6.5 Probability & Two-Way Frequency Tables
Key Takeaways
- Theoretical probability measures the likelihood of an event: $P(E) = \frac{\text{Favorable Outcomes}}{\text{Total Possible Outcomes}}$, bounded strictly within $[0, 1]$.
- The addition rule states $P(A \cup B) = P(A) + P(B) - P(A \cap B)$; subtracting the joint intersection prevents double-counting elements belonging to both sets.
- Conditional probability restricts the sample space to a specific subpopulation: $P(A \mid B) = \frac{n(A \cap B)}{n(B)}$, where the denominator must equal the marginal total of the conditioning group $B$.
- Key conditioning phrases such as 'given that', 'of the participants who...', or 'selected from those with...' indicate that the denominator is a row or column subtotal, rather than the grand table total.
- Two events $A$ and $B$ are statistically independent if and only if $P(A \mid B) = P(A)$ (or equivalently $P(A \cap B) = P(A) \cdot P(B)$).
6.5 Probability & Two-Way Frequency Tables
Probability questions on the Digital SAT test your ability to calculate simple, joint, compound, and conditional probabilities. The vast majority of these questions present data formatted inside a two-way frequency (contingency) table.
The single most critical skill is identifying the correct denominator by carefully reading how the problem constrains the sample space.
1. Probability Axioms & Basic Formulas
The probability of an event $E$ within a finite sample space $S$ with equally likely outcomes is:
Fundamental Properties
- Range Bounds: $0 \le P(E) \le 1$.
- $P(E) = 0 \implies$ Impossible event.
- $P(E) = 1 \implies$ Certain event.
- Complement Rule: The probability that $E$ does not occur is:
2. The Addition Rule (Union of Events)
When evaluating the probability that event $A$ OR event $B$ occurs, we use the General Addition Rule:
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| THE ADDITION RULE & VENN DIAGRAM |
| |
| +-------------------+ +-------------------+ |
| | Set A | | Set B | |
| | ( A ∩ B ) | |
| | ( Overlap) | |
| +-------------------+ +-------------------+ |
| |
| P(A or B) = P(A) + P(B) - P(A and B) |
| We MUST subtract P(A and B) once to prevent double-counting the overlap! |
| |
| [ MUTUALLY EXCLUSIVE (DISJOINT) EVENTS ]: |
| If A and B cannot occur simultaneously, P(A and B) = 0: |
| --> P(A or B) = P(A) + P(B) |
+-----------------------------------------------------------------------------+
Worked Example 1: AP Course Overlap
Problem: In a senior class of $300$ students, $140$ are enrolled in AP Chemistry, $180$ are enrolled in AP Calculus, and $70$ are enrolled in both courses. If a senior is selected at random, what is the probability that the student is enrolled in AP Chemistry, AP Calculus, or both?
Step-by-Step Solution:
- Identify the individual probabilities:
- Apply the addition rule:
3. Two-Way Frequency Tables: Anatomy & Navigation
A two-way table categorizes individuals according to two distinct qualitative variables.
+-----------------------------------------------------------------------------+
| TWO-WAY CONTINGENCY TABLE ANATOMY |
| |
| Variable Y: Outcome 1 Variable Y: Outcome 2 Total |
| Variable X: Cat 1 Cell (1,1) Cell (1,2) Row 1 |
| Variable X: Cat 2 Cell (2,1) Cell (2,2) Row 2 |
| Total Col 1 Total Col 2 Total GRAND TOT |
| |
| - Joint Frequencies: The interior 4 cells (e.g. Cat 1 AND Outcome 1) |
| - Marginal Frequencies: The row totals and column totals |
| - Grand Total (N): Bottom-right cell (sum of all observations) |
+-----------------------------------------------------------------------------+
4. Conditional Probability: Restricting the Denominator
Conditional probability calculates the likelihood of an event occurring given that another condition has already been met.
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| THE THREE SAT PROBABILITY QUESTION TYPES |
| |
| Sample Table: Exam Performance & Review Session Attendance |
| |
| Category Passed Exam Did Not Pass Row Total |
| ------------------------- ------------ ------------- ----------- |
| Attended Review Session 72 8 80 |
| Did Not Attend Review 48 32 80 |
| Column Total 120 40 160 |
| |
| [ TYPE 1: OVERALL / JOINT PROBABILITY ] |
| "What is the probability that a randomly chosen student attended review |
| AND passed the exam?" |
| --> Denominator = GRAND TOTAL (160) --> 72 / 160 = 0.45 |
| |
| [ TYPE 2: CONDITIONAL ON ROW ] |
| "Given that a student ATTENDED REVIEW, what is the prob they passed?" |
| --> Denominator = ROW TOTAL (80) --> 72 / 80 = 0.90 |
| |
| [ TYPE 3: CONDITIONAL ON COLUMN ] |
| "If a student who PASSED is selected at random, prob they attended?" |
| --> Denominator = COLUMN TOTAL (120) --> 72 / 120 = 0.60 |
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Denominator Selection Guide for SAT Word Problems
| English Phrasing | Condition / Sample Space | Denominator to Use |
|---|---|---|
| "If a person is selected at random from the entire group..." | No condition (all subjects) | Grand Total ($N$) |
| "Given that the person is in Group A..." | Condition is Group A | Row or Column Total for Group A |
| "Of the participants who selected Choice B..." | Condition is Choice B | Row or Column Total for Choice B |
| "If a Female participant is selected at random..." | Condition is Female | Total Females |
5. Independent vs. Dependent Probability
Two events $A$ and $B$ are independent if the occurrence of $B$ does not affect the probability of $A$:
Sequential Probability: Replacement Rules
- With Replacement (Independent): The sample space $n$ remains constant on successive draws.
- Without Replacement (Dependent): Both the favorable count and the total sample space decrease on successive draws.
Worked Example 2: Probability Without Replacement
Problem: A jar contains 6 red marbles, 4 blue marbles, and 5 green marbles ($15$ total). Two marbles are selected at random without replacement. What is the probability that both marbles are blue?
Step-by-Step Solution:
- First Draw: $P(\text{Blue}_1) = \frac{4}{15}$.
- Second Draw (given first was blue): $3$ blue marbles remain out of $14$ total: $P(\text{Blue}_2 \mid \text{Blue}_1) = \frac{3}{14}$.
- Apply the general multiplication rule:
6. Desmos Playbook for Complex Probability Calculations
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| DESMOS PROBABILITY SHORTCUTS |
| |
| 1. Exact Fraction Simplification: |
| Type: 72 / 120 --> Click the fraction icon to display 3/5. |
| |
| 2. Multi-Step Dependent Multiplication: |
| Type: (4/15) * (3/14) --> Displays 2/35 directly. |
| |
| 3. Overlapping Union Arithmetic: |
| Type: (140 + 180 - 70) / 300 --> Displays 5/6 directly. |
+-----------------------------------------------------------------------------+
[!TIP] Desmos Fraction Safety: Always enter multi-term conditional probabilities into Desmos as a single consolidated fraction:
(140 + 180 - 70) / 300. Clicking the fraction toggle button ensures your final answer is simplified to lowest terms, matching College Board Multiple-Choice options or SPR input format.
A total of 160 students took a certification exam, with their attendance at a preparatory review session recorded in the table below.
If a student who passed the exam is selected at random, what is the probability that the student attended the review session?Group Passed Exam Did Not Pass Total Attended Review Session 72 8 80 Did Not Attend Review 48 32 80 Total 120 40 160
In a high school senior class of 300 students, 140 students are enrolled in AP Chemistry, 180 students are enrolled in AP Calculus, and 70 students are enrolled in both courses. If a senior is selected at random, what is the probability that the student is enrolled in AP Chemistry, AP Calculus, or both?
A jar contains 6 red marbles, 4 blue marbles, and 5 green marbles (15 marbles total). Two marbles are selected at random from the jar one after the other without replacement. What is the probability that both selected marbles are blue?