Maths by Chance

Let's say we are observing an event that to us seems random. E.g. The number of cars passing through a road. Suppose we learn that on average, there are m cars passing the road within 1 hour. Goal: Can we make a model to suggest what is the probability of seeing exactly r cars passing the road in an hour? Let's represent one hour as a line, and on the line there are marks to represent the time on which a car passes. Let's suppose there are m marks, to represent the average case of seeing m cars on the road in an hour. Let's divide the 1 hour line into n uniform time segments, n large enough such that in one time segment there is at most 1 car. Out of the n time segments, m of them contains a car. Then it is reasonable to say that the probability of a time segment to contain a car is m/n. Let's analyse as n goes to infinity. We can say P(seeing exactly r cars in an hour) =  P(exactly r cars contained in n segments) = \( \binom{n}{r} \left( \frac{m}{n} \right)^r \left

Prove that \( \left( 1 + \frac{1}{n} \right)^n \) converges. Solution Let \( G(n) =  \left( 1 + \frac{1}{n} \right)^n \). Binomial expansion, \( G(n) = \sum_{k = 0}^{n} \frac{ \binom{n}{k} }{n^k} \). By simple inequality argument, \( \frac{ \binom{n}{k} }{n^k} < \frac{1}{k!} \). So if we let \( F(n) = \sum_{k = 0}^{n} \frac{1}{k!} \), we have \( G(n) < F(n) \). By simple inequality argument, \( F(n) < 1 + \sum_{k = 1}^{n} \frac{1}{2^{k-1}} < 3 \). So G and F are bounded above. By AM-GM, \( \frac{G(n-1)}{G(n)} = \left( \frac{ n^2 }{ n^2 - 1} \right)^{(n-1)} \frac{ n}{n+1} < \left(  \frac{ \frac{ n^2 }{ n^2 - 1} (n-1) + \frac{ n}{n+1} }{n} \right)^n = 1 \). So \( G(n-1) < G(n) \). Since G is monotonic increasing and bounded above, G converges. Similarly F converges. Let's also prove that G and F converges to the same limit. By studying individual terms in \( F(n) - G(n) \) and making simple inequality arguments, we can show that for each \( \epsilon > 0 \) we can

Problem:  Given a circle of radius 1, pick a chord at random. What is the chance that the length of the chord is less than the radius? Solution: Depending on how you pick it at random, the probability can be different! 1. If you pick it by randomly choosing 2 points on the circle's circumference, then the chance would be 1/3. 2. Every chord has a midpoint which is also a perpendicular bisector. If we pick this midpoint randomly within the circle, its chord length will be less than radius as long as it is outside of a smaller circle with radius $\frac{\sqrt{3}}{2}$. So the chance is 1/4 based on the ratio of area. 3. From the centre, randomly choose a radial direction and randomly pick a point along this radius, then draw a radially perpendicular line through this point, we'll get a chord. Then the chance would be \( 1 - \frac{\sqrt{3}}{2} \). Surprising, huh. In fact, you can compute the distribution of the chord length given the picking method. Is there a way to pick a chord &

Problem You and your friend play a game involving coin-flips. Each time the coin lands head, your friend pays you $ 1. But each time the coin lands tail, you pay your friend $ 1. The game goes on until either you win $ 100 or you lose $ 50. What is the probability of you winning? Solution 3 solutions, two of them are algebraic, one of them is argumentative. Algebraic Solution 1 Let \( p_i \) be the probability of winning $ 100 given that we already accumulated \( i \) dollars (can be negative, meaning we are losing money). Then we have \( p_i = \frac{1}{2} p_{i-1} + \frac{1}{2} p_{i+1} \), with special value \( p_{100} = 1 \) and \( p_{-50} = 0 \). It can be shown that \( p_i = \frac{i + 50}{150} \) the only solution that satisfies the system of linear equations. This approach can be extended to a more general problem: what is the probability of winning if we stop at either winning $n or losing $m? It should be \( \frac{m}{n+m} \). Algebraic Solution 2 This approach is inspired by the