Probability that a page has at least three figures A books with 120 pages has 6 figures. Every figure could be on a page with the same probability. What is the probability that a randomly selected page has at least three figures?
Expanding on lulu's comment, the formula for binomial probability is:

$$P(x) = {n \choose x}p^x q^{n-x}$$

where $n$ is the number of trials, $x$ is the number of successes, $p$ is the success probability of one trial, and $q = 1 - p$ or the failure probability of one trial.

In your case, $n = 6$, $x = 3$, and $p = \frac{1}{120}$, so the probability should be:

$${6 \choose 3} \left(\frac{1}{120}\right)^3 \left(1-\frac{1}{120} \right)^{3}$$

$$\approx 1.13 \cdot 10^{-5}$$
Probability of finding a figure on a page is $P(A) = 6/120 = 0.05$. So finding 3 such images in the same page would mean $P(A) * P(A) * P(A) = 0.000125$