已知某种特定疾病的感染率约为万分之一（每 10 000 人中约有 1 人感染）。针对该疾病的一项检测准确率为 99.9%，即检测结果与真实情况相符的概率为 99.9%。求在已知某人检测结果呈阳性的条件下，此人确实患有该疾病的概率。 研究人员正在调查戴口罩是否有助于减少某种特定疾病的传播。他们抽取了一个包含 50 人的代表性样本。在该样本中，有 7 名戴口罩的人病倒了，6 名没戴口罩的人病倒了。样本中共有 31 人戴了口罩。 事件 M 和 C 定义如下： M：随机选择的一名人员戴了口罩。
C：随机选择的一名人员感染了病毒。 判断 M 和 C 是否相互独立。 在查看了收集到的样本结果后，研究人员评论说：“戴口罩是毫无意义的，因为生病的人中戴口罩的比没戴口罩的更多。” 通过比较适当的概率，论证你是否同意这一评论。

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.