周五晚上，班长给他的 20 位同学发邮件，提议周末一起去观看新上映的动作惊悚片。随机变量 $X$ 表示该周末对他的邮件给出积极回应的同学人数。 假设 $X$ 服从分布 $B(20,p)$，其中 $0 < p \le 0.5$。 在每个工作日的早晨，班长都会从一个装有全班所有学生名字的袋子中抽取一个名字。班长记录下名字，然后将其放回袋子中。被抽中名字的学生将负责当天的教室卫生。该班由 12 名女生和 9 名男生组成。随机变量 $Y$ 表示在一周 5 天内，女生被选值日的天数。

1. Formulate the equation for $p$ using the binomial probability formula.

Given $P(X = 0 \text{ or } 19) = 0.1$
Since $X = 0$ and $X = 19$ are mutually exclusive events:
$P(X = 0) + P(X = 19) = 0.1$

Using the binomial formula $P(X = x) = \binom{n}{x}p^x(1-p)^{n-x}$:
$\binom{20}{0}p^0(1-p)^{20} + \binom{20}{19}p^{19}(1-p)^1 = 0.1$
$(1-p)^{20} + 20p^{19}(1-p) = 0.1$

2. Solve the equation for $p$ using a graphing calculator.

Using a graphing calculator to solve $(1-p)^{20} + 20p^{19}(1-p) - 0.1 = 0$ for the given domain $0 < p \le 0.5$:
$p = 0.108749\dots$
$p \approx 0.109 \text{ (to 3 s.f.)}$

1. Explain why $Y$ satisfies the conditions for a binomial distribution in context.

For $Y$, since the drawn name is put back into the bag (with replacement), the events are independent and the probability of picking a girl remains constant at $\frac{12}{21}$ for each of the 5 days.

2. Explain why $X$ does not perfectly satisfy the conditions for a binomial distribution in context.

For $X$, the responses of the 20 classmates are likely not independent, as friends might influence each other's decision to attend the outing together. Additionally, the probability of a positive response is not identical for every student, as individuals have different preferences for action thrillers or different weekend commitments.

1. Establish the probability distribution for $Y$.

The total number of students is $12 + 9 = 21$.
The probability of choosing a girl on any given day is $p = \frac{12}{21} = \frac{4}{7}$.
Since the days are independent and names are replaced, $Y \sim \text{B}\left(5, \frac{4}{7}\right)$.

2. Determine the condition for a girl to be on duty more often than a boy.

In a 5-day week, for a girl to be on duty more often than a boy, she must be on duty for $3$, $4$, or $5$ days.
Therefore, we need to find $P(Y \ge 3)$.

3. Calculate the required probability.

$P(Y \ge 3) = 1 - P(Y \le 2)$
Using a graphing calculator,
$P(Y \ge 3) = \frac{10624}{16807} \approx 0.63212$
$\approx 0.632 \text{ (to 3 s.f.)}$

1. Calculate the probability from part (iii) and define the binomial distribution for the semester.

From part (iii), let $Y \sim \text{B}\left(5, \frac{12}{21}\right) = \text{B}\left(5, \frac{4}{7}\right)$ be the number of days a girl is on duty in a 5-day week.
For a girl to be on duty more often than a boy, she must be on duty for at least 3 days:
$P(Y \ge 3) = P(Y=3) + P(Y=4) + P(Y=5)$
$P(Y \ge 3) = \binom{5}{3}\left(\frac{4}{7}\right)^3\left(\frac{3}{7}\right)^2 + \binom{5}{4}\left(\frac{4}{7}\right)^4\left(\frac{3}{7}\right) + \binom{5}{5}\left(\frac{4}{7}\right)^5 = \frac{10624}{16807} \approx 0.63212$
Let $W$ be the number of weeks in a 20-week semester where a girl is on duty more often than a boy.
$W \sim \text{B}\left(20, \frac{10624}{16807}\right)$

2. Verify the conditions for a normal approximation and calculate its mean and variance.

We check if a normal approximation is suitable by computing $np$ and $n(1-p)$:
$np = 20 \left(\frac{10624}{16807}\right) \approx 12.64 > 5$
$n(1-p) = 20 \left(1 - \frac{10624}{16807}\right) \approx 7.36 > 5$
Since both values are greater than $5$, $W$ can be approximated by a normal distribution $W_{\text{norm}} \sim \text{N}(\mu, \sigma^2)$ where:
$\mu = np = \frac{212480}{16807} \approx 12.642$
$\sigma^2 = np(1-p) = 12.64235 \times \frac{6183}{16807} \approx 4.6509$

3. Apply continuity correction and find the approximate probability.

We want the probability that there are less than 11 such weeks, which is $P(W < 11) = P(W \le 10)$.
Applying the continuity correction for the upper bound of a discrete interval, we use $10.5$:
$P(W \le 10) \approx P(W_{\text{norm}} \le 10.5)$
Using the normal distribution $\text{N}(12.642, 4.6509)$:
$P(W_{\text{norm}} \le 10.5) \approx 0.16026$
The required probability is $0.160$ (to 3 s.f.).