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 a true positive and a false positive test.

Let $D$ be the event that a person has the disease, and $T^+$ be the event of testing positive.
$P(D \cap T^+) = P(D) \times P(T^+ \mid D) = 0.0001 \times 0.999 = 0.0000999$
$P(D' \cap T^+) = P(D') \times P(T^+ \mid D') = 0.9999 \times (1 - 0.999) = 0.0009999$

2. Calculate the total probability of testing positive.

$P(T^+) = P(D \cap T^+) + P(D' \cap T^+)$
$P(T^+) = 0.0000999 + 0.0009999 = 0.0010998$

3. Calculate the probability that the person has the disease given a positive test.

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

1. Find the probabilities of events $M$, $C$, and their intersection $M \cap C$ from the given sample counts.

$\mathrm{P}(M) = \frac{31}{50}$
$\mathrm{P}(C) = \frac{7 + 6}{50} = \frac{13}{50}$
$\mathrm{P}(M \cap C) = \frac{7}{50}$

2. Test the condition for independent events: $\mathrm{P}(M \cap C) = \mathrm{P}(M) \times \mathrm{P}(C)$.

$\mathrm{P}(M) \times \mathrm{P}(C) = \frac{31}{50} \times \frac{13}{50} = \frac{403}{2500} = 0.1612$
$\mathrm{P}(M \cap C) = \frac{7}{50} = \frac{350}{2500} = 0.14$
Since $\mathrm{P}(M \cap C) \neq \mathrm{P}(M) \times \mathrm{P}(C)$, the events $M$ and $C$ are not independent.

1. Calculate the conditional probability of catching the virus given that the person wore a mask.

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

2. Calculate the conditional probability of catching the virus given that the person did not wear a mask, and form a conclusion.

Number of people who did not wear a mask $= 50 - 31 = 19$.
$P(C|M') = \frac{n(C \cap M')}{n(M')} = \frac{6}{19} \approx 0.316$

Since $P(C|M) < P(C|M')$, the probability of getting sick is lower if one wears a mask. The researcher is ignoring the base rate of mask-wearers; therefore, we disagree with the comment.