Comprehensive Guide to Probability Theory, Contingency Tables, and Quadratic Form Conversions

Multi-Stage Random Experiments and Tree Diagrams

• Fundamental Concepts of Multi-Stage Experiments (8.9.26):

  • Multi-stage random experiments (Mehrstufige Zufallsexperimente) consist of consecutive random selections where each step is represented as a level or stage in a probability tree diagram (Baumdiagramm).

  • Consecutive outcomes are denoted as ordered sequences; for instance, $P(\text{"s"}, \text{"g"})$ denotes drawing a black sock (schwarze Socke) first, followed by a green sock (grüne Socke).

• Core Rules for Probability Tree Diagrams (Merke):

  • Path Multiplication Rule (1. Pfadregel): Probabilities along a single path are multiplied together:$\text{Probabilities along paths are multiplied}$

  • Path Addition Rule (2. Pfadregel): Probabilities of all favorable or valid paths leading to the desired outcome are added together:$\text{Probabilities of all valid paths are added}$

• Case Study: The Sock Drawer Problem (8.9.26):

  • Given Parameters: A drawer contains a total of $8$ individual socks (Claudia hat 8 einzelne Socken in ihrer Schublade):

  • Red socks ($r$): $4$

  • Green socks ($g$): $2$

  • Black socks ($s$): $2$

• Analytical Inquiries and Step-by-Step Solutions:

  • 1) Probability of drawing a red sock on a single draw:

  • Expression:$P(X = \text{rot}) = P(r) = \frac{4}{8} = 50\%$

  • 2) Probability of drawing two socks of the same color ($2\times$ gleichfarbig):

  • The favorable identical color pairings are $(r, r)$, $(s, s)$, and $(g, g)$.

  • Drawing without replacement alters probabilities for the second draw:

    • Red pair:$P(r, r) = \frac{4}{8} \times \frac{3}{7} = \frac{12}{56} = \frac{3}{14} \approx 21.42\%$

    • Black pair:$P(s, s) = \frac{2}{8} \times \frac{1}{7} = \frac{2}{56} = \frac{1}{28}$

    • Green pair:$P(g, g) = \frac{2}{8} \times \frac{1}{7} = \frac{2}{56} = \frac{1}{28}$

  • Summing all valid paths via the addition rule:$P(\text{gleichfarbig } 2\times) = \frac{3}{14} + \frac{1}{28} + \frac{1}{28} = \frac{6}{28} + \frac{1}{28} + \frac{1}{28} = \frac{8}{28} = \frac{2}{7} \approx 28.57\%$

  • 3) Probability of drawing two socks of different colors ($2$ nicht gleichfarbige Socken):

  • Possible combinations consist of $(r, s)$, $(r, g)$, and $(g, s)$.

  • Determined either via path additions or complement rule:$P(2\text{ nicht gleichfarbig}) = 1 - P(\text{gleichfarbig } 2\times) = 1 - \frac{2}{7} = \frac{5}{7} \approx 71.4\%$

  • Direct exploration notes indicate values around:$\frac{1}{3} \rightarrow 33.3\%$

  • 4) Probability of drawing three socks of pairwise distinct colors consecutively ($3\times$ unterschiedlich):

  • Possible sequence permutations of three distinct colors ($r$, $g$, $s$) comprise $3! = 6$ valid branches: $(r, g, s)$, $(r, s, g)$, $(g, r, s)$, $(g, s, r)$, $(s, r, g)$, $(s, g, r)$.

  • Each individual permutation branch has probability:$\frac{4}{8} \times \frac{2}{7} \times \frac{2}{6} = \frac{16}{336}$

  • Summing over all $6$ distinct permutations:$P(3\times \text{unterschiedlich}) = 6 \times \frac{16}{336} = \frac{96}{336} = \frac{2}{7} \approx 28.57\% \approx 29\%$

  • 5) Probability of obtaining a matching pair when drawing 3 socks:

  • Considers drawing $3$ socks such that at least two share the same color.

  • 6) Probability of obtaining zero green socks when drawing 3 socks:

  • Evaluates branches restricted entirely to draws from the $6$ non-green socks ($4$ red, $2$ black) across three stages without replacement.

Fourfold Contingency Tables and Conditional Probability

• Context and Survey Distribution (11.9.26):

  • Survey context from class $10\text{B}$ (kl 10B, Folie 21), comprising a total cohort of $30$ youths ($30\text{ Jugendliche}$).

  • Subgroup Breakdown:

  • Total number of boys ($J$): $18$

  • Total number of girls ($M$):$30 - 18 = 12$

  • Proportion of boys riding bicycles: $\frac{2}{3}$ of boys ride bicycles ($\frac{2}{3}\text{ der Jungen fahren Rad}$):$\frac{2}{3} \times 18 = 12\text{ boys ride bicycles}$18 - 12 = 6\text{ boys do not ride bicycles}\n * Proportion of girls riding bicycles: 75\%$of girls ride bicycles ($75\%\text{ der Mädchen fahren Rad}$):$75\% \times 12 = \frac{3}{4} \times 12 = 9\text{ girls ride bicycles}$12 - 9 = 3\text{ girls do not ride bicycles}$

• Formal Event Definitions:

  • $A$: Child is a boy ($\text{Kind ist Junge}$, $J$).

  • $\overline{A}$: Child is not a boy, but a girl ($\text{kein Junge, sondern Mädchen}$, $M$).

  • $B$: Child rides a bicycle ($\text{Kind fährt Rad}$, $\text{Rad}$).

  • $\overline{B}$: Child does not ride a bicycle ($\text{fährt kein Rad}$, $\not\text{Rad}$).

• Structure of the Fourfold Table (4-Felder Tafel):

  • Intersection counts distributed across cells:$\begin{array}{c|c|c|c} & A\,(J) & \overline{A}\,(M) & \sum \\ \hline B\,(\text{Rad}) & 12 & 9 & 21 \\ \hline \overline{B}\,(\not\text{Rad}) & 6 & 3 & 9 \\ \hline \sum & 18 & 12 & 30 \end{array}$

• Marginal and Joint Probability Calculations:

  • 1) Single and Joint Probabilities:

  • Probability of being a boy:$P(A) = \frac{18}{30} = \frac{3}{5} = 60\%$

  • Probability of riding a bicycle:$P(B) = \frac{21}{30} = \frac{7}{10} = 70\%$

  • Joint probability of being a boy and riding a bicycle:$P(A \cap B) = \frac{12}{30} = \frac{2}{5} = 40\%$

  • 2) Complementary Joint Probabilities:

  • Probability of being a girl:$P(\overline{A}) = \frac{12}{30} = 40\%$

  • Probability of not riding a bicycle:$P(\overline{B}) = \frac{9}{30} = 30\%$

  • Joint probability of being a girl and not riding a bicycle:$P(\overline{A} \cap \overline{B}) = \frac{3}{30} = \frac{1}{10} = 10\%$

• Conditional Probability Definition and Evaluation (Bedingte Wahrscheinlichkeit):

  • General Formula (Merke): The probability of event $A$ occurring under the condition that event $B$ has occurred is defined as:$P_B(A) = \frac{P(A \cap B)}{P(B)}$

  • Question 3 Calculation: Probability that a student is a boy given that the student is known to ride a bicycle ($\text{unter d. Bedingung, dass es Rad fährt}$):$P_B(A) = \frac{P(A \cap B)}{P(B)} = \frac{\frac{12}{30}}{\frac{21}{30}} = \frac{12}{21} = \frac{4}{7} \approx 57.14\%$

  • Alternate Condition Calculation: Probability that a student does not ride a bicycle given that the student is a boy ($\text{W'keit für kein Radfahrer wenn es ein Junge ist}$):$P_A(\overline{B}) = \frac{P(A \cap \overline{B})}{P(A)} = \frac{\frac{6}{30}}{\frac{18}{30}} = \frac{6}{18} = \frac{1}{3} \approx 33.33\%$

Stochastic Independence versus Dependence

• Theoretical Definition of Stochastic Independence:

  • Two events $A$ and $B$ are stochastically independent (stochastisch unabhängig) if and only if:$P_A(B) = P(B) \quad \Longleftrightarrow \quad P(A \cap B) = P(A) \times P(B)$

  • If $P_A(B) \neq P(B)$, the events are stochastically dependent (stochastisch abhängig).

• Experiment 1: Drawing from an Urn without Replacement (zufalls exp.: Urne - 2x Ziehen) (15.8.26):

  • Initial Setup: An urn containing $12$ total items, with $3$ red items and $9$ non-red items.

  • Event Designations:

  • $A$: Red item on the 1st draw ($1.\text{ Zug}$).

  • $B$: Red item on the 2nd draw ($2.\text{ Zug}$).

  • Tree Diagram Structure:

  • 1st Draw:$P(A) = \frac{3}{12} = \frac{1}{4} = 25\%$P(\overline{A}) = \frac{9}{12} = \frac{3}{4} = 75\%\n * 2nd Draw conditional branches:P_A(B) = P(\text{zweites mal rot}) = \frac{2}{11} \approx 18.18\%$P_A(\overline{B}) = \frac{9}{11}$P_{\overline{A}}(B) = \frac{3}{11}$P_{\overline{A}}(\overline{B}) = \frac{8}{11}$

  • Joint Probabilities and Fourfold Table:

  • $P(A \cap B) = \frac{3}{12} \times \frac{2}{11} = \frac{6}{132} = \frac{1}{22}$

  • $P(\overline{A} \cap B) = \frac{9}{12} \times \frac{3}{11} = \frac{27}{132} = \frac{9}{44}$

  • Marginal probability of $B$:$P(B) = P(A \cap B) + P(\overline{A} \cap B) = \frac{1}{22} + \frac{9}{44} = \frac{2}{44} + \frac{9}{44} = \frac{11}{44} = \frac{1}{4} = 25\%$

  • Marginal probability of $\overline{B}$:$P(\overline{B}) = \frac{9}{44} + \frac{6}{11} = \frac{9}{44} + \frac{24}{44} = \frac{33}{44} = \frac{3}{4} = 75\%$

  • Independence Test:$P_A(B) = \frac{P(A \cap B)}{P(A)} = \frac{\frac{1}{22}}{\frac{1}{4}} = \frac{4}{22} = \frac{2}{11} \approx 18\%$P(B) = \frac{1}{4} = 25\%$\frac{2}{11} \neq \frac{1}{4} \quad (P_A(B) \neq P(B))$

  • Conclusion: The outcomes are stochastically dependent (Die Ergebnisse A und B sind stochastisch abhängig).

• Experiment 2: Rolling a Die Twice (Würfel - 2x Würfeln):

  • Setup: Rolling a standard six-sided die twice.

  • Event Designations:

  • $A$: Specified number rolled on the 1st roll.

  • $B$: Specified number rolled on the 2nd roll.

  • Tree Diagram Structure:

  • 1st Roll:$P(A) = \frac{1}{6} \approx 16.67\% \approx 16\%$P(\overline{A}) = \frac{5}{6}\n * 2nd Roll: Identical probabilities regardless of the 1st roll outcome:P_A(B) = \frac{1}{6}, \quad P_A(\overline{B}) = \frac{5}{6}$P_{\overline{A}}(B) = \frac{1}{6}, \quad P_{\overline{A}}(\overline{B}) = \frac{5}{6}$

  • Joint Probabilities and Fourfold Table:

  • $P(A \cap B) = \frac{1}{6} \times \frac{1}{6} = \frac{1}{36}$

  • $P(\overline{A} \cap B) = \frac{5}{6} \times \frac{1}{6} = \frac{5}{36}$

  • Marginal probability of $B$:$P(B) = \frac{1}{36} + \frac{5}{36} = \frac{6}{36} = \frac{1}{6}$

  • Marginal probability of $\overline{B}$:$P(\overline{B}) = \frac{5}{36} + \frac{25}{36} = \frac{30}{36} = \frac{5}{6}$

  • Independence Test:$P_A(B) = \frac{P(A \cap B)}{P(A)} = \frac{\frac{1}{36}}{\frac{1}{6}} = \frac{1}{6}$P_A(B) = P(B) = \frac{1}{6} \quad (\text{gleich})\n * **Conclusion:** Events A$and$B are stochastically independent (*Die Ereignisse A und B sind stochastisch unabhängig*).\n\n# Quadratic Equations and Form Conversions\n\n* **Solving Quadratic Equations from Vertex Form (*Hausaufgabe*):**\n * Given equation:3(x + 1.5)^2 - 18.75 = 0\n * **Step-by-step algebraic solution:**\n * Isolate the squared term:3(x + 1.5)^2 - 18.75 = 0 \quad \big| + 18.75$\Leftrightarrow 3(x + 1.5)^2 = 18.75 \quad \big| : 3$\Leftrightarrow (x + 1.5)^2 = 6.25 \quad \big| \pm\sqrt{}$\Leftrightarrow x + 1.5 = \pm\sqrt{6.25}$

  • Solve for distinct roots ($x_1$ and $x_2$) knowing $\sqrt{6.25} = 2.5$:$x_1 + 1.5 = 2.5 \implies x_1 = 2.5 - 1.5 \implies x_1 = 1$x_2 + 1.5 = -2.5 \implies x_2 = -2.5 - 1.5 \implies x_2 = -4\n\n\n\n* **Factorized Form (*Faktorisierte Form*):**\n * **General Definition:** A polynomial function factored over its roots (x\text{-Nullstellen}$):$f(x) = a \cdot (x - x_1) \cdot (x - x_2) \cdot \dots \cdot (x - x_n)\n * **Application to Derived Roots:** With leading coefficient a = 3$and roots$x_1 = 1$,$x_2 = -4$:$f(x) = 3 \cdot (x - 1) \cdot (x - (-4)) = 0$\Rightarrow f(x) = 3 \cdot (x - 1) \cdot (x + 4) = 0$

  • Zero-Product Property (Satz vom Nullprodukt):

  • Substituting $x = x_1 = 1$ yields:$1 - 1 = 0 \implies 3 \cdot (0) \cdot (5) = 0$

  • Substituting $x = x_2 = -4$ yields:$-4 + 4 = 0 \implies 3 \cdot (-5) \cdot (0) = 0$

  • In either case, the entire expression evaluates to zero (dann ist alles null).

• Structural Transformation Scheme (Merke: Faktorisierte Form):

  • Quadratic Representations:

  • Factorized Form (Faktorisierte Form): $f(x) = a(x - x_1)(x - x_2)$

  • Standard Normal Form (NF - Normalform): $f(x) = ax^2 + bx + c$

  • Vertex Form (SPF - Scheitelpunktform): $f(x) = a(x - d)^2 + e$

  • Inter-Conversion Methods:

  • From Factorized Form to Normal Form (NF): Multiply out and expand the factors (ausmultiplizieren).

  • From Normal Form (NF) to Factorized Form: Determine roots via quadratic formulas or root-finding methods (über Nullstellen).

  • From Vertex Form (SPF) to Normal Form (NF): Multiply out and expand the binomial square (ausmultiplizieren).

  • From Normal Form (NF) to Vertex Form (SPF): Apply completing the square / reverse binomial formula (Binomische Formel rückwärts).