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Von fünf verschiedenen Paar Schuhen werden vier Schuhe zufällig weggenommen. Mit welcher Wahrscheinlichkeit ist wenigstens ein Paar darunter?

Short Answer

Expert verified
\[ P(\text{at least one pair}) = 1 - \frac{\binom{5}{4} \times 2^4}{\binom{10}{4}} \]

Step by step solution

01

- Understanding the problem

We need to determine the probability that at least one pair of shoes is among the four shoes randomly taken from five different pairs.
02

- Calculate total possible outcomes

Calculate the total number of ways to choose 4 shoes from the 10 available shoes. This is given by the binomial coefficient \(\binom{10}{4}\).
03

- Calculate favorable outcomes

Now, calculate the number of ways in which no pair is chosen. First, choose 4 out of the 5 pairs (\(\binom{5}{4}\)). Then, choose one shoe from each pair (\(2^4\)). Thus, the number of favorable outcomes is: \(\binom{5}{4} \times 2^4\).
04

- Calculate probability

Divide the number of favorable outcomes by the total number of outcomes to get the probability that no pair is chosen: \[ P(\text{no pair}) = \frac{\binom{5}{4} \times 2^4}{\binom{10}{4}} \]
05

- Find the complement

To find the probability that at least one pair is chosen, subtract the probability of no pair being chosen from 1: \[ P(\text{at least one pair}) = 1 - P(\text{no pair}) \]

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Key Concepts

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

Combinatorics
Combinatorics is a branch of mathematics that deals with counting, arranging, and combining objects. It is useful in determining the number of possible outcomes in various scenarios. In problems involving probability, combinatorics helps us figure out how many ways something can happen.
In the given problem, we are determining how many ways we can choose 4 shoes out of 10. This involves calculating combinations and permutations based on the problem's constraints.
Understanding combinatorics allows us to break down a complex problem into manageable parts. We can then find the total possible outcomes and the number of favorable outcomes.
Binomial Coefficient
The binomial coefficient is a key element in combinatorics that helps us calculate the number of ways to choose a certain number of objects from a larger set, without regard to the order of selection. It is denoted as \(\binom{n}{k}\), where \(n\) is the total number of objects and \(k\) is the number of objects to choose.
The formula for the binomial coefficient is:
\[\binom{n}{k} = \frac{n!}{k!(n-k)!}\]
In our problem, the total number of ways to choose 4 shoes from 10 is calculated using \(\binom{10}{4}\). This tells us how many different groups of 4 shoes we can form from the 10 available shoes.
Next, we use another binomial coefficient \(\binom{5}{4}\) to calculate the number of ways to choose 4 pairs out of 5. Once we have our 4 pairs, we choose one shoe from each, giving us the total favorable outcomes when no pair is chosen.
Complement Rule
The complement rule is an essential concept in probability theory. It is used to find the probability of the complement of an event. The complement of an event \(A\) is basically everything that is not \(A\). The probability of the complement of \(A\) can be calculated by subtracting the probability of \(A\) from 1:
\[ P(A^c) = 1 - P(A) \]
In our problem, instead of directly calculating the probability that at least one pair of shoes is chosen, we first calculate the probability that no pairs are chosen. This is P(no pair). Then, we use the complement rule to find the probability of the event we are interested in (at least one pair being chosen).
Using the complement rule simplifies the problem and often makes calculations easier. By focusing on the simpler scenario first (no pairs chosen), we can more easily determine the probability of the more complex scenario (at least one pair chosen).

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Most popular questions from this chapter

Wenn auf der Speisekarte in \(\$ 3.1\) drei Sorten Eis und 2 Sorten Mus zur Auswahl stehen, wieviele verschiedene Menus gibt es dann? Wenn wir berücksichtigen, daß der Gast das Gemüse oder das Dessert oder beides weglassen kann, wieviele verschiedene Menus ergeben sich dann?

Drei Würfel werden zweimal geworfen. Mit welcher Wahrscheinlichkeit zeigen sie beidesmal dieselben Zahlen (a) wenn die Würfel unterscheidbar sind, (b) wenn sie es nicht sind ? ( Hinweis: Machen Sie eine Fallunterscheidung nach dem Muster des ersten Wurfs: ein Paar, ein Tripel, oder alle verschieden; nehmen Sie dann den zweiten Wurf entsprechend hinzu.)

Auf wieviele Weisen können sich 4 Jungen und 4 Mädchen zu Paaren anordnen? Auf wieviele Weisen können sie eine Reihe bilden, in der Jungen und Mädchen abwechseln?

In einem kleinen See fängt man 100 Forellen und setzt sie nach Markierung wieder ein. Später fängt man wieder 100 und findet darunter 7 markierte. Wie groß ist die Wahrscheinlichkeit dafür, falls der See n Forellen enthält? (Was ist Ihre beste Schätzung für den wahren Wert von n ? Letzteres ist eine Frage von der Art, wie sie die Statistik stellt.)

Bei einer Einberufungslotterie werden die 366 Tage (einschließlich des 29. Februar) des Jahres erfaßt. Mit welcher Wahrscheinlichkeit sind die ersten 180 (natürlich ohne Zurücklegen) ausgelosten Tage gleichmäßig auf die 12 Monate verteilt? Mit welcher Wahrscheinlichkeit enthalten die ersten 30 ausgelosten Tage keinen im August oder September?

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