Section 1.1 Probability Rules Notes

Probability notation and the basic model

  • In probability, p(E) denotes the probability of an event E. The event is a set of outcomes we care about.
  • The probability is set up as a fraction:
    • Numerator: number of successes (outcomes that realize the event E).
    • Denominator: number in the sample space (the total number of possible outcomes).
  • Sample space = the set of all possible outcomes; an outcome is a single result; an event is the subset of outcomes we’re interested in.
  • Examples:
    • Coin toss with a fair coin: sample space has 2 outcomes. Probability of heads =
      P( ext{Heads}) = \frac{1}{2}.
      Numerator = 1 (one way to get heads), Denominator = 2 (two possible outcomes).
    • Dice roll with a fair six-sided die: sample space has 6 outcomes. Probability of rolling a 5 =
      P(5) = \frac{1}{6}.
      Numerator = 1 (one five on the die), Denominator = 6.
  • Probability values lie in the interval
    0 \le P(E) \le 1.
    There are no negative probabilities.
  • Do not always reduce fractions when doing raw calculations. Keeping raw data (denominator showing the sample size) helps you interpret the results. Some questions later will require reduced fractions, but often raw form is preferred for clarity.
  • When answers involve decimals, fractions, or percents, you will encounter all three forms in this course.
  • If the entire probability space sums to 1 (or 100%), the rule is:
    \sum P(Ei) = 1 \quad\text{over all mutually exclusive outcomes } Ei.
    Example: Heads and tails on a fair coin: P(Heads)+P(Tails) = \tfrac{1}{2}+\tfrac{1}{2}=1.

Key concepts: notations, sample space, and events

  • Event vs outcome:
    • Outcome: a single result within the sample space (e.g., H on a coin toss, 4 on a die).
    • Event: a set of outcomes we’re interested in (e.g., "rolling an even number", or "getting heads").
  • Complement: the event not happening, denoted with a prime (or superscript c):
    • If E is an event, its complement is E^c (not E).
    • Probability complement rule:
      P(E^c) = 1 - P(E)
      or equivalently, P(E) = 1 - P(E^c).
  • Practical way to compute complements:
    • Use the formula: P(E^c) = 1 - P(E).
    • Or, add up probabilities of all outcomes not in E (common sense): for example, not blue in marbles is the sum of red, green, and yellow probabilities which should total the complement.
  • Sum rule intuition: the probabilities of all mutually exclusive outcomes from a given experiment must sum to 1 (or 100%).
    • Example with marbles: blue 4/10, red 2/10, green 1/10, yellow 3/10; total probability =
      \frac{4}{10}+\frac{2}{10}+\frac{1}{10}+\frac{3}{10}=\frac{10}{10}=1.

Worked examples: coins and dice

  • Coin basics:
    • A coin has 2 outcomes: heads (H) and tails (T).
    • Probability of heads on a single flip: P(H)=\frac{1}{2}; Probability of tails: P(T)=\frac{1}{2}.
    • Notation: heads or tails together cover all possibilities, so P(H) + P(T) = 1 (100%).
  • Rolling a die:
    • A standard die has 6 sides. Probability of rolling any specific number (e.g., 4) is P(4)=\frac{1}{6}.
    • Not rolling a 4 (complement): P(\text{not }4) = 1 - P(4) = 1 - \frac{1}{6} = \frac{5}{6}.
    • If you convert to decimal: \frac{5}{6} \approx 0.833\overline{3}.
  • Notation cautions:
    • 1/6 as a decimal is 0.1666… (repeating). When exact form is requested, prefer \dfrac{1}{6} rather than decimals with repeating bars on computer input.
    • If asked for a percent (nearest whole percent), you would round accordingly (e.g., \approx 16.7\% \to 17\%).
  • Two coins (order matters):
    • Outcomes: HH, HT, TH, TT (4 outcomes).
    • The order of flips matters unless the problem says otherwise.
    • In general, for independent trials with two outcomes per trial, the total number of outcomes for n flips is 2^n.
  • Two dice or more complex experiments:
    • For n dice, each die has 6 outcomes, so total outcomes = 6^n.
    • When combining different experiments (e.g., coins and dice), multiply the number of outcomes: e.g., 2^{\text{#coins}} \cdot 6^{\text{#dice}}.
  • Practical counting rule: base = number of possibilities per trial; exponent = number of trials (the action/verb).
    • Examples:
    • 4 coins: base 2, exponent 4 → total outcomes = 2^4 = 16.
    • 5 dice: base 6, exponent 5 → total outcomes = 6^5.

Tree diagrams and sample spaces for multiple tosses

  • Tree diagrams visually display outcomes by branching at each trial.
  • Two coins example on the board shows: HH, HT, TH, TT as four distinct outcomes; HT and TH are different because the first coin and second coin are labeled differently.
  • Three coins example yields 8 outcomes (two choices per coin, repeated three times):
    • HH H, HH T, H T H, H T T, T H H, T H T, T T H, T T T (order matters).
  • Takeaway: order matters unless the problem explicitly says that it does not.
  • Summary for independent trials:
    • If you flip n coins: total outcomes = 2^n.
    • If you roll n dice: total outcomes = 6^n.
    • If you mix trials (coins and dice), multiply: 2^{\text{coins}} \cdot 6^{\text{dice}}.

Decks of cards and color-based sample spaces

  • Card basics:
    • A standard deck has 52 cards: 4 suits (spades, clubs, hearts, diamonds).
    • Suits: spades and clubs are black; hearts and diamonds are red.
    • Each suit has 13 cards: Ace through 10 (counted as numbers 1–10) and 3 face cards (J, Q, K).
    • Therefore, 4 suits × 13 cards per suit = 52 cards total.
  • Observing color of two drawn cards (without specifying replacement):
    • Possible color sequences (order matters) are: RR, RB, BR, BB.
    • The long sample space for two draws is these 4 outcomes, reflecting both the color of the first card and the color of the second card.
    • If you were drawing only one card, the sample space would be simply {red, black} (2 outcomes).
  • Practical takeaway: for two draws (two cards), you use a sample space with 4 outcomes when focusing on color; the total count grows with the number of draws.

Blood type probability model and complement example

  • Blood types considered: O, A, B, AB (positive/negative not included here).
  • The probabilities for each blood type should sum to 1 (or 100%). If P(O) = 0.45, then:
    • Probability of not having type O:
      P(O^c) = 1 - P(O) = 1 - 0.45 = 0.55.
    • Alternatively, by summing the remaining types (A, B, AB):
      P(A) + P(B) + P(AB) = 0.55.
  • Practical notes:
    • The instructor demonstrates two ways to compute the complement: using the rule (1 - P(O)) or via summing the other categories.
    • The math palette in the platform can convert between decimals, fractions, and percentages as needed.

In-class assignment navigation and MyOpenMath specifics

  • Course structure:
    • Unit 1 -> Chapter 1 sections 1.1 to 1.3 contain notes; Homework 1 corresponds to these notes.
    • In-class assignment 1.1 is four questions; completed together in class.
  • Question 1 (blood type example):
    • The probability model lists all blood types with their probabilities, summing to 1.
    • Not O method: compute 1 - P(O); Common-sense method: sum the probabilities of A, B, and AB.
    • The instructor notes the use of a palette to switch between formats (fraction, decimal, exponent forms).
  • Question 2 (odds):
    • Odds can be in favor or against. Here, "odds against" means the first number is losses, the second is wins.
    • Denominator for odds is the total number of outcomes: Losses + Wins.
    • If odds against are 1 : 23 (losses : wins), then the probability of winning is \frac{23}{24}.
    • The instructor shows how to enter these values in MyOpenMath (expressions vs. decimals).
  • Question 3 (deck of cards):
    • Textbook provides a digital deck; review the card structure (52 cards; 4 suits; 10 numbered cards per suit; 3 face cards per suit).
    • When selecting two cards and observing color, the sample space is the color sequences across both cards (RR, RB, BR, BB).
    • If only observing color of one card, the sample space would be {red, black} (2 outcomes).
  • Question 4 (counting outcomes with dice and coins):
    • Rolling three dice (base = 6) and flipping a certain number of coins (base = 2) yields total outcomes calculated as 6^{3} \cdot 2^{c} where c is the number of coins.
    • In the example, the total given was 6,912 outcomes for that particular configuration; the key idea is to multiply the base outcomes for each independent component.
  • How to navigate the textbook in Brightspace:
    • Example problems: Example 2 (sample space for one coin), Example 3, Example 4 (in-class reference).
    • Problems 32–34 (and nearby problems) refer back to Example 4 when dealing with two or three coin flips (e.g., exactly two heads, at least one head, at most one head).
  • Quick guidance for homework problems:
    • Two-card color problem: use the long sample space for two cards (RR, RB, BR, BB).
    • Three-coin problems: list all 8 possible sequences to count favorable outcomes and the total outcomes as 2^3.
    • For problems asking for "at most" or "at least" a certain number of heads, count the corresponding subset of outcomes from the 2^n total outcomes.

Quick reference: formulas and rules to memorize

  • Probability model notation:
    P(E) = \frac{\text{# of successes}}{\text{# in sample space}}.
  • Probability range:
    0 \le P(E) \le 1.
  • Complement rule:
    P(E^c) = 1 - P(E).
  • Sum rule (total probability):
    \sum{i} P(Ei) = 1,
    where the E_i are all mutually exclusive outcomes covering the entire sample space.
  • Independent trials counting:
    • Two outcomes per trial (coins): total outcomes for n flips = 2^n.
    • Six outcomes per trial (dice): total outcomes for n dice = 6^n.
    • For mixed experiments, multiply: \text{Total} = \prodj (\text{base}j^{\text{exponent}_j}).
  • Notation for readability on exams/online platforms:
    • Fractions, decimals, and percents are interchangeable forms of the same probability value.
    • On MyOpenMath, you may enter expressions directly (e.g., \dfrac{1}{6} or 0.1666\overline{6}) and convert as required; use reduced fractions when asked.

Philosophical and practical implications

  • Understanding probability notation reduces confusion: P(E) is not just a number; it encodes how many favorable outcomes exist relative to the total possibilities in the experiment.
  • Complementarity reframes “not happening” scenarios in a simple, intuitive way; it emphasizes that uncertainty splits into a chosen event and its complement.
  • The counting principle (base and exponent) helps model real-world processes with multiple trials, reinforcing how independence leads to multiplicative counting. It also highlights the power of simple rules (e.g., 2^n for coin flips, 6^n for dice) to scale to complex problems.
  • Real-world relevance: blood type probabilities, odds vs. probability, and deck-of-cards problems model common decision-making and risk assessment tasks.

Summary checklist for section 1.1

  • Read P(E) as a fraction with numerator = number of successes and denominator = sample space size.
  • Identify the sample space and the event; distinguish outcomes vs events.
  • Remember that 0 ≤ P(E) ≤ 1 and that the probabilities across all outcomes sum to 1.
  • Use the complement to handle "not" questions: P(E^c) = 1 - P(E).
  • For independent trials, count total outcomes by multiplying per-trial possibilities; use 2^n for coins, 6^n for dice, and multiply when combining different trials.
  • For two-card color questions, two draws yield 4 color-outcome sequences: RR, RB, BR, BB (order matters).
  • Deck basics: 52 cards, 4 suits, 13 cards per suit; color split 26 red / 26 black; color-outcome sample spaces depend on the number of draws.
  • Use textbook examples and the MyOpenMath interface (palette) to switch between fractions, decimals, and exponents when needed.
  • Be comfortable with both rule-based (complement and sum) and intuitive (common-sense) approaches to probability questions.