9.3 Compound Probability: Addition Rule, Multiplication Rule & Conditional Probability

9.3 Compound Probability: Addition Rule, Multiplication Rule & Conditional Probability

Compound probability examines situations where two or more individual events combine. On the CLEP College Mathematics examination, mastering compound events requires distinguishing between unions ("$A$ OR $B$"), which use the Addition Rule, and intersections ("$A$ AND $B$"), which use the Multiplication Rule.

Furthermore, understanding how previous occurrences alter the available sample space—known as Conditional Probability ($P(B|A)$)—is essential for analyzing two-way contingency tables and multi-stage tree diagrams.

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1. The Addition Rule for Unions ($A$ OR $B$)

The union of two events $A$ and $B$, denoted $A \cup B$, represents the event that event $A$ occurs, event $B$ occurs, or both occur.

+-----------------------------------------------------------------------------+
|                        THE GENERAL ADDITION RULE                            |
|                                                                             |
|   +---------------------------------------------------------------------+   |
|   |  SAMPLE SPACE S                                                     |   |
|   |                                                                     |   |
|   |        +-----------------+     +-----------------+                  |   |
|   |        |     EVENT A     |     |     EVENT B     |                  |   |
|   |        |                 |     |                 |                  |   |
|   |        |            +----+-----+---+             |                  |   |
|   |        |            | A INTERSECT B|             |                  |   |
|   |        |            | P(A and B)   |             |                  |   |
|   |        +------------+--------------+-------------+                  |   |
|   |                                                                     |   |
|   +---------------------------------------------------------------------+   |
|                                                                             |
|             P(A U B) = P(A) + P(B) - P(A INTERSECT B)                       |
+-----------------------------------------------------------------------------+

The General Addition Rule Formula

For any two events $A$ and $B$ within sample space $S$:

$$P(A \cup B) = P(A) + P(B) - P(A \cap B)$$

Why subtract $P(A \cap B)$? When adding $P(A)$ and $P(B)$, the overlapping outcomes that belong to both $A$ and $B$ ($A \cap B$) are counted twice. Subtracting $P(A \cap B)$ once corrects for this double counting.

Mutually Exclusive (Disjoint) Events

Two events $A$ and $B$ are mutually exclusive (disjoint) if they cannot occur simultaneously ($A \cap B = \emptyset$). In this case, their intersection has probability zero:

$$P(A \cap B) = 0 \implies P(A \cup B) = P(A) + P(B)$$

Worked Example 1: Non-Mutually Exclusive Deck Selection

Problem: A card is drawn from a standard 52-card deck. What is the probability that the card is a Queen or a Diamond?

Solution:

1. Identify individual event probabilities:
   • Let $Q$ = event card is a Queen $\implies n(Q) = 4 \implies P(Q) = \frac{4}{52}$.
   • Let $D$ = event card is a Diamond $\implies n(D) = 13 \implies P(D) = \frac{13}{52}$.
2. Identify the intersection (overlap):
   • $Q \cap D$ = Queen of Diamonds $\implies n(Q \cap D) = 1 \implies P(Q \cap D) = \frac{1}{52}$.
3. Apply the General Addition Rule: $$P(Q \cup D) = P(Q) + P(D) - P(Q \cap D) = \frac{4}{52} + \frac{13}{52} - \frac{1}{52} = \frac{16}{52} = \frac{4}{13} \approx 0.3077$$

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2. Independent vs. Dependent Events & The Multiplication Rule ($A$ AND $B$)

The intersection of two events $A$ and $B$, denoted $A \cap B$, represents the joint event that both event $A$ and event $B$ occur simultaneously or sequentially.

+-----------------------------------------------------------------------------+
|                      INDEPENDENT VS. DEPENDENT EVENTS                       |
|                                                                             |
|   INDEPENDENT EVENTS:                                                       |
|   - The occurrence of event A does NOT change the probability of event B.   |
|   - Formal Condition: P(B | A) = P(B)  or  P(A n B) = P(A) * P(B)           |
|   - Common context: Sampling WITH replacement, separate dice/coin tosses.   |
|                                                                             |
|   DEPENDENT EVENTS:                                                         |
|   - The occurrence of event A ALTERS the probability of event B.            |
|   - Formal Condition: P(B | A) != P(B)  =>  P(A n B) = P(A) * P(B | A)      |
|   - Common context: Sampling WITHOUT replacement, sequential draws.         |
+-----------------------------------------------------------------------------+

The Multiplication Rule Formulas

• For Independent Events: $$P(A \cap B) = P(A) \times P(B)$$
• For Dependent Events (General Multiplication Rule): $$P(A \cap B) = P(A) \times P(B \mid A)$$

Where $P(B \mid A)$ represents the conditional probability of event $B$ occurring given that event $A$ has already occurred.

Worked Example 2: Sampling With vs. Without Replacement

Problem: A jar contains $6$ red marbles and $4$ blue marbles ($10$ marbles total). Two marbles are drawn sequentially. Find the probability that both marbles are red:

1. If the first marble is replaced before drawing the second marble (with replacement).
2. If the first marble is not replaced (without replacement).

Solution:

1. With Replacement (Independent):
   • First draw: $P(R_1) = \frac{6}{10} = \frac{3}{5}$.
   • Second draw: $P(R_2) = \frac{6}{10} = \frac{3}{5}$.
   • $P(R_1 \cap R_2) = P(R_1) \times P(R_2) = \frac{6}{10} \times \frac{6}{10} = \frac{36}{100} = \frac{9}{25} = 0.36$.
2. Without Replacement (Dependent):
   • First draw: $P(R_1) = \frac{6}{10} = \frac{3}{5}$.
   • Second draw: Given that $1$ red marble was removed, $5$ red marbles remain out of $9$ total marbles: $P(R_2 \mid R_1) = \frac{5}{9}$.
   • $P(R_1 \cap R_2) = P(R_1) \times P(R_2 \mid R_1) = \frac{6}{10} \times \frac{5}{9} = \frac{30}{90} = \frac{1}{3} \approx 0.3333$.

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3. Conditional Probability Mechanics

Conditional probability calculates the likelihood of an event $B$ occurring under the restrictive condition that another event $A$ has already occurred. In essence, conditioning on $A$ shrinks the sample space from the universal space $S$ down to the restricted subset $A$.

+-----------------------------------------------------------------------------+
|                        CONDITIONAL PROBABILITY MODEL                        |
|                                                                             |
|   +---------------------------------------------------------------------+   |
|   |  ORIGINAL SAMPLE SPACE S                                            |   |
|   |                                                                     |   |
|   |        +-----------------------------------+                        |   |
|   |        |   NEW REDUCED SAMPLE SPACE: A     |                        |   |
|   |        |                                   |                        |   |
|   |        |           +-----------------------+--+                     |   |
|   |        |           |  FAVORABLE REGION:    |  |                     |   |
|   |        |           |   A INTERSECT B       |  |  (Outside of A      |   |
|   |        |           +-----------------------+--+   is discarded)     |   |
|   |        +-----------------------------------+                        |   |
|   +---------------------------------------------------------------------+   |
|                                                                             |
|                       P(B | A) = P(A n B) / P(A)                            |
+-----------------------------------------------------------------------------+

The Conditional Probability Formula

For any two events $A$ and $B$ with $P(A) > 0$:

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

Worked Example 3: Die Roll Conditioning

Problem: A standard fair 6-sided die is rolled once. Given that the roll resulted in an odd number, what is the conditional probability that the number rolled is a prime number?

Solution:

1. Define the full sample space: $S = \{1, 2, 3, 4, 5, 6\}$ ($n(S) = 6$).
2. Define the given condition event $A$ (Odd number): $A = \{1, 3, 5\} \implies n(A) = 3$.
3. Define the target event $B$ (Prime number): $B = \{2, 3, 5\}$.
4. Find the intersection $A \cap B$ (Odd and Prime): $A \cap B = \{3, 5\} \implies n(A \cap B) = 2$.
5. Compute conditional probability: $$P(\text{Prime} \mid \text{Odd}) = \frac{n(A \cap B)}{n(A)} = \frac{2}{3} \approx 0.6667$$

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4. Two-Way Contingency Table Analysis

A two-way contingency table organizes observed bivariate data across two categorical variables, providing the clearest format for evaluating marginal, joint, and conditional probabilities.

Sample Contingency Table: Academic Major vs. Internship Status

A survey evaluates $200$ university seniors regarding their degree program and whether they completed a professional internship:

Types of Probabilities Derived from Contingency Tables

1. Marginal Probability (Single Variable Total / Grand Total):
   • Probability a randomly selected student is a Business major: $$P(B) = \frac{\text{Row Total for } B}{\text{Grand Total}} = \frac{100}{200} = 0.50$$
   • Probability a randomly selected student completed an internship: $$P(I) = \frac{\text{Column Total for } I}{\text{Grand Total}} = \frac{140}{200} = 0.70$$
2. Joint Probability (Intersection Cell / Grand Total):
   • Probability a student is a STEM major AND completed an internship: $$P(T \cap I) = \frac{\text{Intersection Cell }(T, I)}{\text{Grand Total}} = \frac{50}{200} = 0.25$$
3. Conditional Probability (Intersection Cell / Conditioning Row or Column Total):
   • Given that a student is a Business major, what is the probability they completed an internship? $$P(I \mid B) = \frac{n(B \cap I)}{n(B)} = \frac{70}{100} = 0.70$$
   • Given that a student completed an internship, what is the probability they are a STEM major? $$P(T \mid I) = \frac{n(T \cap I)}{n(I)} = \frac{50}{140} = \frac{5}{14} \approx 0.3571$$

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5. Multi-Stage Modeling with Tree Diagrams

A tree diagram visualizes multi-stage sequential probabilistic processes. By displaying branches for each stage:

• Branch probabilities represent conditional probabilities $P(B \mid A)$.
• Multiplying along a branch path calculates the joint intersection probability: $P(A \cap B) = P(A) \times P(B \mid A)$.
• Summing distinct path probabilities calculates the total probability of an outcome via the Law of Total Probability.

+-----------------------------------------------------------------------------+
|                        TREE DIAGRAM BRANCH MECHANICS                        |
|                                                                             |
|                         +---> Defective (0.02)  ===> P(M1 n D) = 0.012      |
|                         |                                                   |
|   +---> Machine 1 (0.60)+                                                   |
|   |                     +---> Non-Defect (0.98) ===> P(M1 n D') = 0.588     |
|   |                                                                         |
|   |                                                                         |
|   +---> Machine 2 (0.40)+---> Defective (0.05)  ===> P(M2 n D) = 0.020      |
|                         |                                                   |
|                         +---> Non-Defect (0.95) ===> P(M2 n D') = 0.380     |
|                                                                             |
|             Total Defective Probability P(D) = 0.012 + 0.020 = 0.032        |
+-----------------------------------------------------------------------------+

Worked Example 4: Factory Quality Control & Bayesian Inference

Problem: A manufacturing plant produces parts using two machines. Machine 1 produces $60\%$ of total output ($P(M_1) = 0.60$) with a $2\%$ defect rate ($P(D \mid M_1) = 0.02$). Machine 2 produces $40\%$ of output ($P(M_2) = 0.40$) with a $5\%$ defect rate ($P(D \mid M_2) = 0.05$).

1. What is the overall probability that a randomly chosen manufactured part is defective ($P(D)$)?
2. If a randomly selected part is found to be defective, what is the probability that it was produced by Machine 2 ($P(M_2 \mid D)$)?

Solution:

1. Total Probability of Defect ($P(D)$):
   • Path 1 (Machine 1 and Defective): $P(M_1 \cap D) = 0.60 \times 0.02 = 0.012$.
   • Path 2 (Machine 2 and Defective): $P(M_2 \cap D) = 0.40 \times 0.05 = 0.020$.
   • $P(D) = P(M_1 \cap D) + P(M_2 \cap D) = 0.012 + 0.020 = 0.032 = 3.2\%$.
2. Conditional Bayesian Probability ($P(M_2 \mid D)$): $$P(M_2 \mid D) = \frac{P(M_2 \cap D)}{P(D)} = \frac{0.020}{0.032} = \frac{20}{32} = \frac{5}{8} = 0.625 = 62.5\%$$

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6. TI-30XS MultiView Calculator Keystrokes

+-----------------------------------------------------------------------------+
|                 COMPOUND PROBABILITY EVALUATION ON TI-30XS                  |
|                                                                             |
|   - Addition Rule: 120 / 200 + 90 / 200 - 50 / 200 [enter]                  |
|   - Tree Multiplication: 0.40 * 0.05 [enter]  -> yields 0.02                |
|   - Conditional Ratio: ( 0.020 ) / ( 0.032 ) [enter] [<>] -> yields 5/8     |
+-----------------------------------------------------------------------------+

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7. Common CLEP Traps & Strategic Checkpoints

• Trap 1: Confusing Mutually Exclusive with Independent: Mutually exclusive events cannot occur together ($P(A \cap B) = 0$). Independent events satisfy $P(A \cap B) = P(A) \cdot P(B)$. Two events with positive probabilities cannot be both mutually exclusive and independent!
• Trap 2: Forgetting to Subtract the Intersection in Unions: In $P(A \cup B)$, never write $P(A) + P(B)$ unless you have verified that the events are mutually exclusive.
• Trap 3: Inverting the Conditional Probability Denominator: In $P(B \mid A)$, the conditioning event $A$ (after the vertical bar) must be in the denominator: $P(B \mid A) = \frac{P(A \cap B)}{P(A)}$. Writing $\frac{P(A \cap B)}{P(B)}$ evaluates $P(A \mid B)$, which is a completely different quantity.