The question is ill-posed because there are many different ways of choosing two spots “at random”. BUT if we assume that the cuts are made independently with each chosen according to a uniform probability per unit length then the answer is the fraction of the big square that is not shaded in the diagram below . (ie p = 1-(17/20)^2)

Source: (3) Alan Cooper’s answer to Question: A bar of length ℓ is broken into three pieces at two random spots. What is the probability that the length of at least one piece is less than ℓ/20? Can anyone hlp me? Thks a lot – Quora

Since this is tagged with “Puzzles and Trick Questions” it may be that I am missing something. But my answer would be ${\mathrm{\Sigma }}_{m=0}^{n-1}{p}^{2n-1-m}(1-p{)}^m$ .

This follows the pattern of the best 2 out of 3 case where Alice has to win either two or three games – which happens in cases lww,wlw,wwl or www with probability $3p^2(1–p)+p^3=3p^2–2p^3$ (where the fact that the game may be stopped when she wins twice just corresponds to the fact that $pp(1-p)+ppp=p^2$ , and the same answer is obtained by taking the complement of the cases where Bob wins either 2 or 3 games).

Source: (3) Alan Cooper’s answer to Alice and Bob flip a biased coin, best $n$ out of $2n-1$ win. If the probability of Alice winning a flip is $p$, what is her chance of winning the series? – Quora

This question has been around for a while and has some decent answers. But I want to suggest a simpler and more intuitive version. (And it will be easier to follow if I replace the variable $x$ by $t$ so as not to confuse it with the real part of the complex function value.)

First, to define $f(z)=e^z$ without using trigonometry or ever mentioning trig functions, we can use either the power series or the complex differential equation $f^{\prime }=f$ with $f(0)=1$. And either way we get $\frac{d}{dt}{e}^{it}=i{e}^{it}$.

Now multiplication by $i$ just rotates the complex plane by a right angle, so the curve in the plane given parametrically by $(x(t),y(t))$ with $x(t)+iy(t)=e^{it}$ has a tangential velocity vector which is always perpendicular to its position vector and equal in magnitude.

Since it starts at $t=0$ at $(x,y)=(1,0)$ the curve is just the unit circle centred at the origin.

And since its velocity vector is always of length 1, if we think of the parameter $t$ as representing time, then the point moves with speed 1 and so the time taken to complete a circuit, ie the period of $e^{it}$, is just the same as the circumference of the unit circle (commonly denoted by $2\pi$ ).

Source: (3) Alan Cooper’s answer to How do I find the period of $e^{ix}$ without using trigonometry? – Quora