Comprehensive Guide to Probability Rules and Calculations

Fundamentals of Probability

• Probability Definition:

  • A probability is a quantitative value inclusive between $0$ and $1$ describing the chance or likelihood of a specific event occurring.
  • Mathematically, for any event $E$, its probability $P(E)$ must satisfy:     $0 \le P(E) \le 1$

• The Law of Large Numbers:

  • As the number of repetitions of a probability experiment increases, the proportion with which a certain outcome is observed gets progressively closer to the theoretical probability of that outcome.
  • Coin Flipping Example:
  • As a coin is flicked/tossed an increasing number of times, the ratio of heads or tails relative to the total number of tosses approaches $\frac{1}{2}$ ($0.5$), which is the true theoretical probability of landing a head or a tail.          

• Core Terminology:

  • Experiment: A structured process that leads to the occurrence of one, and only one, of several possible observations.
  • Outcome: The specific result obtained from a single trial of an experiment.
  • Sample Space ($S$): The set containing all possible outcomes of a random experiment.
  • Single coin toss: $S = \{H, T\}$
  • Single standard die roll: $S = \{1, 2, 3, 4, 5, 6\}$
  • Two coin toss: $S = \{(H,H), (H,T), (T,H), (T,T)\}$
  • Probability Event ($E$): A subset of the sample space $S$. If the result of performing an experiment produces an outcome contained within event $E$, then event $E$ has occurred.

Probability Rules and Probability Models

• Fundamental Rules of Probability:

  1. The probability of any individual event $E$ must strictly be between $0$ and $1$, inclusive:      $0 \le P(E) \le 1$
  2. The sum of the probabilities of all simple outcomes comprising the entire sample space $S = \{e_1, e_2, \dots, e_n\}$ must equal exactly $1$:      $P(e_1) + P(e_2) + \dots + P(e_n) = 1$

• Probability Model:

  • A probability model is a comprehensive list of all possible outcomes of a probability experiment paired with each outcome's corresponding probability.
  • Any valid probability model must simultaneously fulfill both fundamental rules of probability.

• Verification of a Probability Model (Peanut M&M Chocolate Candies Example):

  • Consider selecting a single candy at random from a bag of peanut M&M milk chocolates containing brown, yellow, red, blue, orange, and green colors.      
  • Color Probability Distribution:
  • Brown: $0.12$
  • Yellow: $0.15$
  • Red: $0.12$
  • Blue: $0.23$
  • Orange: $0.23$
  • Green: $0.15$
  • Model Validation:
  • Rule 1 Verification: Every individual outcome probability lies within the range $0 \le P(E) \le 1$.
  • Rule 2 Verification: Summing all probabilities yields:       $0.12 + 0.15 + 0.12 + 0.23 + 0.23 + 0.15 = 1.00$
  • Because both rules are strictly met, this table represents a valid probability model.

• Special Event Classifications:

  • Impossible Event: An event with a probability of $0$ ($P(E) = 0$). It can never occur.
  • Certainty Event: An event with a probability of $1$ ($P(E) = 1$). It occurs in every single trial.
  • Unusual Event: An event that has a significantly low likelihood of occurrence, standardly defined as any event where:     $P(E) < 0.05$

Methods for Computing Probabilities

• Empirical (Relative Frequency) Method:

  • Uses empirical data collected directly through physical observation or evidence rather than pure theoretical reasoning or logic.
  • Formula for empirical probability:     $P(E) \approx \frac{\text{Frequency of } E}{\text{Total number of repetitions of experiment}} = \frac{f}{n} = \text{Relative Frequency}$
  • Dining Out Survey Example (April 2010):
  • A sample of $n = 521$ adults was surveyed regarding how often they dine out.
  • Survey Results & Empirical Probability Model:              
    • Several times a week: Frequency = $103$; $P(\text{Several times a week}) = \frac{103}{521} \approx 0.1977$
    • Once or twice a week: Frequency = $204$; $P(\text{Once or twice a week}) = \frac{204}{521} \approx 0.3916$
    • A few times a month: Frequency = $130$; $P(\text{A few times a month}) = \frac{130}{521} \approx 0.2495$
    • Very rarely: Frequency = $79$; $P(\text{Very rarely}) = \frac{79}{521} \approx 0.1516$
    • Never: Frequency = $5$; $P(\text{Never}) = \frac{5}{521} \approx 0.0096$
  • Interpretations:
    • Probability an adult dines out a few times per month: $0.2495$
    • Evaluating if "Never" dining out is unusual: $P(\text{Never}) = 0.0096$. Since $0.0096 < 0.05$, it is classified as an unusual event.

• Classical Method:

  • Applies exclusively to experiments with equally likely outcomes (where every simple event possesses the exact same probability of occurrence).
  • If an experiment contains $n$ equally likely outcomes and event $E$ consists of $m$ outcomes:     $P(E) = \frac{\text{Number of ways } E \text{ can occur}}{\text{Total number of possible outcomes}} = \frac{N(E)}{N(S)} = \frac{m}{n}$
  • Fun-Size M&M Bag Example:
  • Contents: $9$ brown, $6$ yellow, $7$ red, $4$ orange, $2$ blue, $2$ green.
  • Total candies $N(S) = 9 + 6 + 7 + 4 + 2 + 2 = 30$
  • Probability of selecting Yellow: $P(\text{Yellow}) = \frac{6}{30} = 0.2$
  • Probability of selecting Blue: $P(\text{Blue}) = \frac{2}{30} \approx 0.067$
  • Likelihood comparison: Since $P(\text{Yellow}) = \frac{6}{30}$ and $P(\text{Blue}) = \frac{2}{30}$, selecting a yellow candy is exactly $3$ times as likely as selecting a blue candy.

• Subjective Method:

  • A subjective probability is determined based on personal judgment, experience, intuition, or belief rather than mathematical counting or empirical observational data.

• Classification Identification Examples:

  • "The next toss of a fair coin will land on heads" $\rightarrow$ Classical Probability (based on symmetric, equally likely outcomes).
  • "Italy will win soccer's World Cup the next time the competition is held" $\rightarrow$ Subjective Probability (based on opinion/estimation).
  • "The probability that a family of three children has two boys and one girl is approximately $0.36$ based on a survey of $500$ families" $\rightarrow$ Empirical Probability (calculated from observational survey data: $\frac{180}{500} = 0.36$).

Addition Rules for Disjoint and Non-Disjoint Events

• Disjoint (Mutually Exclusive) Events:

  • Two events $E$ and $F$ are disjoint (mutually exclusive) if and only if they share no outcomes in common ($E \cap F = \emptyset$). They cannot occur simultaneously.
  • Addition Rule for Disjoint Events:     $P(E \text{ or } F) = P(E) + P(F)$
  • Extended Addition Rule for Disjoint Events:     $P(E \text{ or } F \text{ or } G \dots) = P(E) + P(F) + P(G) + \dots$
  • Venn Diagram Representation:          
  • Example: Sample space $S = \{0, 1, 2, 3, 4, 5, 6, 7, 8, 9\}$ ($N(S) = 10$).
  • Let $E$ be "choose a number $\le 2$" ($E = \{0, 1, 2\}$, $N(E) = 3$).
  • Let $F$ be "choose a number $\ge 8$" ($F = \{8, 9\}$, $N(F) = 2$).
  • $P(E) = \frac{3}{10} = 0.3$, $P(F) = \frac{2}{10} = 0.2$
  • Since $E$ and $F$ are disjoint:       $P(E \text{ or } F) = P(E) + P(F) = 0.3 + 0.2 = 0.5$

• General Addition Rule (Non-Disjoint Events):

  • For any two events $E$ and $F$ (whether overlapping or disjoint):     $P(E \text{ or } F) = P(E) + P(F) - P(E \text{ and } F)$
  • Set Theory Notation:     $P(E \cup F) = P(E) + P(F) - P(E \cap F)$
  • $P(E)$: Probability of event $E$ occurring.
  • $P(F)$: Probability of event $F$ occurring.
  • $P(E \cup F)$: Probability of event $E$ or event $F$ occurring.
  • $P(E \cap F)$: Probability of both events $E$ and $F$ occurring at the same time.          

• Application Examples of the General Addition Rule:

  • School Sports Participation:
  • In a school of $100$ students, $50$ play football ($F$), $20$ play basketball ($B$), and $10$ play both.
  • $P(F) = \frac{50}{100} = 0.5$, $P(B) = \frac{20}{100} = 0.2$, $P(F \cap B) = \frac{10}{100} = 0.1$
  • Probability that a randomly chosen student plays at least one sport:       $P(F \cup B) = 0.5 + 0.2 - 0.1 = 0.6$
  • Card Selection from a Standard Deck (52 Cards):
  • Drawing a Two ($T$) or a Five ($F$):       $P(T) = \frac{4}{52} = \frac{1}{13}$, $P(F) = \frac{4}{52} = \frac{1}{13}$       Events are mutually exclusive:       $P(T \cup F) = \frac{1}{13} + \frac{1}{13} = \frac{2}{13}$
  • Drawing an Eight ($E$) or a Heart ($H$):       $P(E) = \frac{4}{52} = \frac{1}{13}$, $P(H) = \frac{13}{52} = \frac{1}{4}$, $P(E \cap H) = \frac{1}{52}$$P(E \cup H) = \frac{4}{52} + \frac{13}{52} - \frac{1}{52} = \frac{16}{52} = \frac{4}{13}$
  • Drawing a Queen ($Q$) or a Red card ($R$):       $P(Q) = \frac{4}{52} = \frac{1}{13}$, $P(R) = \frac{26}{52} = \frac{1}{2}$, $P(Q \cap R) = \frac{2}{52} = \frac{1}{26}$$P(Q \cup R) = \frac{4}{52} + \frac{26}{52} - \frac{2}{52} = \frac{28}{52} = \frac{7}{13}$

• Contingency Table Example: Cigar Smoking and Cancer Mortality:

  • Study data for $10,000$ men:
  • Never smoked cigars: $5$ Died from Cancer, $9250$ Did Not Die from Cancer (Row Total = $9255$)
  • Former cigar smoker: $14$ Died from Cancer, $352$ Did Not Die from Cancer (Row Total = $366$)
  • Current cigar smoker: $66$ Died from Cancer, $313$ Did Not Die from Cancer (Row Total = $379$)
  • Column Totals: Died from Cancer = $85$; Did Not Die from Cancer = $9915$; Grand Total = $10,000$
  • Calculations:
  • Probability a randomly selected individual died from cancer $P(D)$, where total died = $5 + 14 + 66 = 85$:       $P(D) = \frac{85}{10000} = 0.0085$
  • Probability individual was a current cigar smoker $P(C)$, where total current smokers = $66 + 313 = 379$:       $P(C) = \frac{379}{10000} = 0.0379$
  • Probability individual died from cancer AND was a current cigar smoker $P(D \cap C)$, given by the intersection cell ($66$):       $P(D \cap C) = \frac{66}{10000} = 0.0066$
  • Probability individual died from cancer OR was a current cigar smoker $P(D \cup C)$, using the General Addition Rule:       $P(D \cup C) = P(D) + P(C) - P(D \cap C) = 0.0085 + 0.0379 - 0.0066 = 0.0398$

Complement Rule

• Complement Definition:
  • The complement of an event $A$, denoted as $A^c$ or $A'$, consists of all outcomes in the sample space $S$ that are not contained in event A$.\n  \n  \n\n* **Complement Rule Formula**:\n * If E$is an event, the probability that$E$does not occur is$1$minus the probability that$E does occur:\n    P(A^c) = 1 - P(A)\n\n# Independence and the Multiplication Rule\n\n* **Definitions**:\n * **Independent Events**: Two events A$and$B are independent if the occurrence or non-occurrence of one event does not affect or alter the probability of the other event occurring.\n * **Dependent Events**: Two events are dependent if the occurrence of event A$influences or changes the probability of event$B occurring.\n\n* **Multiplication Rule for Independent Events**:\n * If events A$and$B are independent:\n    P(A \cap B) = P(A) \times P(B)\n\n* **Comparing Mutually Exclusive and Independent Events**:\n * Mutually exclusive and independent events are fundamentally distinct concepts.\n * **Mutually Exclusive Events**: Cannot happen together (P(A \cap B) = 0).\n * **Independent Events**: Occurrence of one does not change the probability of the other (P(A \cap B) = P(A) \times P(B)).\n  \n  \n\n* **Independence Identification Examples**:\n * "Roll a die and get a 3" and "toss a coin and get heads" \rightarrow **Independent**.\n * "Earned a bachelor's degree" and "earn more than $100,000 per year" \rightarrow **Dependent**.\n\n* **Solving "At Least One" Problems Using Complements**:\n * Problem: Find the probability of obtaining at least one head in three tosses of a fair coin.\n * Let P(H) represent the probability of getting at least one head.\n * The complement H^c$is obtaining no heads, which corresponds solely to three consecutive tails:$\frac{1}{2} \times \frac{1}{2} \times \frac{1}{2} = \frac{1}{8}.\n * Using the Complement Rule:\n    P(H) = 1 - P(H^c) = 1 - \frac{1}{8} = \frac{7}{8}\n * **Probability Tree Diagram**:\n * A diagram that visually structures sequential events and outcomes alongside their calculated probabilities.\n    \n    \n\n* **Multiplication Rule Example (Independent Survival)**:\n * According to vital statistics, the probability that a 60-year-old female survives the year is 0.99186$($99.186\%).\n * Assuming survival between two chosen females is independent:\n    P(\text{First survives and second survives}) = P(\text{First survives}) \times P(\text{Second survives})\n    P(\text{Both survive}) = (0.99186) \times (0.99186) = 0.9838\n\n# Conditional Probability and General Multiplication Rule\n\n* **Conditional Probability**:\n * Notation P(A \mid B)$reads as "the probability of event$A$given event$B".\n * Represents the probability that event A$occurs under the condition that event$B has already occurred.\n * **Conditional Probability Formula**:\n    P(A \mid B) = \frac{P(A \cap B)}{P(B)}\n    \n    \n\n* **Exam Performance Example**:\n * Given: P(\text{Pass Statistics}) = 0.62$,$P(\text{Pass Physics}) = 0.48$,$P(\text{Pass Both}) = 0.29.\n * Probability Sue passes Physics given that she passed Statistics:\n    P(P \mid S) = \frac{P(P \cap S)}{P(S)} = \frac{0.29}{0.62} \approx 0.4677\n\n* **Formal Proof of Independence via Conditional Probability**:\n * Events A$and$B are independent if and only if:\n    P(A \mid B) = \frac{P(A \cap B)}{P(B)} = \frac{P(A) \times P(B)}{P(B)} = P(A)\n\n* **In-Hospital Cardiac Arrest Study (Contingency Table Analysis)**:\n * Study investigating 86,748 hospital cardiac arrest patients categorized by shift and outcome:\n * **Day or Evening Shift**: Survived = 11,604$; Did Not Survive =$46,989$; Shift Total =$58,593\n * **Graveyard Shift (After 11 PM)**: Survived = 4,139$; Did Not Survive =$24,016$; Shift Total =$28,155\n * **Total**: Survived = 15,743$; Did Not Survive =$71,005$; Grand Total =$86,748\n * **Calculations**:\n * Probability cardiac arrest occurred during Graveyard Shift P(G)$, using$\frac{28155}{86748}:\n      P(G) \approx 0.3246\n * Probability patient survived for discharge P(S)$, using$\frac{15743}{86748}:\n      P(S) \approx 0.1815\n * Probability patient survived given cardiac arrest was on Graveyard Shift P(S \mid G)$, using$\frac{4139}{28155}:\n      P(S \mid G) \approx 0.1470\n * Probability cardiac arrest occurred on Graveyard Shift given patient survived P(G \mid S)$, using$\frac{4139}{15743}:\n      P(G \mid S) \approx 0.2629\n * **Independence Evaluation & Hospital Recommendations**:\n * Are "Survived for Discharge" (S$) and "Graveyard Shift" ($G) independent?\n * No, because P(S) \approx 0.1815 eq P(S \mid G) \approx 0.1470$(and$P(G) \approx 0.3246 eq P(G \mid S) \approx 0.2629).\n * Recommendation: Survival rates are noticeably lower during the graveyard shift (14.70\%$versus$19.80\% during day/evening shifts). Hospitals should re-evaluate staffing levels, nocturnal medical coverage, and immediate emergency response procedures after 11 PM.\n\n* **General Multiplication Rule**:\n * Calculates the joint probability of two events occurring together (E \cap F) across independent and dependent scenarios.\n * For **Dependent Events**:\n    P(E \cap F) = P(E) \times P(F \mid E)\n * For **Independent Events**:\n    P(E \cap F) = P(E) \times P(F)\n * **Urn Ball Selection Example (Sampling Without Replacement)**:\n * An urn contains 4$red balls and$6$white balls (total =$10 balls). Two balls are selected sequentially without replacement.\n * *a) Probability 1st ball is Red and 2nd ball is White (P(R_1 \cap W_2)$)*:       $P(R_1) = \frac{4}{10}$$P(W_2 \mid R_1) = \frac{6}{9}$$P(R_1 \cap W_2) = \left(\frac{4}{10}\right) \times \left(\frac{6}{9}\right) = \frac{24}{90} \approx 0.267$
  • *b) Probability both balls drawn are White (P(W_1 \cap W_2)$)*:\n      P(W_1) = \frac{6}{10}\n      P(W_2 \mid W_1) = \frac{5}{9}\n      P(W_1 \cap W_2) = \left(\frac{6}{10}\right) \times \left(\frac{5}{9}\right) = \frac{30}{90} \approx 0.333$$