Let $X$ be the waiting time in minutes. Since the arrival is completely random within the 15-minute window, $X$ follows a Uniform distribution on the interval $[0, 15]$.

Parameters:
- $a = 0$
- $b = 15$

The probability density function is $f(x) = \frac{1}{15 - 0} = \frac{1}{15}$.

We want to find the probability of waiting more than 10 minutes: $P(X > 10)$.
This corresponds to the interval from 10 to 15.

$P(X > 10) = \text{Area of rectangle} = \text{Width} \times \text{Height}$
$P(X > 10) = (15 - 10) \times \frac{1}{15} = 5 \times \frac{1}{15} = \frac{1}{3} \approx 0.333$

There is a 33.3% chance the passenger waits more than 10 minutes.

Let $X$ be the total waiting time, $X \sim U(0, 15)$.
We are given that $X > 5$ (they already waited 5 minutes).
We want to find the probability that they wait at least 5 more minutes, which means $X > 10$, given $X > 5$.

Using the conditional probability formula:
$P(X > 10 \mid X > 5) = \frac{P(X > 10 \cap X > 5)}{P(X > 5)}$

Since $X > 10$ is a stricter condition, the intersection is just $X > 10$.
$P(X > 10 \mid X > 5) = \frac{P(X > 10)}{P(X > 5)}$

Calculate the individual probabilities:
$P(X > 10) = (15 - 10) \times \frac{1}{15} = \frac{5}{15} = \frac{1}{3}$
$P(X > 5) = (15 - 5) \times \frac{1}{15} = \frac{10}{15} = \frac{2}{3}$

Now, divide them:
$P(X > 10 \mid X > 5) = \frac{1/3}{2/3} = \frac{1}{2} = 0.5$

Given they have already waited 5 minutes, there is a 50% chance they will wait at least 5 more minutes.