When Mr. Lee sends a text message to any of his students over the weekend, he gets a reply, on average, about 6 out of 10 times. On a particular weekend, Mr. Lee sends a text message to 25 students.

1. State the assumption regarding independence of trials in context.

Whether any student replies to the text message is independent of whether any other student replies.

2. State the assumption regarding constant probability of success in context.

The probability of a student replying is constant for each of the 25 students.

1. Provide a real-world reason why the assumption of independence or constant probability fails in this context.

Explain the failure of independence: Students may communicate with one another about the text message, so one student's decision to reply could affect another's.
Explain the failure of constant probability: Different students have different levels of availability or phone usage habits over the weekend.

1. State the probability distribution clearly.

Let $X$ be the number of students who reply out of $25$.
$X \sim \mathrm{B}(25, 0.6)$

2. Translate 'at least half' into a probability expression and evaluate it.

Half of 25 students is $12.5$. Since the number of students must be an integer, 'at least half' means 13 or more students.
$\mathrm{P}(X \ge 13) = 1 - \mathrm{P}(X \le 12)$
$\mathrm{P}(X \ge 13) \approx 0.84627 \approx 0.846 \text{ (to 3 s.f.)}$

1. Define the distribution and formulate the required probability inequality.

Let $X$ be the number of students who reply out of 25.
$X \sim \mathrm{B}(25, p)$

We require the probability of more than 20 students replying to be at least $0.90$:
$\mathrm{P}(X > 20) \ge 0.9$
$\mathrm{P}(X \ge 21) \ge 0.9$
$1 - \mathrm{P}(X \le 20) \ge 0.9$

2. Solve for the critical value of $p$ using a Graphing Calculator (GC).

Using a graphing calculator, we can plot $Y_1 = 1 - \mathrm{P}(X \le 20)$ (where $p$ is the variable $X$ on the GC) and $Y_2 = 0.9$.
Finding the point of intersection gives:
$p = 0.90377\dots$

3. State the least value of $p$.

Since $\mathrm{P}(X \ge 21)$ is an increasing function of $p$, we require:
$p \ge 0.90377$

The least value of $p$ is $0.904$ (to 3 s.f.).