11.3 Conditional Probability, Two-Way Tables, and Independence

11.3 Conditional Probability, Two-Way Tables, and Independence

Quick Answer: The conditional probability of event $A$ occurring given that event $B$ has already occurred is denoted $P(A|B)$ and calculated as $P(A|B) = \frac{P(A \cap B)}{P(B)} = \frac{n(A \cap B)}{n(B)}$. In a two-way frequency table, the condition restricts the denominator to the row or column total corresponding to event $B$. Two events $A$ and $B$ are independent if and only if knowing that $B$ occurred does not change the probability of $A$. To prove independence on the Regents exam, you must mathematically verify that $P(A|B) = P(A)$ or that $P(A \cap B) = P(A) \cdot P(B)$.

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1. Anatomy of Two-Way Frequency Tables (AII-S.CP.4)

A two-way frequency table (contingency table) organizes sample data according to two distinct categorical variables. One variable defines the rows and the other defines the columns.

Consider a survey of $200$ high school seniors categorized by whether they participate in interscholastic athletics and whether they are enrolled in an Advanced Placement (AP) course:

Types of Frequencies

1. Joint Frequencies: The counts located in the interior cells where row and column categories intersect. For example, $n(E \cap A) = 36$ represents the joint frequency of seniors who are both enrolled in an AP course and participate in athletics.
2. Marginal Frequencies: The sums located along the right and bottom margins of the table, representing the total count for a single category irrespective of the other variable. For example, $n(E) = 90$ is the marginal frequency of all AP students, and $n(A) = 80$ is the marginal frequency of all student athletes.
3. Grand Total ($N$): The bottom-right corner value ($N = 200$), representing the total number of individuals surveyed.

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2. Calculating Basic, Joint, and Union Probabilities

Using the grand total $N$, probabilities are calculated as relative frequencies:

• Marginal Probability: The probability of a single categorical event. Divide the marginal frequency by the grand total:

  $$P(E) = \frac{n(E)}{N} = \frac{90}{200} = 0.45$$ $$P(A) = \frac{n(A)}{N} = \frac{80}{200} = 0.40$$

• Joint Probability (Intersection $\cap$, "and"): The probability that an individual satisfies both conditions simultaneously. Divide the interior cell count by the grand total:

  $$P(E \cap A) = \frac{n(E \cap A)}{N} = \frac{36}{200} = 0.18$$

• Union Probability ($\cup$, "or"): The probability that an individual belongs to category $E$, category $A$, or both. By the General Addition Rule:

  $$P(E \cup A) = P(E) + P(A) - P(E \cap A) = \frac{90}{200} + \frac{80}{200} - \frac{36}{200} = \frac{134}{200} = 0.67$$

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3. Conditional Probability: Concept and Formulas (AII-S.CP.4)

Conditional probability calculates the likelihood of an event occurring given that another event is already known to have occurred. It is denoted:

$$P(A|B) \quad (\text{read as: the probability of } A \text{ given } B)$$

Sample Space Restriction

The condition following the vertical bar ("given $B$") restricts the sample space. Instead of dividing by the grand total $N$, the denominator is restricted strictly to the subgroup defined by condition $B$.

          Grand Sample Space N = 200
     ┌──────────────────────────────────────────────────┐
     │                                                  │
     │        Restricted Condition B: n(B)              │
     │      ┌─────────────────────────────┐             │
     │      │            Intersection     │             │
     │      │            n(A ∩ B)         │             │
     │      │            [ Numerator ]    │             │
     │      │                             │             │
     │      │     [ New Denominator ]     │             │
     │      └─────────────────────────────┘             │
     │                                                  │
     └──────────────────────────────────────────────────┘

The Mathematical Formulas

1. From Raw Frequency Counts:

   $$P(A|B) = \frac{n(A \cap B)}{n(B)}$$

2. From Relative Probabilities:

   $$P(A|B) = \frac{P(A \cap B)}{P(B)} \quad \text{provided } P(B) > 0$$

Practical Table Reading: Row vs. Column Conditioning

Using our senior class table:

• "Find the probability that a student is in an AP course, given that the student is an athlete:" The condition is Athlete ($A$). Look only at the Athlete column total ($80$). Of those $80$ athletes, $36$ take an AP course:

  $$P(E|A) = \frac{n(E \cap A)}{n(A)} = \frac{36}{80} = 0.45$$

• "Find the probability that a student is an athlete, given that the student is in an AP course:" The condition is AP Course ($E$). Look only at the AP Course row total ($90$). Of those $90$ AP students, $36$ are athletes:

  $$P(A|E) = \frac{n(A \cap E)}{n(E)} = \frac{36}{90} = 0.40$$

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4. Mathematical Definitions and Tests for Independence (AII-S.CP.4)

Two events $A$ and $B$ are statistically independent if the occurrence of one event does not alter the probability that the other event occurs.

The Three Equivalent Independence Criteria

On the Regents exam, you can prove or disprove independence using any one of three equivalent mathematical tests:

1. Test 1 (Conditional Test on $A$):

   $$P(A|B) = P(A)$$

   "The probability of $A$ occurring given that $B$ occurred is identical to the unconditional probability of $A$."

2. Test 2 (Conditional Test on $B$):

   $$P(B|A) = P(B)$$

   "The probability of $B$ occurring given that $A$ occurred is identical to the unconditional probability of $B$."

3. Test 3 (Multiplication Rule for Independent Events):

   $$P(A \cap B) = P(A) \cdot P(B)$$

   "The joint probability of both events occurring equals the product of their individual marginal probabilities."

Testing the Senior Class Example for Independence

Let us test whether taking an AP course ($E$) and participating in athletics ($A$) are independent in our sample of $200$ seniors:

• Method 1: Comparing $P(E|A)$ to $P(E)$:

  $$P(E|A) = \frac{36}{80} = 0.45$$ $$P(E) = \frac{90}{200} = 0.45$$

  Since $P(E|A) = P(E) = 0.45$, events $E$ and $A$ are independent!

• Method 2: Comparing $P(E \cap A)$ to $P(E) \cdot P(A)$:

  $$P(E \cap A) = \frac{36}{200} = 0.18$$ $$P(E) \cdot P(A) = (0.45)(0.40) = 0.18$$

  Since $P(E \cap A) = P(E) \cdot P(A)$, the events are verified to be independent.

Summary Table of Independence Tests

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5. Worked Problems

Worked Problem 1: Navigating a Medical Diagnostic Table

Problem: A clinical study evaluates a rapid diagnostic test for a seasonal virus across a trial cohort of $1,000$ patients:

1. Determine the probability that a randomly chosen patient tests positive.
2. Determine the conditional probability that a patient is genuinely infected with the virus, given that they received a positive test result ($P(V|+)$). Express as a fraction and a decimal rounded to the nearest thousandth.
3. Determine whether testing positive ($+$) and being infected ($V$) are independent events. Justify your answer mathematically.

• Step 1: Calculate $P(+)$.

  $$P(+) = \frac{n(+)}{N} = \frac{130}{1000} = 0.130$$

• Step 2: Calculate $P(V|+)$. The condition is a positive test result ($+$). Restrict the denominator to the column total of positive tests ($130$):

  $$P(V|+) = \frac{n(V \cap +)}{n(+)} = \frac{85}{130} = \frac{17}{26} \approx 0.654$$

• Step 3: Test for independence. Compute the marginal probability of being infected:

  $$P(V) = \frac{n(V)}{N} = \frac{100}{1000} = 0.100$$

  Compare $P(V|+)$ to $P(V)$:

  $$P(V|+) = 0.654 \quad \text{and} \quad P(V) = 0.100$$

  Since $P(V|+) \neq P(V)$ ($0.654 \neq 0.100$), testing positive and being infected are dependent (not independent) events. Receiving a positive test result significantly raises the probability of infection from $10\%$ to $65.4\%$.

Worked Problem 2: Testing Independence from Survey Probabilities

Problem: In a survey of music streaming subscribers, the probability that a subscriber listens to classical playlists is $P(C) = 0.25$, the probability that a subscriber is over 40 years old is $P(A) = 0.40$, and the probability that a subscriber is both over 40 and listens to classical playlists is $P(A \cap C) = 0.14$. Algebraically determine whether listening to classical playlists and being over 40 years old are independent events.

• Step 1: State the independence condition to test. Events $A$ and $C$ are independent if and only if:

  $$P(A \cap C) = P(A) \cdot P(C)$$

• Step 2: Evaluate both sides of the equation. Left side:

  $$P(A \cap C) = 0.14$$

  Right side:

  $$P(A) \cdot P(C) = (0.40)(0.25) = 0.10$$

• Step 3: Compare values and write conclusion.

  $$0.14 \neq 0.10 \implies P(A \cap C) \neq P(A) \cdot P(C)$$

  Because the joint probability ($0.14$) does not equal the product of the individual marginal probabilities ($0.10$), the events are not independent (they are dependent).

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6. Common Regents Pitfalls & Exam Strategies

• Pitfall 1: Dividing by the Grand Total in Conditional Problems. On questions asking for $P(A|B)$, students frequently divide by $N$ (calculating $P(A \cap B)$) instead of dividing by the marginal total $n(B)$. Always circle the condition word ("given") and use that category total as the denominator.
• Pitfall 2: Confusing Independence with Mutually Exclusive (Disjoint) Events. Mutually exclusive events cannot happen together ($P(A \cap B) = 0$). If two events with positive probabilities are mutually exclusive, they can never be independent because $P(A \cap B) = 0 \neq P(A) \cdot P(B)$.
• Pitfall 3: Giving Subjective or Verbal Justifications. On questions asking "Are events A and B independent? Justify your answer," students who write verbal explanations like "No, because older people naturally enjoy classical music more" receive 0 credits. You must show the mathematical comparison (e.g., $P(A|B) \neq P(A)$) with exact fractions or decimals.