Statistical model for hitting an object on a path I am searching for a probability model for estimating the probability for hitting an object on a defined path. 
For example: A ship is travelling from Portugal to Cuba (a distance of about $d = 7,000 km$) with a speed of $v = 3 m/s$. On it's journey it might hit a lost container. Let's estimate that about $n = 2,000$ containers are floating in the sea worldwide (see here for annual numbers of newly lost containers). As we want to solve this parametric we might take other dangers like whales into account later. The ship is $l_s = 8 m$ long and $w_s = 3 m$ wide. The rectangular containers are $l_c = 13m$ long and $w_c = 3m$ wide. 
Assuming we simplify that to "All containers are evenly distributed and stay at the same position" and "The ship has a rectangular area of $l_s \cdot w_s$". We also say that any intersection of that area and a container counts as hitting. The world's oceans have a combined area of $\sim 3.6 \cdot 10^8 km^2$ (according to Wikipedia). 
What is the probability of a ship hitting a container or what model can we use to quantify this in general? 
 A: This question has a famous counterpart.

“Buddha in the Chiggala Sutta of the Samyutta Nikaya... states that human life is a very precious opportunity to practice the Dharma because it is very hard to come by. He describes the situation of a blind sea-turtle who lives in the ocean and only rises to the surface once every hundred years. A small yoke is floating on the surface of the vast ocean. The Buddha asks his hearers to contemplate the odds against the turtle happening to surface at the exact spot in the ocean where the yoke is floating, so that the turtle’s head is encircled by it. He stresses that the chances of gaining human rebirth are as remote as this”.
Vessantara, “The Five Female Buddhas”

But let’s continue our hand-waving reasoning. Stick to the point

Assuming we simplify that to "All containers are evenly distributed and stay at the same position"

Then the hitting probability is a probability that a container will intersect the ship trajectory, which we simplify to a strip of length $d$ and width $w_s$. Let $l$ be the trajectory of the center of the ship. The intersection depends on the distance $h$ of the container center to the line $l$ and an angle $\alpha$ ($0\le \alpha\le\pi/2$) between the longest side of the container and $l$. The intersection happens iff $h\le w_s/2+(l_c/2)\sin\alpha+(w_c/2)\cos\alpha$. Its probability $p(\alpha)$ is equal to is the area of the strip of width $w_s+l_c\sin\alpha+w_c\cos\alpha$ and length $d$ divided by the combined area $A_O$ of the world's oceans, that is $p(\alpha)=d(w_s+l_c\sin\alpha+w_c\cos\alpha)/A_O$. Assuming that all $\alpha$’s are evenly distributed, the probability $p$ of hitting the ship by a chosen container is
$$\frac 2\pi\int_0^{\pi/2} p(\alpha)d\alpha=$$
$$\frac 2\pi\int_0^{\pi/2} \frac{d(w_s+l_c\sin\alpha+w_c\cos\alpha)}{A_O}  d\alpha=$$
$$\frac {2d}{\pi A_O}(\pi w_s/2+l_c+w_c)\sim$$
$$2.56\cdot 10^{-7}.$$
At last, assuming that containers are distributed independently, the probability that the ship will be hit is $1-(1-p)^n\simeq 5.12\cdot 10^{-4}$. This small probability looks quite travel safe, but we still have to remember about “Titanic”. :-)
