It is known that 8% of the population of a large city use a particular web browser called Voyager. A researcher wishes to interview people from the city who use Voyager and selects people at random, one at a time.

1. Find the probability of a randomly selected person not using Voyager.

$\mathrm{P}(\text{does not use Voyager}) = 1 - 0.08 = 0.92$

2. Calculate the probability that the first two people do not use Voyager and the third person does.

$\text{Required probability} = 0.92 \times 0.92 \times 0.08$
$= 0.067712$
$\approx 0.0677 \text{ (to 3 s.f.)}$

1. State the first condition in the context of the question (Independence).

Whether a person uses Voyager is independent of whether any other person uses Voyager.

2. State the second condition in the context of the question (Constant probability).

The probability that a randomly selected person uses Voyager is constant.

1. Define the random variable and calculate the mean and variance of the binomial distribution.

Let $X$ be the number of people out of $80$ who use Voyager.
$X \sim \text{B}(80, 0.08)$
$\text{E}(X) = np = 80 \times 0.08 = 6.4$
$\text{Var}(X) = np(1-p) = 6.4 \times 0.92 = 5.888$

2. State the suitable normal approximation distribution.

Since $np = 6.4 > 5$ and $n(1-p) = 73.6 > 5$, the binomial distribution can be suitably approximated by a normal distribution:
$X \sim \text{N}(6.4, 5.888)$ approximately.

3. Apply continuity correction and calculate the required probability.

We need the probability that fewer than $10$ people use Voyager, which is $P(X < 10)$ or $P(X \le 9)$.
Since we are using a continuous distribution to approximate a discrete one, we apply a continuity correction:
$P(X \le 9) \approx P(X_{\text{norm}} \le 9.5)$

Using a graphing calculator with $\mu = 6.4$ and $\sigma = \sqrt{5.888}$:
$P(X_{\text{norm}} \le 9.5) \approx 0.89928$
$\approx 0.899 \text{ (to 3 s.f.)}$

1. Define the random variable and set up the probability inequality.

Let $X$ be the number of people out of a sample of $n$ who use Voyager.
$X \sim \mathrm{B}(n, 0.08)$
We require the probability that at least $10$ people use Voyager to be more than $0.2$:
$\mathrm{P}(X \ge 10) > 0.2$

2. Rewrite the inequality in terms of cumulative probabilities suitable for a graphing calculator.

$1 - \mathrm{P}(X \le 9) > 0.2$
$\mathrm{P}(X \le 9) < 0.8$

3. Use the table function on a graphing calculator to find the least value of $n$.

Using the graphing calculator to generate a table of values for $\mathrm{P}(X \le 9)$ given different integer values of $n$:
When $n = 92$, $\mathrm{P}(X \le 9) \approx 0.8013 > 0.8$
When $n = 93$, $\mathrm{P}(X \le 9) \approx 0.7930 < 0.8$

Therefore, the least value of $n$ is $93$.