Multi-Stage Chance Experiments and Tree Diagrams

Multi-Stage Chance Experiments and Tree Diagrams

  • Experiment Overview:

    • Two balls are drawn sequentially from a container.

    • Initial container composition: 33 total balls consisting of 11 blue ball and 22 red balls.

    • Replacement condition: The first drawn ball is not replaced back into the container before drawing the second ball (sampling without replacement).

    • Representation method: A shortened/condensed tree diagram (Verkürztes Baumdiagramm) is used to represent the experiment steps efficiently.


Condensed tree diagram for a two-stage chance experiment without replacement

Probability Formulas and Stage Breakdown

  • General Probability Calculation Formula:

    • Probability P=Favorable OutcomesPossible Outcomes\text{Probability } P = \frac{\text{Favorable Outcomes}}{\text{Possible Outcomes}}

    • Probability P=Number of Balls of Specific ColorTotal Number of Balls\text{Probability } P = \frac{\text{Number of Balls of Specific Color}}{\text{Total Number of Balls}}

  • First Stage (First Draw):

    • Probability of drawing a blue ball:     P(blue)=13P(\text{blue}) = \frac{1}{3}

    • Probability of drawing a red ball:     P(red)=23P(\text{red}) = \frac{2}{3}

  • Second Stage (Second Draw without Replacement):

    • Branch following a blue ball on the first draw:

    • Remaining container composition: 00 blue balls and 22 red balls (total 22 balls).

    • Probability of drawing a red ball:       P(red∣blue)=22=1P(\text{red} \mid \text{blue}) = \frac{2}{2} = 1

    • Probability of drawing a blue ball:       P(blue∣blue)=0P(\text{blue} \mid \text{blue}) = 0

    • Branches following a red ball on the first draw:

    • Remaining container composition: 11 blue ball and 11 red ball (total 22 balls).

    • Probability of drawing a red ball:       P(red∣red)=12P(\text{red} \mid \text{red}) = \frac{1}{2}

    • Probability of drawing a blue ball:       P(blue∣red)=12P(\text{blue} \mid \text{red}) = \frac{1}{2}

Calculating Event Probabilities

  • Individual Path Calculations:

    • Probability of drawing red then blue:     P(red,blue)=23×12=13P(\text{red}, \text{blue}) = \frac{2}{3} \times \frac{1}{2} = \frac{1}{3}

    • Probability of drawing blue then blue:     P(blue,blue)=0P(\text{blue}, \text{blue}) = 0

    • Probability of drawing blue then red:     P(blue,red)=13×1=13P(\text{blue}, \text{red}) = \frac{1}{3} \times 1 = \frac{1}{3}

  • Combined Event Calculations:

    • Combined probability where color order does not matter (drawing one red ball and one blue ball in any order):     P(blue,red or red,blue)=13×1+23×12=23P(\text{blue}, \text{red} \text{ or } \text{red}, \text{blue}) = \frac{1}{3} \times 1 + \frac{2}{3} \times \frac{1}{2} = \frac{2}{3}

    • Probability of drawing a red ball on the second draw (P(?;red)P(? ; \text{red})):     P(?;red)=13×1+23×12=23P(? ; \text{red}) = \frac{1}{3} \times 1 + \frac{2}{3} \times \frac{1}{2} = \frac{2}{3}

Fundamental Path Rules

  • Rule 1 (Path Multiplication Rule / Pfadmultiplikationsregel):

    • When following a single continuous path through a tree diagram, the probabilities along the individual branches of that path are multiplied together.

  • Rule 2 (Path Addition Rule / Pfadadditionsregel):

    • When calculating the overall probability of an event that consists of multiple distinct branches/paths, the individual path probabilities must be added together.