On a Friday evening, the class chairperson emails his 20 classmates to propose an outing to watch a newly released action thriller. The random variable $X$ is the number of classmates who respond positively to his email that weekend. It is assumed that $X$ has the distribution $B(20,p)$ such that $0 < p \le 0.5$. On each weekday morning, the class chairperson draws a name from a bag containing the names of all the students in the class. The chairperson notes the name and then puts it back into the bag. The person whose name is drawn will be responsible for keeping the homeroom tidy for that day. The class consists of 12 girls and 9 boys. The random variable $Y$ is the number of days, out of a 5-day week, in which a girl is picked to be on duty.

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.).