Question

(a) Find the probability that in two tosses of a coin, one is heads and one tails. That in six tosses of a die, all six of the faces show up. That in 12 tosses of a 12-sided die, all 12 faces show up. That in \(n\) tosses of an \(n\) -sided die, all \(n\) faces show up. (b) The last problem in part (a) is equivalent to finding the probability that, when \(n\) balls are distributed at random into \(n\) boxes, each box contains exactly one ball. Show that for large \(n,\) this is approximately \(e^{-n} \sqrt{2 \pi n}\)

Short answer

For large n, the probability is approximately \[ e^{-n} \sqrt{2 \pi n} \].

Step-by-step solution

Find the Probability of One Head and One Tail in Two Coin Tosses

First, list all possible outcomes when a coin is tossed twice: HH, HT, TH, TT. There are a total of 4 possible outcomes. Now, find the outcomes with one head and one tail. These outcomes are HT and TH. There are 2 such favorable outcomes. The probability is the number of favorable outcomes divided by the total number of outcomes. Hence, the probability is \[ P_1 = \frac{2}{4} = \frac{1}{2} \]

Find Probability that All Six Faces Show Up in Six Tosses of a Die

Calculate the number of favorable arrangements. Each arrangement must have one of each face (1, 2, 3, 4, 5, 6) exactly once. Since the order does matter, there are \[6!\] possible arrangements. Each toss of the die can result in any one of the six faces. Thus, there are a total of \[6^6\] possible outcomes. The probability is given by\[ P_2 = \frac{6!}{6^6} \]

Find Probability that All 12 Faces Show Up in 12 Tosses of a 12-sided Die

Similarly, for a 12-sided die, each face must appear exactly once. Thus, the number of favorable arrangements is \[12!\]. The total number of possible outcomes is \[ 12^{12} \]. Therefore, the probability is given by\[ P_3 = \frac{12!}{12^{12}} \]

Generalize for an n-sided Die

To generalize, in n tosses of an n-sided die, each face must appear exactly once. Therefore, the number of favorable arrangements is \[ n! \] and the total number of possible outcomes is \[ n^n \]. Thus, the probability is given by\[ P_n = \frac{n!}{n^n} \]

Use Stirling's Approximation for Large n

Using Stirling's approximation for large n, \[ n! \] is approximately given by\[ n! \sim \sqrt{2 \pi n} \left( \frac{n}{e} \right)^n \]. Thus,\[ P_n = \frac{n!}{n^n} \approx \frac{\sqrt{2 \pi n} \left( \frac{n}{e} \right)^n}{n^n} = \frac{\sqrt{2 \pi n}}{e^n} = e^{-n} \sqrt{2 \pi n} \]

Relate to Distributing Balls into Boxes

This final result is equivalent to distributing n balls into n boxes such that each box contains exactly one ball. Thus, for large n, the probability that when n balls are distributed at random into n boxes, each box contains exactly one ball, is approximately\[ e^{-n} \sqrt{2 \pi n} \]

Key concepts

Coin Toss Probability

Understanding the probability of different outcomes when a coin is tossed is fundamental in probability theory. When you toss a coin, there are two possible outcomes: heads (H) or tails (T). In a scenario of two coin tosses, the total number of possible outcomes is calculated by measuring the permutations, which gives us four (HH, HT, TH, TT). To find the probability of obtaining one heads and one tails, we focus on the favorable outcomes (HT and TH), which are two. The probability formula is the number of favorable outcomes divided by the total number of possible outcomes. Therefore, the probability is: \[ P_1 = \frac{2}{4} = \frac{1}{2} \].Calculating probabilities for multiple coin tosses follows similar principles, with each additional toss exponentially increasing the number of possible outcomes.

Die Toss Probability

Calculating probabilities with a die toss involves similar principles as with coin tosses but with more outcomes per individual roll. For a single roll of a six-sided die, there are six possible outcomes (1, 2, 3, 4, 5, 6). If we extend this to six rolls and seek the probability that each face appears exactly once, we need to consider the permutations of the six faces. The number of favorable arrangements is the number of permutations, which is \[ 6! \].Each die roll is independent, and the total number of possible outcomes is \[ 6^6 \].Thus, the probability is calculated using the formula: \[ P_2 = \frac{6!}{6^6} \].As the number of faces and rolls increases, such as with a twelve-sided die or an n-sided die, similar calculations apply using factorials and exponentiation.

Stirling's Approximation

Stirling's approximation is a powerful mathematical tool used to estimate factorials for large numbers, which is particularly useful in probability theory. The approximation is given by: \[ n! \sim \sqrt{2 \pi n} \left( \frac{n}{e} \right)^n \].When applied to the generalized probability problem where an n-sided die is tossed n times, and each face appears exactly once, Stirling's approximation helps simplify the factorial in the numerator: \[ P_n = \frac{n!}{n^n} \approx \frac{\sqrt{2 \pi n} \left( \frac{n}{e} \right)^n}{n^n} \].This further simplifies to: \[ P_n \approx \frac{\sqrt{2 \pi n}}{e^n} = e^{-n} \sqrt{2 \pi n} \].Such approximations allow for easier computations and understanding of probabilities in cases involving large numbers.

Combinatorics

Combinatorics is the branch of mathematics dealing with counting, arrangement, and combination of objects. It is crucial for solving problems related to permutations and probabilities. For instance, in the die toss probability problems mentioned, calculating \[ 6! \] or \[ 12! \] involves understanding permutations, where the order of outcomes matters. Combinatorics provides tools like factorials to count arrangements when order matters and combinations when order doesn't matter. For instance, in the problem of distributing balls into boxes, combinatorics helps determine how many different ways balls can be placed in boxes. Permutation and combination calculations are foundational for more complex probability problems in various fields.

Random Distribution Model

A random distribution model is used to describe the probability and behavior of distributing a set of objects randomly. In the generalized problem, where n balls are placed into n boxes such that each box contains exactly one ball, we can use a random distribution model to understand this process. The key idea is to calculate the probability that each box gets exactly one ball. For large values of n, this probability is approximated using: \[ P_n \approx e^{-n} \sqrt{2 \pi n} \].This model is not only helpful in theoretical calculations but also has practical applications in fields like computer science, logistics, and operations research. Understanding random distribution models allows for predictions and optimizations in scenarios involving randomness and multiple possibilities.