A manufacturer produces screen protectors for handphones. On average, 4% of the screen protectors are cracked. The screen protectors are sold in boxes of $n$.

1. State the assumption of independence in context

Whether a screen protector is cracked must be independent of whether any other screen protector in the box is cracked.

2. State the assumption of constant probability in context

The probability that any given screen protector is cracked must be constant (at $0.04$).

1. Set up the inequality to express that the probability of exactly 2 cracked protectors is strictly greater than the probability of 1.

Let $X \sim \mathrm{B}(n, 0.04)$. Since 2 is the unique most likely number, $P(X=2) > P(X=1)$.

$\binom{n}{2} (0.04)^2 (0.96)^{n-2} > \binom{n}{1} (0.04)^1 (0.96)^{n-1}$

2. Simplify the first inequality to find the lower bound for $n$.

$\frac{n(n-1)}{2} (0.04)^2 (0.96)^{n-2} > n(0.04)(0.96)^{n-1}$

Dividing both sides by $n(0.04)(0.96)^{n-2}$ (which is strictly positive for $n \ge 2$):
$\frac{n-1}{2} (0.04) > 0.96$
$n - 1 > \frac{0.96 \times 2}{0.04}$
$n - 1 > 24 \times 2$
$n - 1 > 48 \implies n > 49$

3. Set up and simplify the inequality for the probability of 2 being greater than 3 to find the upper bound.

Similarly, $P(X=2) > P(X=3)$.

$\binom{n}{2} (0.04)^2 (0.96)^{n-2} > \binom{n}{3} (0.04)^3 (0.96)^{n-3}$

$\frac{n(n-1)}{2} (0.04)^2 (0.96)^{n-2} > \frac{n(n-1)(n-2)}{6} (0.04)^3 (0.96)^{n-3}$

Dividing both sides by $\frac{n(n-1)}{2} (0.04)^2 (0.96)^{n-3}$:
$0.96 > \frac{n-2}{3} (0.04)$
$\frac{0.96 \times 3}{0.04} > n - 2$
$24 \times 3 > n - 2$
$72 > n - 2 \implies n < 74$

4. Combine the bounds to state the integer range of $n$.

Combining the two bounds gives:
$49 < n < 74$

Since $n$ must be an integer, the range of possible values is:
$50 \le n \le 73$

1. Define the random variable and its distribution

Let $X$ be the number of cracked screen protectors in a box of 50.
$X \sim \text{B}(50, 0.04)$

2. Calculate the probability using a graphing calculator

$\text{P}(X \le 3) = 0.8608688... \approx 0.861$ (to 3 s.f.)

1. Set up the required conditional probability.

Let $X$ be the number of cracked screen protectors in a box. We are given $X \sim \mathrm{B}(50, 0.04)$.

We need to find $\mathrm{P}(X > 2 \mid X \le 6)$.
$\mathrm{P}(X > 2 \mid X \le 6) = \frac{\mathrm{P}(X > 2 \text{ and } X \le 6)}{\mathrm{P}(X \le 6)}$
$= \frac{\mathrm{P}(3 \le X \le 6)}{\mathrm{P}(X \le 6)}$

2. Express the numerator in terms of cumulative probabilities.

$\mathrm{P}(3 \le X \le 6) = \mathrm{P}(X \le 6) - \mathrm{P}(X \le 2)$

3. Evaluate the probabilities using a graphing calculator and find the final answer.

Using the binomial cumulative distribution function:
$\mathrm{P}(X \le 6) \approx 0.996148$
$\mathrm{P}(X \le 2) \approx 0.676714$

Substituting these values into the conditional probability expression:
$\mathrm{P}(X > 2 \mid X \le 6) = \frac{0.996148 - 0.676714}{0.996148}$
$= 0.320669 \dots$
$\approx 0.321 \text{ (to 3 s.f.)}$

1. Calculate the probability that a single box contains at most 3 cracked screen protectors.

Let $X$ be the number of cracked screen protectors in a box of 50.
$X \sim \mathrm{B}(50, 0.04)$
$P(X \le 3) = P(X=0) + P(X=1) + P(X=2) + P(X=3)$
$P(X \le 3) \approx 0.86086$

2. Formulate and calculate the required probability using the multiplication of probabilities.

For the 7th box to be the 4th box with at most 3 cracked screen protectors, there must be exactly 3 such boxes among the first 6, AND the 7th box must also be one.
Let $p = 0.86086$.
$\text{Required probability} = \binom{6}{3} p^3 (1-p)^3 \times p$
$= 20 \times (0.86086)^4 \times (1 - 0.86086)^3$
$= 20 \times 0.54956 \times (0.13914)^3$
$= 0.029607...$
$\approx 0.0296 \text{ (to 3 s.f.)}$