A start: Draw the two diagonals of the square, and also the two lines that join midpoints of opposite sides. These $4$ lines divide the square into $8$ congruent right-angled isosceles triangles. By symmetry, it is enough to determine the probability that a randomly chosen point within a specific one of these triangles is closer to the centre of the square than it is to one of the sides of the square.
Assume (it makes no difference) that the original square has vertices $(2,2)$, $(-2,2)$, $(-2,-2)$, and $(2,-2)$. Consider the triangle $T$ with vertices $(0,0)$, $(2,0)$, and $(2,2)$. This triangle has area $2$. So for our probability, we will determine the area of the region $K$ in $T$ consisting of all points $(x,y)$ such that the distance from $(x,y)$ to the origin is $\le$ the distance from $(x,y)$ to the nearest side of the original square. Then our required probability is the area of $K$ divided by $2$,
So let us determine the locus of all points $(x,y)$ in $T$ that are equidistant from the origin and the nearest side of the original square.
The distance to the origin is $\sqrt{x^2+y^2}$, and the distance to the nearest side of the original square is $2-x$. So our locus is $\sqrt{x^2+y^2}=2-x$. Square and simplify. We get $x=\frac{1}{4}(4-y^2)$, part of a parabola.
Our region $K$ is the region bounded by the $x$-axis, the parabola $x=\frac{1}{4}(4-y^2)$, and the line $y=x$. We find the area of $K$ by integrating with respect to $y$.
Setting up the integral is slightly tricky. We will be integrating from $y=0$ to where $x=\frac{1}{4}(4-y^2)$ meets $y=x$. That gives the equation $y=\frac{1}{4}(4-y^2)$, which has the solution $y=2\sqrt{2}-2$. Thus our required area is
$$\int_0^{2\sqrt{2}-2} \left(\frac{1}{4}(4-y^2)-y\right)\,dy.$$
After calculating the area, don't forget to divide by $2$.
Your solution is correct. You may also give an alternative justification for your simplification in the second part by applying Bayes Theorem.
$P(X\geq 2 \vert X\geq 1) = \frac{P(X\geq 1 \vert X\geq 2)P(X\geq 2)}{P(X\geq 1)}$
Observing that $P(X\geq 1 \vert X\geq 2) = 1$ we obtain
$P(X\geq 2 \vert X\geq 1) = \frac{P(X\geq 2)}{P(X\geq 1)}$
The details of which you have already worked out.
Best Answer
Write your inequality $\ \frac{1}{6}>\left(\frac{2}{3}\right)^n\ $ as $$ 3^{n-1}>2^{n+1}\ . $$ The highest powers of $2$ and $3$ you need to check to find out which of the alternatives you're given is the correct one are $\ 2^7=128\ $ and $\ 3^5=243\ $, respectively. These are small enough for the check to carried out reasonably quickly without any artificial aids.