4.2: And, Or, Independent, and Mutually Exclusive - Study Notes

Page 1: 4.2 - And, Or, Independent, and Mutually Exclusive

  • Coffee order example (Lois Lane): In the Daily Planet cafeteria, employees can purchase either hot or iced coffee, in sizes small, medium, and large.

    • Total number of choices for a coffee order = 2 \times 3 = 6.
    • Lois’ coffee order is an example of an event. (An event is any outcome or set of outcomes from a random process.)

  • Key takeaway from this page: The coffee order illustrates the idea of an event being a specific outcome or a set of outcomes from a sample space.

Page 2: AND vs. OR; Basic Set Operations; Sample Space Example

  • Let (A = {1,2,3,4,5}) and (B = {4,5,6,7,8}).

    • The event (A \cup B) (A OR B) is any outcome that belongs to (A) or to (B) (or both).
    • The event (A \cap B) (A AND B) is any outcome that belongs to both (A) and (B).
    • Thus:
    • (A \cup B = {1,2,3,4,5,6,7,8}).
    • (A \cap B = {4,5}).

  • Item-pull from a box (Santa Ana College Stat C1000, 4.2) – Sample Space:

    • Box contains:
    • 2 sweaters (S) and 2 hats (H); one item knitted (K) and the other crocheted (C).
    • The item pulled at random has the sample space:
    • ({S\,K), (S\,C), (H\,K), (H\,C)}).

  • Questions and answers:

    • (b) In how many ways can the item be a sweater OR knitted?
    • Outcomes that are sweaters: (S\,K) and (S\,C).
    • Outcomes that are knitted: (S\,K) and (H\,K).
    • Union yields: {(S\,K), (S\,C), (H\,K)}) → 3 ways.
    • (c) What is the probability that the item is a sweater OR knitted?
    • (P(S\, OR\, K) = \dfrac{3}{4} = 0.75).
    • (d) In how many ways can the item be a sweater AND knitted?
    • Only ((S\,K)) matches both criteria → 1 way.
    • (e) What is the probability that the item is a sweater AND knitted?
    • (P(S\,\text{AND}\,K) = \dfrac{1}{4} = 0.25).

  • Summary: This page reinforces basic set operations (union vs intersection) and applies them to a small, concrete sample space.

Page 3: Continued - Independent vs. Dependent; Independence Check

  • Independent vs. Dependent:

    • Two events are independent if (P(A \cap B) = P(A)P(B)).
    • Otherwise, they are dependent.
    • In words: For independent events, the occurrence of one event does not change the probability of the other.

  • Example: One card drawn from a standard 52-card deck.

    • Which pairs of events are independent?
    • Pair 1: Drawing a spade ((A)) and drawing a Jack ((B)).
      • (P(A) = \dfrac{13}{52} = \dfrac{1}{4}).
      • (P(B) = \dfrac{4}{52} = \dfrac{1}{13}).
      • (P(A \cap B) = P(\text{Jack of Spades}) = \dfrac{1}{52}).
      • Check: (P(A)P(B) = \dfrac{1}{4} \cdot \dfrac{1}{13} = \dfrac{1}{52} = P(A \cap B)).
      • Therefore, independent.
    • Pair 2: Drawing a spade (A) and drawing a black card (B).
      • (P(B) = \dfrac{26}{52} = \dfrac{1}{2}).
      • (P(A) = \dfrac{13}{52} = \dfrac{1}{4}).
      • (P(A \cap B) = P(\text{spade}) = \dfrac{13}{52} = \dfrac{1}{4}) (since all spades are black).
      • Check: (P(A)P(B) = \dfrac{1}{4} \cdot \dfrac{1}{2} = \dfrac{1}{8} \neq P(A \cap B) = \dfrac{1}{4}).
      • Therefore, not independent.

  • Real-world takeaway: Some event pairs may look related (e.g., suit and color), and independence must be checked via probabilities, not just intuition.

Page 4: Independence Recall and 4.1 Recall

  • Quick recap of test ideas from section 4.1 (a preview for 4.2):

    • A, B, C, D, F categories (grading) vs. Studied vs. Did Not Study can be analyzed for independence.
    • Example data table (given):
    • Studied: A=5, B=10, C=5, D=1, F=0; total Studied = 21.
    • Did Not Study: A=0, B=2, C=6, D=11, F=10; total Did Not Study = 29.
    • Column totals: A=5, B=12, C=11, D=12, F=10; Overall total = 50.
    • (a) Determine if earning an A and having studied are independent events using the data:
    • P(A) = total A / total overall = (5/50 = 0.10).
    • P(Studied) = total Studied / total = (21/50 = 0.42).
    • P(A and Studied) = cells with A and Studied = (5/50 = 0.10).
    • If independent, (P(A \cap \text{Studied}) = P(A)P(\text{Studied}) = 0.10 \times 0.42 = 0.042).
    • Since (0.10 \neq 0.042), they are not independent.
    • (b) Brief memory lane: Correlation (association) vs. causation; recognizing that association does not imply causation.

  • Practical takeaway: In real data, independence is a property of the underlying probabilistic model, not a guaranteed feature of observed counts; correlation may exist without implying causation.

Page 5: Replacement vs. No Replacement – Probabilities with Cards (Part 1)

  • Replacement (with replacement): Monica and Rachel draw cards; the deck is reset after each draw.

  • Rachel draws a queen (Q).

    • After drawing her card, if the card is replaced, the next draw is from the full deck again.
    • If she had kept the queen (no replacement), the next draw would be from the remaining 51 cards.
    • Answers (assuming aces are low and replacement scenario):
    • (a) P(first card is a queen) = \dfrac{4}{52} = \dfrac{1}{13}.
    • (b) P(next card is higher than a queen) = \dfrac{4}{52} = \dfrac{1}{13}.
    • Note: “Higher than a queen” means rank higher than Q. If aces are low, the only rank higher than Queen is King (K). There are 4 kings in the deck, hence 4 good outcomes out of 52.

  • Insight: With replacement, probabilities reset between draws; without replacement, probabilities change as the composition of the deck changes.

Page 6: Replacement vs. No Replacement – Probabilities without Replacement

  • Had the first card been drawn without replacement, find the following probabilities (assume aces low):

    • (a) P(first card is a queen) = \dfrac{4}{52} = \dfrac{1}{13}.
    • (b) P(next card is higher than a queen) =
    • Unconditional interpretation (second card is a King): \dfrac{4}{52} = \dfrac{1}{13}.
    • Conditional interpretation (given the first card was a queen): \dfrac{4}{51}.
    • Note: By symmetry, the unconditional probability that the second card is a King equals 4/52 = 1/13. However, if you explicitly condition on the first card being a queen, the probability for the second card to be higher than a queen becomes 4/51.

  • The 5% Rule of Independence (a practical guideline):

    • If the sample is drawn from a finite population and the sampling fraction is small (typically ≤ 5%), independence is a reasonable assumption.
    • Formal idea: If the population size is N and sample size is n, and if n/N ≤ 0.05, then sampling without replacement may be treated as approximately independent.

Page 7: CSULB Example – Independence Thresholds

  • CSULB has over 37,000 students.
    • (a) If you sample 1,500 randomly selected students, can you assume independence zwischen trials? Yes, because
    • Sampling fraction = (1500 / 37000 \approx 0.0405 < 0.05).
    • Conclusion: Trials can be treated as approximately independent.
    • (b) How many students would have to be in the sample so that you could no longer assume independence?
    • Threshold for independence: when (n/N > 0.05).
    • Solve for n: (n > 0.05 \times 37000 = 1850).
    • Therefore, a sample larger than 1,850 would typically violate the 5% rule for this population.

Page 8: Mutually Exclusive (Disjoint) Events

  • Mutually Exclusive (Disjoint) Events:

    • Two events that cannot occur simultaneously are disjoint (mutually exclusive).

    • A card is drawn from a standard 52-card deck:

    • (a) Probability that the card is red AND a diamond:

      • Diamonds are red; this is the event “diamond.”
      • There are 13 diamonds in the deck.
      • (P( ext{red } \text{AND } \text{diamond}) = \dfrac{13}{52} = \dfrac{1}{4}.

    • (b) Probability that the card is red AND a spade:

      • A card cannot be both red and a spade (spades are black).
      • (P( ext{red AND spade}) = 0.

  • Practical note: For mutually exclusive events A and B, P(A ∪ B) = P(A) + P(B) and P(A ∩ B) = 0.

Page 9: Summary of Disjointness and Basic Formulas

  • Two events that are disjoint are either A and B such that P(A ∪ B) = P(A) + P(B) when P(A ∩ B) = 0.

  • In general, for any events A and B:

    • P(A ∪ B) = P(A) + P(B) − P(A ∩ B).

  • If A and B are disjoint (mutually exclusive):

    • P(A ∩ B) = 0 and thus P(A ∪ B) = P(A) + P(B).

  • If A and B are independent:

    • P(A ∩ B) = P(A)P(B).

  • Important relationships:

    • E and its complement E^c are always disjoint (can’t happen and not happen at the same time).
    • The disjointness property implies simple addition for probabilities of unions when there is no overlap.
    • Dependent vs. independent is about whether the occurrence of one event changes the probability of the other.

Page 10: Quick Reference — Terminology and Key Formulas

  • Terminology:

    • Mutually Exclusive (Disjoint) Events: A and B cannot occur together.
    • Independent Events: The occurrence of A does not affect the probability of B.
    • A and its complement A^c are always disjoint.

  • Key formulas:

    • Union-Intersection relationship (general):
    • (P(A \cup B) = P(A) + P(B) - P(A \cap B)).
    • Disjoint case (A and B are disjoint):
    • (P(A \cap B) = 0) and (P(A \cup B) = P(A) + P(B)).
    • Independent case:
    • (P(A \cap B) = P(A)P(B)).

  • Quick checks from page examples:

    • 52-card deck: independence of (spade) and (Jack) holds because
    • (P(A)=1/4), (P(B)=1/13), (P(A\cap B)=1/52) and (P(A)P(B)=1/52).
    • (spade) and (black card) are not independent because (P(A\cap B)=1/4\neq 1/8 = P(A)P(B)).

  • Ethical/practical implications:

    • Distinguishing correlation from causation (as noted in Page 5) is crucial in interpreting real-world data; correlation does not imply causation.
    • When applying independence assumptions in statistics, ensure the sampling design justifies the assumption (e.g., using the 5% rule for sampling without replacement).