已知某种特定疾病在每 10,000 人中大约会感染 1 人。针对这种疾病的检测准确率为 99.9%，即检测结果有 99.9% 的时间与真实结果相符。求一个人在检测结果呈阳性的条件下，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。选取了一个由 50 人组成的代表性样本。在该样本中，有 7 个戴口罩的人病倒了，有 6 个没戴口罩的人病倒了。样本中总共有 31 人戴了口罩。 事件 $M$ 和 $C$ 的定义如下： $M$：随机选择的人戴了口罩这一事件。 $C$：随机选择的人感染了病毒这一事件。 判断 $M$ 和 $C$ 是否独立。 在查看了收集到的样本结果后，研究人员评论说：“戴口罩是没有意义的，因为病倒的人中戴口罩的人比没戴的人多。” 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 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.