在一个周五晚上，班长给他的 20 名同学发送了电子邮件，提议周末一起去观看一部新上映的动作惊悚片。随机变量 X 表示那个周末对他的邮件给予积极回复的同学人数。 假设 X 服从二项分布 $B(20,p)$，其中 $0 < p \le 0.5$。 在每个工作日的早晨，班长会从一个装有全班所有学生名字的袋子中抽取一个名字。班长记录下名字后，再将其放回袋中。名字被抽中的人将负责当天教室的卫生。该班由 12 名女生和 9 名男生组成。随机变量 Y 表示在一个包含 5 个工作日的星期中，女生被抽中值日的次数（天数）。

1. Formulate the equation in terms of $p$

We are given that $X \sim \mathrm{B}(20, p)$.
The events $X = 0$ and $X = 19$ are mutually exclusive, so $P(X = 0 \text{ or } 19) = P(X = 0) + P(X = 19)$.
Using the binomial probability formula:
$P(X = 0) = \binom{20}{0}p^0(1-p)^{20} = (1-p)^{20}$
$P(X = 19) = \binom{20}{19}p^{19}(1-p)^1 = 20p^{19}(1-p)$

Thus, equating the sum to $0.1$, the equation is:
$(1-p)^{20} + 20p^{19}(1-p) = 0.1$

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 $0 < p \le 0.5$:
$p \approx 0.108749$
$p = 0.109 \text{ (to 3 s.f.)}$

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

For $Y$, since the drawn name is put back into the bag each time (sampling with replacement), the probability of picking a girl remains constant every day, and the outcome of each day's draw is completely independent of the other days.

2. Explain why X is less likely to satisfy the binomial conditions.

Conversely, for $X$, the classmates are likely to discuss the outing among themselves. Thus, their decisions to respond positively are not independent of one another (they may influence each other). Furthermore, the probability of a positive response may naturally vary from person to person.

1. Identify the parameters of the binomial distribution for $Y$.

Let $Y$ be the number of days a girl is picked to be on duty in a 5-day week.
Since the name is drawn and put back (with replacement), the probability of picking a girl each day is constant: $p = \frac{12}{12+9} = \frac{12}{21} = \frac{4}{7}$.
Therefore, $Y \sim \text{B}\left(5, \frac{4}{7}\right)$.

2. Translate the condition 'girl is on duty more often than a boy' into a mathematical inequality.

In a 5-day week, for a girl to be on duty more often than a boy, the number of days a girl is on duty must be strictly greater than the number of days a boy is on duty.
$Y > 5 - Y$
$2Y > 5 \implies Y > 2.5$
Since $Y$ must be an integer, this requires $Y \ge 3$.

3. Calculate the probability $P(Y \ge 3)$.

We need to evaluate $P(Y \ge 3)$:
$P(Y \ge 3) = 1 - P(Y \le 2)$
Using a graphing calculator (or the binomial probability formula),
$P(Y \ge 3) \approx 0.632117$
$P(Y \ge 3) = 0.632 \text{ (to 3 s.f.)}$

1. Retrieve the probability that a girl is on duty more often than a boy in a week (from part iii).

From part (iii), let $Y \sim \text{B}(5, \frac{4}{7})$ be the number of days a girl is on duty in a 5-day week.
The probability that a girl is on duty more often than a boy is $P(Y \ge 3)$.
$P(Y \ge 3) = P(Y=3) + P(Y=4) + P(Y=5)$
$= \binom{5}{3}\left(\frac{4}{7}\right)^3\left(\frac{3}{7}\right)^2 + \binom{5}{4}\left(\frac{4}{7}\right)^4\left(\frac{3}{7}\right) + \binom{5}{5}\left(\frac{4}{7}\right)^5$
$= \frac{10624}{16807} \approx 0.63212$

2. State the binomial distribution for the number of weeks and establish the normal approximation parameters.

Let $W$ be the number of weeks (out of 20) where a girl is on duty more often than a boy. Then $W \sim \text{B}(20, 0.63212)$.
Since $np = 20 \times 0.63212 \approx 12.6 > 5$ and $n(1-p) = 20(1 - 0.63212) \approx 7.36 > 5$, $W$ can be suitably approximated by a normal distribution.
$\mu = np = 20 \times \frac{10624}{16807} \approx 12.642$
$\sigma^2 = np(1-p) = 12.642 \times \left(1 - \frac{10624}{16807}\right) \approx 4.6509$
Hence, $W \sim \text{N}(12.642, 4.6509)$ approximately.

3. Apply continuity correction and find the required probability.

We need the probability that there are less than 11 such weeks, which is $P(W < 11)$ or $P(W \le 10)$.
Applying continuity correction for the continuous normal approximation:
$P(W \le 10) \approx P(W_{\text{cont}} \le 10.5)$
Using a graphing calculator with $\mu = 12.642$ and $\sigma = \sqrt{4.6509}$:
$P(W_{\text{cont}} \le 10.5) \approx 0.16026$
$\approx 0.160 \text{ (to 3 s.f.)}$