Statistics Flashcards: Using Two Way Tables For Probability

Study Using Two Way Tables For Probability in Statistics with focused flashcards that help you recognize the idea, recall the key rule, and apply it in practice-style prompts.

Statistics

Using Two Way Tables For Probability

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QUESTION
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What is the definition of a two-way frequency table?

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ANSWER

A table of counts for data classified by two categorical variables. Shows frequencies for combinations of two categorical variables.

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Flashcard 1: What is the definition of a two-way frequency table?

Answer: A table of counts for data classified by two categorical variables. Shows frequencies for combinations of two categorical variables.

Flashcard 2: Find P(AB)P(A\cap B) if the cell count for (A,B)(A,B) is 1818 out of a grand total of 120120.

Answer: 18120=0.15\frac{18}{120}=0.15. Apply joint probability formula: cell count over grand total.

Flashcard 3: Decide if AA and BB are independent when P(A)=0.4P(A)=0.4, P(B)=0.5P(B)=0.5, P(AB)=0.2P(A\cap B)=0.2.

Answer: Independent, since 0.4×0.5=0.20.4\times 0.5=0.2. Check if P(A)P(B)=P(AB)P(A)P(B) = P(A \cap B).

Flashcard 4: What equation must hold for events AA and BB to be independent using probabilities?

Answer: P(AB)=P(A)P(B)P(A\cap B)=P(A)P(B). Independence means joint probability equals product of marginals.

Flashcard 5: Find P(BA)P(B\mid A) if the count in (A,B)(A,B) is 99 and the total for AA is 3636.

Answer: 936=0.25\frac{9}{36}=0.25. Apply conditional formula with AA as the condition.

Flashcard 6: Find P(BA)P(B\mid A) if count(AB)=9\text{count}(A\cap B)=9, count(A)=30\text{count}(A)=30.

Answer: 930=0.30\frac{9}{30}=0.30. Divide joint count by condition's marginal.

Flashcard 7: What conditional-probability equality indicates AA and BB are independent (when defined)?

Answer: P(AB)=P(A)P(A\mid B)=P(A). Independence means conditioning doesn't change the probability.

Flashcard 8: What is the complement rule for a conditional probability P(AB)P(A\mid B)?

Answer: P(AcB)=1P(AB)P(A^c\mid B)=1-P(A\mid B). Complement probabilities sum to 1 within the conditional space.

Flashcard 9: Find P(AB)P(A\mid B) if count(AB)=12\text{count}(A\cap B)=12 and count(B)=40\text{count}(B)=40.

Answer: 1240=0.30\frac{12}{40}=0.30. Apply conditional probability formula.

Flashcard 10: Find P(A)P(A) if the marginal total for AA is 4545 out of a grand total of 150150.

Answer: 45150=0.30\frac{45}{150}=0.30. Apply marginal probability formula.

Flashcard 11: What is the formula for the probability of an event using a marginal total?

Answer: P(A)=marginal total for Agrand totalP(A)=\frac{\text{marginal total for }A}{\text{grand total}}. Divide row/column sum by total count.

Flashcard 12: Identify the sample space when using a two-way table to model outcomes.

Answer: All table cells (each category pair) plus the grand total as the denominator. Each cell represents an outcome; grand total normalizes probabilities.

Flashcard 13: What is a row total in a two-way frequency table?

Answer: The sum of counts across a single row. Adds all values in that row to find the row's marginal total.

Flashcard 14: Compute P(AcB)P(A^c\mid B) if P(AB)=0.72P(A\mid B)=0.72.

Answer: 10.72=0.281-0.72=0.28. Use complement rule: probabilities sum to 1.

Flashcard 15: What is the difference between joint and conditional probability in a two-way table?

Answer: Joint uses grand total; conditional uses the given category total. Joint divides by all outcomes; conditional divides by given category.

Flashcard 16: What formula gives the conditional probability P(AB)P(A\mid B) from a two-way table?

Answer: P(AB)=count in A and Btotal count in BP(A\mid B)=\frac{\text{count in }A\text{ and }B}{\text{total count in }B}. Restricts to condition BB by dividing by BB's total.

Flashcard 17: Find P(AB)P(A\mid B) if the count in (A,B)(A,B) is 77 and the total for BB is 2020.

Answer: 720=0.35\frac{7}{20}=0.35. Apply conditional formula: cell count over condition's total.

Flashcard 18: Find P(AB)P(A\mid B) if P(AB)=0.18P(A\cap B)=0.18 and P(B)=0.6P(B)=0.6.

Answer: 0.180.6=0.30\frac{0.18}{0.6}=0.30. Apply P(AB)=P(AB)P(B)P(A|B) = \frac{P(A \cap B)}{P(B)}.

Flashcard 19: What formula gives the joint probability P(AB)P(A \cap B) from a two-way table?

Answer: P(AB)=count in A and Bgrand totalP(A \cap B)=\frac{\text{count in }A\text{ and }B}{\text{grand total}}. Divides the cell count by grand total for joint probability.

Flashcard 20: What is a joint frequency in a two-way table?

Answer: A count in an interior cell for a specific pair of categories. Frequency where two specific categories intersect.

Flashcard 21: Decide if AA and BB are independent if P(A)=0.4P(A)=0.4, P(B)=0.5P(B)=0.5, and P(AB)=0.2P(A\cap B)=0.2.

Answer: Independent, since 0.2=0.4×0.50.2=0.4\times 0.5. Check if P(AB)=P(A)P(B)P(A \cap B) = P(A)P(B): 0.2=0.4×0.50.2 = 0.4 \times 0.5

Flashcard 22: What formula gives the marginal probability P(A)P(A) from a two-way table?

Answer: P(A)=row or column total for Agrand totalP(A)=\frac{\text{row or column total for }A}{\text{grand total}}. Divides marginal total by grand total for single-event probability.

Flashcard 23: What condition using probabilities shows events AA and BB are independent?

Answer: P(AB)=P(A)P(B)P(A\cap B)=P(A)P(B). Independent events satisfy the multiplication rule.

Flashcard 24: Decide if AA and BB are independent when P(A)=0.3P(A)=0.3, P(B)=0.6P(B)=0.6, P(AB)=0.25P(A\cap B)=0.25.

Answer: Not independent, since 0.3×0.6=0.180.250.3\times 0.6=0.18\ne 0.25. Product 0.180.18 doesn't equal joint probability 0.250.25.

Flashcard 25: Which equality using conditional probability shows AA and BB are independent?

Answer: P(AB)=P(A)P(A\mid B)=P(A). Conditioning doesn't change probability when independent.

Flashcard 26: What is a two-way frequency table used for when each object has two categorical variables?

Answer: A table of counts for all combinations of two categorical variables. Organizes counts when objects have two categorical attributes.

Flashcard 27: What is the grand total in a two-way frequency table?

Answer: The sum of all cell counts in the table. Add all interior cells or all marginal totals.

Flashcard 28: Identify the correct comparison to check independence in a table: compare P(AB)P(A\mid B) to what value?

Answer: Compare P(AB)P(A\mid B) to P(A)P(A). Equal values indicate independence.

Flashcard 29: Find P(A)P(A) if the row total for AA is 5454 out of a grand total of 9090.

Answer: 5490=0.6\frac{54}{90}=0.6. Marginal probability uses row total over grand total.

Flashcard 30: What is a marginal total in a two-way table?

Answer: A row or column total showing counts for one variable alone. Sum of all values in that row or column.

Flashcard 31: What is the formula for conditional probability from a two-way table?

Answer: P(AB)=count(AB)count(B)P(A\mid B)=\frac{\text{count}(A\cap B)}{\text{count}(B)}. Divide joint count by condition's total count.

Flashcard 32: Choose the correct interpretation of P(ScienceGrade 10)P(\text{Science}\mid \text{Grade }10) from a two-way table.

Answer: Probability a student prefers Science among only Grade 1010 students. Conditional restricts to Grade 10 subset only.

Flashcard 33: What is the formula for joint probability using a two-way table cell count?

Answer: P(AB)=cell countgrand totalP(A \cap B)=\frac{\text{cell count}}{\text{grand total}}. Divide the cell count by total count for joint probability.

Flashcard 34: What is a column total in a two-way frequency table?

Answer: The sum of counts down a single column. Adds all values in that column to find the column's marginal total.

Flashcard 35: Decide if AA and BB are independent if P(A)=0.3P(A)=0.3, P(B)=0.6P(B)=0.6, and P(AB)=0.25P(A\cap B)=0.25.

Answer: Not independent, since 0.250.3×0.60.25\ne 0.3\times 0.6. Check if P(AB)=P(A)P(B)P(A \cap B) = P(A)P(B): 0.250.180.25 \ne 0.18

Flashcard 36: Find P(AB)P(A\cap B) if the cell count for (A,B)(A,B) is 1818 and the grand total is 120120.

Answer: 18120=0.15\frac{18}{120}=0.15. Apply joint probability formula: cell/total.

Flashcard 37: Identify the sample space when a two-way frequency table is used for probability.

Answer: All individuals in the table; total outcomes equal the grand total. Each person counted is one possible outcome.

Flashcard 38: Find P(AB)P(A\mid B) if the count in (A,B)(A,B) is 1212 and the total for BB is 4848.

Answer: 1248=0.25\frac{12}{48}=0.25. Apply conditional formula: cell count over condition's total.

Flashcard 39: Compute the missing cell: grand total 200200, other three cells in a 2×22\times 2 table sum to 155155.

Answer: 200155=45200-155=45. All cells must sum to grand total.