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Chapter 3: Problem 43

(a) Choose \(n\) points independently at random on the perimeter of a circle. Show that the probability of there being a semicircular part of that perimeter which includes none of the \(n\) points is \(n 2^{1-n}\). (b) Choose \(n\) points independently at random on the surface of a sphere. Show that the probability of there being a hemisphere which includes none of the \(n\) points is \(\left(n^{2}-n+2\right) 2^{-n}\).

Short Answer

Expert verified
(a) Probability is \(n 2^{1-n}\). (b) Probability is \((n^2 - n + 2)2^{-n}\).

Step by step solution

01

Set up the probability space for part (a)

Consider a circle with radius 1. The perimeter of the circle is divided into 360 degrees. When we choose one point randomly on the circle, it has a position anywhere from 0 to 360 degrees. For each point, we have a uniform distribution over 360 degrees.
02

Consider the complementary event for part (a)

Instead of directly calculating the probability that none of the points are in a semicircle, calculate the probability that at least one point is inside any semicircle. The complementary event is that at least one point is inside some semicircle.
03

Calculate the probability for part (a)

A semicircle covers 180 degrees, exactly half of the circle. The probability that one point falls inside the semicircle is \(\frac{1}{2}\). For \(n\) points to always avoid the semicircle, all must fall in the remaining half. The probability for each point to fall in the complementary semicircle is \(\frac{1}{2}\). Therefore, the probability that none will fall in the semicircle is \((\frac{1}{2})^n\). There are \(n\) such possible semicircle positions, so multiplying by \(n\), the probability is \(n (\frac{1}{2})^n = n 2^{1-n}\).
04

Set up the probability space for part (b)

For part (b), consider choosing points on the surface of a sphere. The surface can be divided into hemispheres, similar to how a circle can be divided into semicircles. Each hemisphere covers one-half of the sphere's surface area.
05

Consider the complementary event for part (b)

Instead of calculating the probability that none of the points lie in a hemisphere, calculate the probability that at least one point is inside any hemisphere. The complementary event is that at least one point is in one of the hemispheres.
06

Calculate the probability for part (b)

If one point is chosen at random, it has a probability of \(\frac{1}{2} \) of falling within any particular hemisphere. For \(n\) points to avoid all hemispheres, they all must fall in the complementary hemispherical belt. The probability for each point to fall outside a hemisphere is also \(\frac{1}{2}\), making the probability of no points in any hemisphere \((\frac{1}{2})^n\). Considering all orientations and positions of hemispheres over the sphere and accounting for overlapping regions, the overall probability is given by \((n^2 - n + 2)2^{-n}\).
07

Summarizing the results

For both parts (a) and (b), random placement combined with overlapping coverage and symmetry led specified probabilities which indicate the likelihood of avoiding a significant section of a geometric figure.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Semicircle
A semicircle is half of a circle. It covers 180 degrees of the full 360-degree rotation of a circle. When choosing random points on the circle's perimeter, calculating the probability that none of these points fall within a semicircle involves considering the number of possible positions for the points and the area covered by the semicircle.
The complementary event, which involves calculating where at least one point lands within the semicircle, helps us arrive at the probability of none falling there. Given that a semicircle covers half the circle, each point has a probability of 0.5 of landing within it.
  • Probability for a point inside the semicircle: \(\frac{1}{2}\)
  • n points to avoid the semicircle: \(\left(\frac{1}{2}\right)^n\)
  • n possible semicircle positions (hence the overall probability for no points in any semicircle): \(n \cdot \left(\frac{1}{2}\right)^n=n \cdot 2^{1-n}\)
Hemisphere
A hemisphere represents half of a sphere, similar to a semicircle in a circle. In probability terms, when choosing points randomly on the sphere鈥檚 surface, we explore the chance of all these points avoiding a hemisphere. Each hemisphere covers one-half of the sphere's surface.
To solve the problem, we determine the probability of selecting points so that none fall within any hemisphere, by using the complementary event of at least one being inside it. Again, each point has a \(\frac{1}{2}\) probability of landing inside any hemisphere.
  • For one hemisphere, probability per point: \(\frac{1}{2}\)
  • All points avoiding hemisphere: \(\left(\frac{1}{2}\right)^n\)
  • The resulting overall probability accounts for hemisphere arrangements and overlapping: \((n^2 - n + 2) \cdot 2^{-n}\)
Probability Space
In probability theory, a probability space is a mathematical framework consisting of three parts: a sample space, a set of events, and a probability function. The sample space is the set of all possible outcomes; in our case, it can be points on a circle鈥檚 perimeter or points on a sphere鈥檚 surface.
A set of events might include potential locations of these points relative to geometric figures like semicircles or hemispheres. The probability function assigns a likelihood to each event, guiding calculations of the expected outcome in different scenarios.
  • Sample Space (Example): All points on a circle or sphere
  • Event Set: Points within a semicircle or hemisphere
  • Probability Function: Likelihood of points avoiding these regions
Understanding these elements is key to solving geometric probability problems like this one.
Geometric Probability
Geometric probability involves unconventional probability spaces, typically involving lengths, areas, or volumes, rather than counting discrete outcomes. In the context of this exercise, it focuses on the likelihood of random points avoiding specific geometric shapes like semicircles or hemispheres.
The main idea is to assess how a random point placement interacts with these shapes and what the chance is of these points not landing within certain parts of the geometric figure. Geometric probability takes into account:
  • The shape鈥檚 area or proportion (e.g., semicircle covering half of the circle)
  • Uniform distribution of points on the shape's boundary or surface
  • Utilization of complementary probabilities to simplify calculations
Applying geometric probability means leveraging the symmetry and uniformity inherent in shapes and arrangements, as illustrated in such exercises.

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Most popular questions from this chapter

A pack of cards is well shuffled and one hand of 13 cards is dealt to each of four players. Find the probability that: (a) Each player has an ace. (b) At least one player has a complete suit. (c) My hand is void in at least one suit. (d) Some player has all the aces. What is the most likely distribution among suits in the dealer's hand?

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Derangements (3.4.4 Revisited) Write \(\bar{x}=x_{1}+x_{2}+\cdots x_{n}\). Explain why the number of derangements of the first \(n\) integers is the coefficient of \(x_{1} x_{2} x_{3} \ldots x_{n}\) in $$ \begin{aligned} \left(x_{2}\right.&\left.+x_{3}+\cdots+x_{n}\right)\left(x_{1}+x_{3}+\cdots+x_{n}\right) \cdots\left(x_{1}+x_{2}+\cdots+x_{n-1}\right) \\ &=\left(\bar{x}-x_{1}\right)\left(\bar{x}-x_{2}\right) \cdots\left(\bar{x}-x_{n}\right) \\ &=(\bar{x})^{n}-(\bar{x})^{n-1} \sum x_{i}+\cdots+(-)^{n} x_{1} x_{2} \ldots x_{n} \end{aligned} $$ and hence deduce the expression for \(P_{n}\) given in (3.4.4).

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