The problem: suppose 406 dice are rolled. Find the probability of getting exactly 100 5s.
Binomial definitely applies, since there is a fixed number of trials (406 dice) and there are two distinct outcomes (rolling a 5 = success, not rolling a 5 = failure). Let $X$ be the number of 5s showing, then we are approximating $P(X = 100)$.
First, $n \cdot p \approx 67.7 \geq 5$ and $n \cdot q \approx 338.3 \geq 5$, so it looks like normal can be used. Also, $\mu = n \cdot p \approx 67.5$ and $\sigma = \sqrt{n \cdot p \cdot q} \approx 7.5$, furthermore the random variable $X$ (number of 5s showing) can by made continuous by writing it as an interval: $99.5 < X < 100.5$
Now set up the problem and use tables:
$P(X = 100) \approx P(99.5 < X < 100.5) = P(4.26 < z < 4.4) = 1 - 1 = 0$
Getting exactly 100 5s is a very unlikely event, so this makes sense to me, but the class notes uploaded by our teacher show a different answer (0.00715).
Although the probability you get from the normal distribution is very small, it isn't zero: $$P(4.26 < z < 4.4) \approx 5\times10^{-6}$$ That said, the probability can also be calculated exactly from the binomial distribution: $$ P(100) = \binom{406}{100}\frac{5^{306}}{6^{406}}\approx 11 \times 10^{-6} $$ which can be estimated from Stirling's approximation $n!\approx (n/e)^n\sqrt{2\pi n}$ and a good log table. The normal approximation gets the right order of magnitude, but it's still a factor of 2 off.
What I'm not sure of is where the value $0.00715$ comes from. Are you sure you're looking at the right answer?