Practice Problems: Probability and Contingency Tables

Sample Population and Category Frequencies

  • The sample consists of a categorical distribution of cat coat types and patterns comprising a total count of 9797 individuals:
    • Tabby: 3333
    • Tuxedo: 2424
    • Calico: 1616
    • Lynx Point: 1010
    • Spotted: 88
    • Snowshoe: 66
    • Total Count: 33+24+16+10+8+6=9733 + 24 + 16 + 10 + 8 + 6 = 97

Event Definitions and Marginal Probabilities

  • Primary events are designated based on specific coat patterns within the population:
    • Event AA represents selecting a cat with a Tuxedo pattern.
    • Event BB represents selecting a cat with a Lynx Point pattern.
  • The marginal probabilities of the events are calculated directly from their sample counts relative to the total population size of 9797:
    • P(A)=2497≈0.25P(A) = \frac{24}{97} \approx 0.25
    • P(B)=1097≈0.10P(B) = \frac{10}{97} \approx 0.10
  • The complementary probabilities (the probability of an event not occurring) are determined using the complement rule P(E′)=1−P(E)P(E') = 1 - P(E):
    • Complement of event AA: P(A′)=1−0.25=0.75P(A') = 1 - 0.25 = 0.75
    • Complement of event BB: P(B′)=1−0.10=0.90P(B') = 1 - 0.10 = 0.90

Two-Way Joint Probability Table

Contingency table displaying joint and marginal probabilities for events A and B

  • A two-way probability grid is constructed to model joint events under the operational assumption of independence, where the joint probability of two events XX and YY is evaluated via the multiplication rule: P(X∩Y)=P(X)×P(Y)P(X \cap Y) = P(X) \times P(Y).
  • Joint probability calculations for each cell in the contingency layout:
    • Intersection of AA and BB:
    • P(A∩B)=P(A)×P(B)P(A \cap B) = P(A) \times P(B)
    • P(A∩B)=0.25×0.10=0.025≈0.03P(A \cap B) = 0.25 \times 0.10 = 0.025 \approx 0.03
    • Intersection of A′A' and BB:
    • P(A′∩B)=P(A′)×P(B)P(A' \cap B) = P(A') \times P(B)
    • P(A′∩B)=0.75×0.10=0.075≈0.08P(A' \cap B) = 0.75 \times 0.10 = 0.075 \approx 0.08
    • Intersection of AA and B′B':
    • P(A∩B′)=P(A)×P(B′)P(A \cap B') = P(A) \times P(B')
    • P(A∩B′)=0.25×0.90=0.225≈0.23P(A \cap B') = 0.25 \times 0.90 = 0.225 \approx 0.23
    • Intersection of A′A' and B′B':
    • P(A′∩B′)=P(A′)×P(B′)P(A' \cap B') = P(A') \times P(B')
    • P(A′∩B′)=0.75×0.90=0.675≈0.68P(A' \cap B') = 0.75 \times 0.90 = 0.675 \approx 0.68

Marginal Probability Reconstructions and Rounding Observations

  • Row totals reconstruct the marginal probabilities for event BB and its complement B′B' by summing the calculated joint probabilities:
    • Reconstructed P(B)=P(A∩B)+P(A′∩B)=0.03+0.08=0.11P(B) = P(A \cap B) + P(A' \cap B) = 0.03 + 0.08 = 0.11
    • Reconstructed P(B′)=P(A∩B′)+P(A′∩B′)=0.23+0.68=0.91P(B') = P(A \cap B') + P(A' \cap B') = 0.23 + 0.68 = 0.91
    • Total probability across rows: P(B)+P(B′)=0.11+0.91=1.02P(B) + P(B') = 0.11 + 0.91 = 1.02
  • Column totals reconstruct the marginal probabilities for event AA and its complement A′A' by summing down each column:
    • Reconstructed P(A)=P(A∩B)+P(A∩B′)=0.03+0.23=0.26P(A) = P(A \cap B) + P(A \cap B') = 0.03 + 0.23 = 0.26
    • Reconstructed P(A′)=P(A′∩B)+P(A′∩B′)=0.08+0.68=0.76P(A') = P(A' \cap B) + P(A' \cap B') = 0.08 + 0.68 = 0.76
    • Total probability across columns: P(A)+P(A′)=0.26+0.76=1.02P(A) + P(A') = 0.26 + 0.76 = 1.02
  • Note on rounding discrepancy:
    • Intermediate rounding of each product to two decimal places introduces slight discrepancy between the original marginal probabilities (P(A)=0.25P(A) = 0.25, P(B)=0.10P(B) = 0.10) and the marginal sums reconstructed from rounded joint probabilities (P(A)=0.26P(A) = 0.26, P(B)=0.11P(B) = 0.11), producing total sums of 1.021.02 rather than precisely 1.001.00.