac{13}{50} = \frac{403}{2500} = 0.1612$
Since $P(M \cap C) \neq P(M) \times P(C)$, the events $M$ and $C$ are not independent.

1. Calculate the conditional probability of getting sick given the person wore a mask

$\mathrm{P}(C \mid M) = \frac{n(C \cap M)}{n(M)} = \frac{7}{31} \approx 0.226$

2. Calculate the conditional probability of getting sick given the person did not wear a mask

$n(M') = 50 - 31 = 19$

$\mathrm{P}(C \mid M') = \frac{n(C \cap M')}{n(M')} = \frac{6}{19} \approx 0.316$

3. Compare the probabilities to evaluate the researcher's comment

Since $\mathrm{P}(C \mid M) < \mathrm{P}(C \mid M')$, the probability of catching the virus is lower when a person wears a mask. Therefore, I disagree with the researcher's comment.