Is it safe to go?

The first thing we can deduct is that the radius of $\color{#777777}{S_{i}}$ is equal to $\frac{r}{5}$ , where $r$ is the radius of Galvan. This is because the biggest quakes will occur only when $\color{#777777}{S_{i}}$ is completely buried buried under the surface of Galvan, as shown in the diagram above. Comparing volumes:

$V_{\text{Galvan}}=\color{#BA33D6}{\dfrac{4\pi r^3}{3}}$ , $V_{S_{i}}=\dfrac{4\pi \left(\frac{r}{5}\right)^3}{3}=\color{#BA33D6}{\dfrac{4\pi r^3}{3}}$ $\cdot \dfrac{1}{125}$

Now, to calculate the probability we have to change our way to think about the problem a bit, since thinking in terms of a complete sphere $\color{#777777}{S_{i}}$ is hard. Instead, it is easier to think about points and lines , zero and one-dimensional objects, rather than objects in higher dimensions. So we'll call the center of $\color{#777777}{S_{i}}$ a point $\color{#777777}{C_{i}}$ . We know that a point $\color{#3D99F6}{P}$ is affected by a quake if it is within the radius of $\color{#777777}{S_{i}}$ , which is to say that $\color{#3D99F6}{P}$ is a distance $\color{#20A900}{d}$ away from $\color{#777777}{C_{i}}$ , where $0< \color{#20A900}{d}$ $<\dfrac{r}{5}$ . Now, we know that the probability of a point randomly appearing within a certain expected space, which we'll call $S_{\text{expected}}$ , is given by the quotient of $S_{\text{expected}}$ and the total space where the point can appear, $S_{\text{total}}$ :

$\mathbb{P}(X)=\dfrac{S_{\text{expected}}}{S_{\text{total}}}$ .

It is clear that $S_{\text{total}}$ cannot be the infinitely vast space, because if $S_{\text{total}}\to \infty$ $\implies \mathbb{P}(X)\to 0$ , and clearly this is not true; it is an event that has been happening for years, and somehow this probabilistic model explains it successfully. Notice that if $\color{#777777}{C_{i}}$ appears inside Galvan, then a quake will happen because an intersection is guaranteed. However, outside Galvan $\color{#777777}{C_{i}}$ is limited to appear a distance $\color{#20A900}{d}$ away from the surface of the planet, where $0< \color{#20A900}{d}$ $<\dfrac{r}{5}$ , because there cannot exist a cycle where a quake doesn't happen; with this boundary we ensure that stays true. Calculating $S_{\text{total}}$ is calculating the volume of the sphere of radius $r + \dfrac{r}{5}=\dfrac{6r}{5}$ :

$S_{\text{total}}=\dfrac{4\pi \left(\frac{6r}{5}\right)^3}{3}=\dfrac{288\pi r^3}{125}$

To calculate $S_{\text{expected}}$ , we'll first calculate the probability of getting hit by a quake, plug it into the formula for $\mathbb{P}(X)$ , and subtract that from 1, as it is easier that way. Consider you are located at a point $\color{#EC7300}{K}$ . Then, if $\color{#777777}{C_{i}}$ appears within a spherical "aura" of radius $\dfrac{r}{5}$ centered at $\color{#EC7300}{K}$ , you will feel a quake.

So the probability of getting hit by a quake is given by:

$\mathbb{P}(Q)=\dfrac{S_{\text{expected}}}{S_{\text{total}}}=\dfrac{4\pi r^3}{375}\cdot \dfrac{125}{288\pi r^3}=\dfrac{1}{216}\approx 0.462\%$ (This works because even though you are in motion, travelling to Mike, at every instance in time you will find yourself at some point $\color{#EC7300}{K}$ )

Here we can quickly see that the probability of NOT getting hit by a quake is $1-\mathbb{P}(Q)=\dfrac{215}{216}\approx 99.537\%$ .

Finally, to calculate the probability of reaching Mike without running into a quake, we need to know how many cycles are we going to go through, since one quake always happens each cycle. Dividing the distance we need to travel by our speed will give us the time it will take us, and since a single a cycle is 10 hours long, dividing by 10 is how many cycles we'll go through:

$\text{Cycles}=\dfrac{12000 \text{km}}{220 \frac{\text{km}}{\text{h}}} \cdot \dfrac{1}{10 \text{h}}=\dfrac{60}{11} \approx 5.46$

This number is greater than 5, so we are going through 6 cycles, which is, 6 times the event needs to repeat. So the final answer is:

$\left(1-\mathbb{P}(Q)\right)^6=\left(\dfrac{215}{216}\right)^6\approx \boxed{97.254\%}$

Which I think is safe enough for Galvanian standards.