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Chapter 5: Problem 27

Ein vollkommener W眉rfel wird \(\mathrm{n}\)-mal geworfen. M und \(\mathrm{m}\) seien die maximale bzw. die minimale Punktzahl, die man dabei erh盲lt. Bestimmen Sie \(\mathrm{P}(\mathrm{m}=2, \mathrm{M}=5) !\) [Hinweis: Beginnen Sie mit \(\mathrm{P}(\mathrm{m} \geq 2 ; \mathrm{M} \leqslant 5) !]\)

Short Answer

Expert verified
The probability \(\text{P}(m=2, M=5)\) is given by \[ \bigg(\frac{4}{6}\bigg)^n - 2\bigg(\frac{3}{6}\bigg)^n + 2\bigg(\frac{2}{6}\bigg)^n \]

Step by step solution

01

Interpret the Problem

Given a fair die, we need to find the probability that the minimum roll is 2 and the maximum roll is 5 over rolls.
02

Consider the Event \(\text{P}(m \geq 2, M \leq 5)\)

First, find the probability that all rolls are between 2 and 5. This includes calculating the probability of the minimum value being at least 2 and the maximum value being at most 5.
03

Calculate Probability for Single Roll

The die has 6 faces. Probability of a single roll being between 2 and 5 is \[ \text{P}(2 \leq X \leq 5) = \frac{4}{6} = \frac{2}{3} \]
04

Extend to \(\text{n}\) Rolls

Since there are \(\text{n}\) independent rolls, the probability that each roll is between 2 and 5 is: \[ \bigg(\frac{2}{3}\bigg)^n \]
05

Adjust for \(\text{P}(m > 2, M < 5)\)

To account for m = 2 and M = 5 specifically, subtract cases where the rolls never include 2 or never include 5: \[ \bigg(\frac{2}{3}\bigg)^n - \bigg(\frac{1}{2}\bigg)^n - \bigg(\frac{5}{6}\bigg)^n + \bigg(\frac{4}{6}\bigg)^n \]
06

Apply Inclusion-Exclusion Principle

Using inclusion-exclusion principle, the adjusted term would be \[ \text{P}(m=2, M=5) = \bigg(\frac{4}{6}\bigg)^n - 2\bigg(\frac{3}{6}\bigg)^n + 2\bigg(\frac{2}{6}\bigg)^n \]

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Key Concepts

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

fair die
A fair die is a six-sided die where each face has an equal probability of landing face up. This means that each side has a \(\frac{1}{6}\) chance of appearing on any given roll. Fair dice are used in probability problems to simplify calculations and make predicting outcomes straightforward. In probability theory, using a fair die ensures that all outcomes are equally likely, which is essential for accurate computations.
independent rolls
When rolling a die multiple times, each roll is considered independent. This means the result of one roll does not influence the result of another.
If you roll a die \(\text{n} \) times, each roll has the same probabilities, regardless of previous rolls.
This concept of independence is crucial for correctly calculating probabilities across multiple events.
For example, if you are looking for the probability that all rolls result in numbers between 2 and 5, you multiply the individual probabilities of each roll being within that range.
inclusion-exclusion principle
The inclusion-exclusion principle is a key concept for calculating the probabilities of combined events. It helps to add and subtract probabilities to avoid double-counting. For instance, in our exercise, we aim to find the probability that all rolls are between 2 and 5 without including cases where rolls are never 2 or 5. The formula is: \(\text{P}(2 \leq X \leq 5)^n - \text{P}(3 \leq X \leq 5)^n - \text{P}(2 \leq X \leq 4)^n + \text{P}(2 \leq X \leq 4)^n\).
This principle helps in precisely calculating the probability by considering all necessary sub-events.
minimum and maximum values
In probability problems involving a series of rolls, finding the minimum (\( \text{m} \)) and maximum (\( \text{M} \)) values is a common task.
In our specific problem, we calculate the probability that the minimum roll is 2 and the maximum roll is 5.
This involves calculating cases where the lowest value across \( \text{n} \) rolls is 2 and the highest is 5.
To solve this, we use probabilities of ranges and adjust using inclusion-exclusion to exclude cases where minimum is never 2 or maximum is never 5.
As illustrated before, we use the adjusted formula to get the precise probability: \( P(\text{m=2, M=5}) = \bigg(\frac{4}{6}\bigg)^n - 2\bigg(\frac{3}{6}\bigg)^n + 2\bigg(\frac{2}{6}\bigg)^n \).
This ensures accurate results by including all needed probabilities.

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Most popular questions from this chapter

Bei einer Kernreaktion kann ein gewisses Teilchen in zwei oder in drei Teilchen zerfallen, oder es zerf盲llt 眉berhaupt nicht. Die Wahrscheinlichkeiten f眉r diese M枚glichkeiten sind \(\mathrm{p}_{2}, \mathrm{p}_{3}, \mathrm{p}_{1} .\) Die neuen Teilchen verhalten sich ebenso und sind unabh盲ngig sowohl voneinander wie auch von der vorigen Reaktion. Man bestimme die Verteilung der Anzahl aller Teilchen nach zwei Reaktionen.

Die drei Genotypen \(A A, A a\), aa sollen in den Verh盲ltnissen \(p^{2}: 2 p q: q^{2}\) mit \(p+q=1\) vorkommen. Wenn ein zuf盲llig aus der Bev枚lkerung zusammengekommenes Eltern- paar einen Nachkommen vom Typp AA hat, wie gro脽 ist dann die Wahrscheinlichkeit daf眉r, da脽 ein weiteres Kind von ihnen ebenfalls vom Typ AA ist ? Man beantworte dieselbe Frage f眉r Aa anstelle von \(A A\).

Zeigen Sie, da脽 eine auf \([0, \infty)\) definierte, nichtwachsende Funktion \(\varphi\), die der C a u c hy's ch e n F u n k t i o n a lg l e i ch u ng $$ \varphi(\mathrm{s}+\mathrm{t})=\varphi(\mathrm{s}) \varphi(\mathrm{t}) \quad, \quad \mathrm{s} \geq 0, \mathrm{t} \geq 0 $$ gen眉gt, gleich \(\mathrm{e}^{-\lambda \mathrm{t}}\) fur ein \(\lambda>0\) sein \(\mathrm{mu} \beta .\) Daher hat eine positive zuf盲llige Variable T die Eigenschaft $$ P(T>s+t \mid T>s)=P(T>t) \quad, \quad s \geq 0, t \geq 0 $$ genau dann, wenn sie einer Exponentialverteilung gehorcht. [Hinweis: \(\varphi(0)=1\); \(\varphi(1 / n)=\alpha^{1 / n}\), wobei \(\alpha=\varphi(1) ; \varphi(m / n)=a^{m / n} ;\) wenn \(m / n \leqslant t<(m+1) / n\), dann ist \(\alpha^{(\mathrm{m}+1) / \mathrm{n}} \leqslant \varphi(t) \leqslant \alpha^{\mathrm{m} / \mathrm{n}}\); daraus schlie脽t man die Aussage f眉r allgemeines \(t\), indem man n gegen \(\infty\) gehen l盲脽t.]

Man nehme an, da脽 die a priori Wahrscheinlichkeiten p bei der Aufgabe mit dem Sonnenaufgang ( Beispiel 9 von \(85.2)\) nur die Werte \(k / 100,1 \leqslant k \leqslant 100\) annehmen k枚nnen, und zwar einen jeden mit Wahrscheinlichkeit \(1 / 100\). Man berechne daf眉r \(\mathrm{P}\left(\mathrm{S}^{\mathrm{n}+1} \mid \mathrm{s}^{\mathrm{n}}\right)\); wenn man 100 durch \(\mathrm{N}\) ersetzt und \(\mathrm{N}\) gegen \(\infty\) gehen l盲脽t, welcher Grenzwert ergibt sich dann?

Unter der Voraussetzung, da脽 die kleinere Zahl kleiner als \(\mathrm{x}\) ist, bestimme man die Verteilung der gr枚脽eren. [Hinweis: man betrachte \(\mathrm{P}(\min <\mathrm{x}, \max <\mathrm{y})\) und die beiden F盲lle \(x \leqslant y\) und \(x>y .]\)
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