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Drawing without replacement

This page was translated from German with AI. It has not been checked by a native speaker. Original page in German

An urn contains \(2\) red and \(3\) blue balls. You draw twice and do not put the ball back (ohne Zurücklegen). Then the probabilities change in the second draw.
📄 Open worksheet (PDF, German)

The video

In the video you see how the branches of the second draw change: after red and after blue there are different balls in the urn.

Subtitles are available in ten languages. Choose them at the bottom right of the video.

Contents of the video

  1. 0:00 Introduction
  2. 0:23 The tree without replacement
  3. 0:55 The product rule
  4. 1:31 The sum rule
  5. 2:00 Careful, mistake!
  6. 2:32 Try it yourself

The most important points

Without replacement

After the first draw there is one ball less in the urn. After red: \(\frac{1}{4}\) red, \(\frac{3}{4}\) blue. After blue: \(\frac{2}{4}\) red, \(\frac{2}{4}\) blue.

The denominator in the second draw is 4, not 5.
The denominator in the second draw is \(4\), not \(5\).

The path rules still apply

Multiply along a path: \(\frac{2}{5} \cdot \frac{1}{4} = \frac{2}{20}\). Add between paths: exactly one red ball \(=\) \(\frac{6}{20} + \frac{6}{20} = \frac{12}{20} = 60\,\%\).

2/20 + 6/20 + 6/20 + 6/20 = 1
\(\frac{2}{20} + \frac{6}{20} + \frac{6}{20} + \frac{6}{20} = 1\)
The green paths fit the event “exactly one red ball”. With replacement it was only 48\,\%.
The green paths fit the event “exactly one red ball”. With replacement it was only \(48\,\%\).

The worksheet

Here you can see the worksheet for the video (in German). First it explains everything again, then there are practice tasks at three levels. You can also open it as a PDF and print it.

📄 Open worksheet (PDF, German)

The worksheet – Page 1
Page 1/4
The worksheet – Page 2
Page 2/4
The worksheet – Page 3
Page 3/4
Show the worksheet solutions
The worksheet – Page 4
Page 4/4

Try it yourself

Task

A bag contains \(4\) yellow and \(2\) green sweets. You take two without putting them back. How big is the probability of a) two yellow b) two different colours?

Show solution

a) \(\frac{4}{6} \cdot \frac{3}{5} = \frac{12}{30} = \frac{2}{5}\) b) \(\frac{4}{6} \cdot \frac{2}{5} + \frac{2}{6} \cdot \frac{4}{5} = \frac{16}{30} = \frac{8}{15}\)

Keep learning with a language model

Du bist meine Mathe-Lehrkraft. Ich lerne Deutsch als Zweitsprache. Unser Thema: Ziehen ohne Zurücklegen (Baumdiagramm mit veränderten Wahrscheinlichkeiten). Stelle mir nacheinander fünf kurze Aufgaben. Warte immer auf meine Antwort. Sag mir dann, ob sie richtig ist, und erkläre kurz, warum. Benutze einfache deutsche Sätze.

How this was made: The idea, structure and examples of the series are mine. I made them with the help of AI: a language model (LLM) built the slides of the video following my instructions. The voice was generated with Qwen3-TTS, modelled on my own voice.

The subtitles, the translations, the worksheet and this page were also made with the help of AI. The translations have not been checked by native speakers.

The videos, worksheets, texts and prompts in this series are licensed under CC BY 4.0: you may use, change and share them if you name the source.