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 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 $\sigma = \sqrt{4.6509}$:
$P(W_{\text{cont}} \le 10.5) \approx 0.16026$
$\approx 0.160 \text{ (to 3 s.f.)}$