11.4 Compound Events, Rules of Probability, and Expected Value

11.4 Compound Events, Rules of Probability, and Expected Value

Quick Answer: The probability of compound events is governed by three fundamental rules: the General Addition Rule ($P(A \cup B) = P(A) + P(B) - P(A \cap B)$), the Complement Rule ($P(A^c) = 1 - P(A)$), and the General Multiplication Rule ($P(A \cap B) = P(A) \cdot P(B|A)$). For independent events, the multiplication rule simplifies to $P(A \cap B) = P(A) \cdot P(B)$. The expected value of a discrete random variable $X$ is its theoretical long-run weighted average: $E(X) = \sum [x_i \cdot P(x_i)]$. A game of chance or commercial decision is defined as fair if and only if the expected net gain equals zero ($E(\text{Net Gain}) = E(X) - \text{Cost} = 0$).

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1. Set Notation and Compound Probability Operations (AII-S.CP.1)

[!NOTE] Scope note. The two assessed standards in this section are AII-S.CP.1 (describe events as subsets of a sample space using unions, intersections, and complements) and AII-S.CP.7 (apply the Addition Rule and interpret the answer in context). The Educator Guide sample items for S-CP.A are a union-of-two-sets question and a complement-of-"at least one" question - exactly the two ideas below. Discrete probability distributions and expected value (S-MD.6 and S-MD.7) are plus standards and are not on the Algebra II blueprint; Sections 4 and 5 below are marked as enrichment.

A compound event combines two or more individual events within a sample space $S$. Set theory provides the standard mathematical notation for compound operations:

          Union (A ∪ B)               Intersection (A ∩ B)             Complement (A^c)
         "A OR B or Both"                 "Both A AND B"                  "NOT A"
       ┌─────────────────┐             ┌─────────────────┐             ┌─────────────────┐
       │  █████████████  │             │     ░░░░░░░     │             │ ░░░░░░░         │
       │ █████░░░░░█████ │             │    ░░ ███ ░░    │             │ ░░░░░  ( A )    │
       │  █████████████  │             │     ░░░░░░░     │             │ ░░░░░░░         │
       └─────────────────┘             └─────────────────┘             └─────────────────┘

• Union ($A \cup B$, read as "$A$ or $B$"): The set of all outcomes belonging to event $A$, event $B$, or both. The probability $P(A \cup B)$ represents the probability that at least one of the events occurs.
• Intersection ($A \cap B$, read as "$A$ and $B$"): The set of all outcomes belonging simultaneously to both event $A$ and event $B$.
• Complement ($A^c$ or $A'$, read as "not $A$"): The set of all outcomes in sample space $S$ that do not belong to event $A$.

The Complement Rule & The "At Least One" Strategy

Because an event and its complement partition the entire sample space ($P(A) + P(A^c) = 1$):

$P(A^c) = 1 - P(A)$

In multi-stage trials, calculating the probability that an event occurs "at least once" directly by listing all combinations is tedious. The Complement Shortcut solves this efficiently:

$P(\text{at least one success}) = 1 - P(\text{zero successes})$

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2. The General Addition Rule & Mutually Exclusive Events (AII-S.CP.7 - printed on the reference sheet)

When calculating $P(A \cup B)$, adding $P(A)$ and $P(B)$ directly counts any outcomes in the intersection twice. To correct for this double-counting, subtract the joint probability:

$\text{General Addition Rule: } P(A \cup B) = P(A) + P(B) - P(A \cap B)$

Mutually Exclusive (Disjoint) Events

Two events $A$ and $B$ are mutually exclusive (or disjoint) if they cannot occur at the same time. They have no outcomes in common:

$A \cap B = \emptyset \implies P(A \cap B) = 0$

For mutually exclusive events, the addition rule simplifies to:

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

Addition Rule Summary

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3. The General Multiplication Rule: Dependent vs. Independent Sequences

Rearranging the conditional probability formula $P(B|A) = \frac{P(A \cap B)}{P(A)}$ yields the General Multiplication Rule for the joint probability of sequential events:

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

Independent Events Multiplication Shortcut

If events $A$ and $B$ are independent, knowing that $A$ occurred does not affect the probability of $B$ ($P(B|A) = P(B)$). The rule simplifies to:

$P(A \cap B) = P(A) \cdot P(B) \quad (\text{if and only if independent})$

Sampling With vs. Without Replacement

• With Replacement: Each item drawn is returned to the pool before the next draw. The population remains constant, so consecutive trials are independent.
• Without Replacement: Items drawn are not returned. The total pool and subset counts decrease with each draw, making consecutive trials dependent ($P(B|A) \neq P(B)$).

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4. Discrete Random Variables & Probability Distributions (Enrichment: (+)S-MD.6)

A discrete random variable $X$ is a variable whose numerical value is determined by the outcome of a random phenomenon, taking on distinct, countable values $x_1, x_2, \dots, x_n$.

A probability distribution for a discrete random variable assigns a probability $P(X = x_i)$ to each possible outcome. Every valid probability distribution must satisfy two mathematical conditions:

1. Every probability is non-negative and bounded by 1: $0 \le P(x_i) \le 1 \quad \text{for all } i$
2. The sum of all probabilities over the entire sample space equals exactly 1: $\sum_{i=1}^n P(x_i) = 1$

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5. Expected Value: Theoretical Long-Run Average (Enrichment: (+)S-MD.6, (+)S-MD.7)

The expected value, denoted $E(X)$ or $\mu_X$, of a discrete random variable is the weighted average of all possible values, weighted by their respective probabilities:

$E(X) = \sum_{i=1}^n [x_i \cdot P(x_i)] = x_1 P(x_1) + x_2 P(x_2) + \dots + x_n P(x_n)$

Conceptual Meaning of Expected Value

Expected value does not represent the outcome of a single trial. It represents the long-run mean outcome per trial if the random process were repeated thousands of times under identical conditions (governed by the Law of Large Numbers).

Expected Net Gain and Game Fairness

In gaming, insurance, and investment analysis, participants pay an initial cost (ticket price, premium, entry fee) to play. The net gain is:

$\text{Net Gain} = \text{Gross Payout } (X) - \text{Cost to Play}$

Because cost is a fixed constant:

$E(\text{Net Gain}) = E(X) - \text{Cost}$

                  ┌───────────────────────────────────────────────┐
                  │             Game Fairness Benchmark           │
                  │               E(Net Gain) = 0                 │
                  └───────────────────────┬───────────────────────┘
                                          │
         ┌────────────────────────────────┴────────────────────────────────┐
         ▼                                                                 ▼
   E(Net Gain) < 0                                                   E(Net Gain) > 0
  Unfair to Player                                                  Favorable to Player
(Player loses on average;                                          (Player gains on average;
 standard commercial games)                                         unsustainable for host)

• Fair Game: A game is mathematically fair if and only if the expected net gain equals zero ($E(\text{Net Gain}) = 0$). Neither the player nor the house has an edge.
• Unfair Game: If $E(\text{Net Gain}) < 0$, the player expects to lose money on average in the long run.

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6. Worked Problems

Worked Problem 1: General Addition Rule in a High School Cohort

Problem: In an Algebra II grade of $160$ students, $96$ students participate in school musical clubs ($M$), $64$ students participate in athletic teams ($A$), and $32$ students participate in both ($M \cap A$).

1. Determine the probability that a randomly chosen student participates in musical clubs or athletic teams ($P(M \cup A)$).
2. Determine the probability that a randomly chosen student participates in neither activity.

• Step 1: Calculate $P(M \cup A)$ using the General Addition Rule. $P(M) = \frac{96}{160} = 0.60, \quad P(A) = \frac{64}{160} = 0.40, \quad P(M \cap A) = \frac{32}{160} = 0.20$ $P(M \cup A) = P(M) + P(A) - P(M \cap A)$ $P(M \cup A) = 0.60 + 0.40 - 0.20 = 0.80$

• Step 2: Calculate the probability of neither using the Complement Rule. $\text{Neither} = (M \cup A)^c$ $P(\text{neither}) = 1 - P(M \cup A) = 1 - 0.80 = 0.20$

Worked Problem 2: Sequential Selection Without Replacement

Problem: A grab-bag contains $5$ red marbles, $3$ blue marbles, and $2$ green marbles ($10$ total). Two marbles are drawn sequentially without replacement.

1. Determine the probability that both selected marbles are red.
2. Determine the probability that at least one selected marble is red.

• Step 1: Calculate $P(\text{Red}_1 \cap \text{Red}_2)$ using the multiplication rule. $P(\text{Red}_1) = \frac{5}{10} = \frac{1}{2}$ After drawing one red marble, $4$ red marbles remain out of $9$ total marbles: $P(\text{Red}_2 | \text{Red}_1) = \frac{4}{9}$ $P(\text{Red}_1 \cap \text{Red}_2) = \left(\frac{5}{10}\right) \cdot \left(\frac{4}{9}\right) = \frac{20}{90} = \frac{2}{9} \approx 0.222$

• Step 2: Calculate $P(\text{at least one Red})$ via the complement. The complement of "at least one red" is "zero red marbles" (both are non-red). There are $5$ non-red marbles (3 blue + 2 green): $P(\text{Non-Red}_1) = \frac{5}{10}$ $P(\text{Non-Red}_2 | \text{Non-Red}_1) = \frac{4}{9}$ $P(\text{zero Red}) = \left(\frac{5}{10}\right) \cdot \left(\frac{4}{9}\right) = \frac{20}{90} = \frac{2}{9}$ Applying the complement shortcut: $P(\text{at least one Red}) = 1 - P(\text{zero Red}) = 1 - \frac{2}{9} = \frac{7}{9} \approx 0.778$

Worked Problem 3: Charity Raffle Expected Value & Fairness Analysis

Problem: A charity sells $500$ raffle tickets at $$10$ each. The raffle awards one Grand Prize of $$1,500$, two Second Prizes of $$500$ each, and five Third Prizes of $$100$ each. All other tickets win $$0$.

1. Construct the discrete probability distribution for the gross payout $X$.
2. Calculate the expected value $E(X)$ of a ticket.
3. Determine the expected net gain per ticket and state whether the raffle is a fair game.

• Step 1: Construct the probability distribution table.

  • Grand Prize: $x_1 = 1500$, $P(x_1) = \frac{1}{500} = 0.002$
  • Second Prize: $x_2 = 500$, $P(x_2) = \frac{2}{500} = 0.004$
  • Third Prize: $x_3 = 100$, $P(x_3) = \frac{5}{500} = 0.010$
  • No Prize: $x_4 = 0$, $P(x_4) = \frac{492}{500} = 0.984$ Check: $0.002 + 0.004 + 0.010 + 0.984 = 1.000$.

• Step 2: Calculate $E(X)$. $E(X) = \sum [x_i \cdot P(x_i)]$ $E(X) = (1500)(0.002) + (500)(0.004) + (100)(0.010) + (0)(0.984)$ $E(X) = 3.00 + 2.00 + 1.00 + 0.00 = \$6.00$ The gross theoretical payout is $$6.00$ per ticket.

• Step 3: Calculate expected net gain and evaluate fairness. $E(\text{Net Gain}) = E(X) - \text{Ticket Cost} = 6.00 - 10.00 = -\$4.00$ Because $E(\text{Net Gain}) = -$4.00 \neq 0$, the raffle is not a fair game. Ticket buyers lose an average of $$4.00$ per ticket purchased. (Across all 500 tickets, the charity expects to raise $500 \times $4.00 = $2,000$ in profit).

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7. Common Regents Pitfalls & Exam Strategies

• Pitfall 1: Blindly Adding Probabilities Without Subtracting the Intersection. When a problem asks for $P(A \text{ or } B)$, never write $P(A) + P(B)$ unless the problem explicitly states that the events are mutually exclusive.
• Pitfall 2: Confusing Gross Payout with Net Gain. When evaluating whether a game is fair, remember to subtract the cost of entry ($E(\text{Net Gain}) = E(X) - \text{Cost}$). Stating that a game with an expected payout of $$6.00$ is "profitable" when tickets cost $$10.00$ loses full credit.
• Pitfall 3: Assuming Draws are Independent When Items Are Not Replaced. When calculating probabilities without replacement, remember that both the numerator and the denominator decrease on the second draw.