Probability Rules, Addition and Multiplication Theorems, and Contingency Table Analysis

Rules of Addition and Complement Rule

• Special Rule of Addition:

  • Applies exclusively when the events under consideration are mutually exclusive (disjoint).
  • Mutually exclusive events cannot occur at the same time; there is no overlap between events in a Venn diagram ($P(A \text{ and } B) = 0$).
  • Formula:         $P(A \text{ or } B) = P(A) + P(B)$
  • Blood Type Example: A sample of $60$ students is surveyed regarding their blood type:
    • Type A: $20$ students
    • Type B: $10$ students
    • Type AB: $5$ students
    • Type O: $25$ students
    • Total students = $20 + 10 + 5 + 25 = 60$
    • Probability of having Type A: $P(\text{Type A}) = \frac{20}{60}$
    • Probability of having Type B: $P(\text{Type B}) = \frac{10}{60}$
    • Probability of having Type A or Type B: $P(\text{Type A or Type B}) = P(\text{Type A}) + P(\text{Type B}) = \frac{20}{60} + \frac{10}{60} = \frac{30}{60} = 0.50$
  • Vegetable Packaging Example: A machine fills plastic bags with a mixture of beans, broccoli, and other vegetables. Variations in vegetable sizes cause weight fluctuations. A quality check of $4{,}000$ packages filled in the past month revealed:
    • Underweight (Event A): $100$ packages
    • Satisfactory weight (Event B): $3{,}600$ packages
    • Overweight (Event C): $300$ packages
    • Total packages = $100 + 3600 + 300 = 4{,}000$
    • Probability a randomly selected package is either underweight or overweight: $P(A \text{ or } C) = P(A) + P(C) = \frac{100}{4000} + \frac{300}{4000} = 0.025 + 0.075 = 0.10$
  • Advertising Firms Income Example: A study of $200$ advertising firms analyzed income after taxes:
    • Under $\$1{,}000{,}000$: $102$ firms
    • Between $\$1{,}000{,}000$ and $\$20{,}000{,}000$: $61$ firms
    • $\$20{,}000{,}000$ or more: $37$ firms
    • Total firms = $102 + 61 + 37 = 200$
    • Part A - Probability a selected firm has an income under $\$1{,}000{,}000$: $P(\text{Under } \$1\text{M}) = \frac{102}{200} = 0.51$
    • Part B - Probability a selected firm has an income between $\$1{,}000{,}000$ and $\$20{,}000{,}000$ OR $\$20{,}000{,}000$ or more: $P(B \text{ or } C) = P(B) + P(C) = \frac{61}{200} + \frac{37}{200} = \frac{98}{200} = 0.49$

• Complement Rule:

  • Used to determine the probability of an event happening by subtracting the probability of the event not happening from $1$.
  • Formula:         $P(A) = 1 - P(\text{not } A)$
  • Die Roll Example: Rolling a standard six-sided die ($\{1, 2, 3, 4, 5, 6\}$):
    • Probability of rolling a 1-dot: $P(1) = \frac{1}{6}$
    • Probability of not rolling a 1-dot (rolling 2, 3, 4, 5, or 6): $P(\text{not } 1) = 1 - P(1) = 1 - \frac{1}{6} = \frac{5}{6}$
  • Vegetable Packaging Re-visited via Complement Rule: Finding the probability of a package being underweight or overweight ($P(\text{not } B)$):         $P(\text{not } B) = 1 - P(B) = 1 - \frac{3600}{4000} = 1 - 0.90 = 0.10$

• Worldwide Enterprises Employee Survey Example:

  • A sample of $2{,}000$ employees is surveyed about a new healthcare plan. Job classifications are mutually exclusive.
  • Maintenance (Event B): $50$ employees
  • Secretary (Event E): $68$ employees
  • Management (Event D): $302$ employees
  • Total employees = $2{,}000$
  • Part A (1) - Probability the first person selected is in Maintenance or a Secretary:         $P(B \text{ or } E) = P(B) + P(E) = \frac{50}{2000} + \frac{68}{2000} = \frac{118}{2000} = 0.059$
  • Part A (2) - Probability the first person selected is not in Management:         $P(\text{not } D) = 1 - P(D) = 1 - \frac{302}{2000} = 1 - 0.151 = 0.849$
  • Part B - Venn Diagram Representation:
    • For Maintenance ($B$) and Secretary ($E$), two completely disjoint circles with zero overlap represent the events.
    • For Not Management ($\text{not } D$), the area outside the circle representing Management ($D$) is shaded.
  • Part C - Complementary vs. Mutually Exclusive Event Analysis:
    • Events $B$ and $E$ are mutually exclusive because an employee cannot hold both positions simultaneously ($P(B \text{ and } E) = 0$).
    • Events $B$ and $E$ are not complementary because their combined probabilities do not equal $1$ ($P(B) + P(E) = 0.059 \neq 1$). For two events to be complementary, their sum must encompass the entire sample space ($P(A) + P(B) = 1$).

• Two Coins Tossed Example:

  • Sample space: $\{HH, HT, TH, TT\}$, total $4$ equally likely outcomes.
  • Event A = Two heads ($\{HH\}$); $P(A) = \frac{1}{4}$
  • Event B = Two tails ($\{TT\}$); $P(B) = \frac{1}{4}$
  • Are Event A and Event B mutually exclusive? Yes, because a coin toss resulting in two heads cannot simultaneously result in two tails.
  • Are Event A and Event B complements? No, because $P(A) + P(B) = \frac{1}{4} + \frac{1}{4} = \frac{2}{4} = 0.50 \neq 1$.

General Rule of Addition and Joint Probability

• General Rule of Addition:

  • Applied when the events are not mutually exclusive, meaning they overlap and can occur simultaneously.
  • Formula:         $P(A \text{ or } B) = P(A) + P(B) - P(A \text{ and } B)$
  • The term $P(A \text{ and } B)$ is subtracted to correct for double-counting the intersection area.
  • This rule applies universally; if events happen to be mutually exclusive, $P(A \text{ and } B) = 0$, simplifying the formula back to the Special Rule of Addition.

• Joint Probability:

  • A probability that measures the likelihood that two or more events will happen together ($P(A \text{ and } B)$).

• Classroom Survey Example (Freshmen & Nissler Students):

  • Total class size = $60$ students.
  • Freshmen (Event F): $15$ students; $P(F) = \frac{15}{60}$
  • Nissler Business School students (Event N): $40$ students; $P(N) = \frac{40}{60}$
  • Students who are both Freshmen and in Nissler ($F \text{ and } N$): $8$ students; $P(F \text{ and } N) = \frac{8}{60}$
  • Probability a student selected at random is a Freshman OR a Nissler student:         $P(F \text{ or } N) = P(F) + P(N) - P(F \text{ and } N) = \frac{15}{60} + \frac{40}{60} - \frac{8}{60} = \frac{47}{60}$

• Florida Tourist Survey Example:

  • Sample size = $200$ tourists in Florida.
  • Visited Disney World (Event D): $120$ tourists
  • Visited Busch Gardens (Event B): $100$ tourists
  • Visited both parks (Event D and B): $60$ tourists
  • Probability a selected person visited Disney World OR Busch Gardens:         $P(D \text{ or } B) = P(D) + P(B) - P(D \text{ and } B) = \frac{120}{200} + \frac{100}{200} - \frac{60}{200} = \frac{160}{200} = 0.80$

• Standard Card Deck Example (King or Heart):

  • A standard deck contains $52$ cards ($4$ suits: Hearts, Diamonds, Clubs, Spades; $13$ cards per suit).
  • Kings (Event K): $4$ cards; $P(K) = \frac{4}{52}$
  • Hearts (Event H): $13$ cards; $P(H) = \frac{13}{52}$
  • King of Hearts ($K \text{ and } H$): $1$ card; $P(K \text{ and } H) = \frac{1}{52}$
  • Probability of drawing a King OR a Heart:         $P(K \text{ or } H) = P(K) + P(H) - P(K \text{ and } H) = \frac{4}{52} + \frac{13}{52} - \frac{1}{52} = \frac{16}{52} \approx 0.3077$

• Student Course Passing Example:

  • Probability of passing History ($H$): $P(H) = 0.60$
  • Probability of passing Math ($M$): $P(M) = 0.70$
  • Probability of passing both History and Math ($H \text{ and } M$): $P(H \text{ and } M) = 0.50$
  • Probability of passing at least one course ($H \text{ or } M$):         $P(H \text{ or } M) = P(H) + P(M) - P(H \text{ and } M) = 0.60 + 0.70 - 0.50 = 0.80$

• Sea Critters Fish Pond Example:

  • A pond contains $140$ fish total:
    • Green swordtails ($G$): $80$ fish
    • Male fish ($M$): $60$ fish total ($36$ male green swordtails + $24$ other males)
    • Male and Green swordtails ($M \text{ and } G$): $36$ fish
  • Part B - Probability selected fish is male: $P(M) = \frac{60}{140}$
  • Part C - Probability selected fish is male AND a green swordtail: $P(M \text{ and } G) = \frac{36}{140}$
  • Part D - Probability selected fish is male OR a green swordtail:         $P(M \text{ or } G) = P(M) + P(G) - P(M \text{ and } G) = \frac{60}{140} + \frac{80}{140} - \frac{36}{140} = \frac{104}{140} \approx 0.7429$

• Rocky Mountain Region Visitors Example:

  • Visited Yellowstone ($Y$): $50\%$
  • Visited Grand Teton ($T$): $40\%$
  • Visited both parks ($Y \text{ and } T$): $35\%$
  • Probability a visitor visited at least one of these parks:         $P(Y \text{ or } T) = P(Y) + P(T) - P(Y \text{ and } T) = 0.50 + 0.40 - 0.35 = 0.55$
  • Are the events mutually exclusive? No, because their joint probability is non-zero ($P(Y \text{ and } T) = 0.35 \neq 0$), showing overlap.

Rules of Multiplication and Independence

• Special Rule of Multiplication:

  • Applies when two or more events are independent.
  • Definition of Independence: The occurrence of one event has no effect on the probability of the occurrence of another event.
  • Formula for two independent events:         $P(A \text{ and } B) = P(A) \times P(B)$
  • Formula for four independent events:         $P(A \text{ and } B \text{ and } C \text{ and } D) = P(A) \times P(B) \times P(C) \times P(D)$

• American Automotive Association Survey:

  • $60\%$ of members made airline reservations last year ($P(\text{Reservation}) = 0.60$).
  • Two members are selected at random independently.
  • Probability both made airline reservations:         $P(A \text{ and } B) = P(A) \times P(B) = 0.60 \times 0.60 = 0.36$

• Armco Traffic Light Systems Example:

  • Accelerated life tests show $95\%$ ($0.95$) of newly developed traffic light systems last $3$ years before failing.
  • A city purchases $4$ systems ($A, B, C, D$).
  • Traffic lights on separate roads operate independently.
  • Probability all $4$ systems operate properly for at least $3$ years:         $P(A \text{ and } B \text{ and } C \text{ and } D) = 0.95 \times 0.95 \times 0.95 \times 0.95 = 0.95^4 \approx 0.8145$
  • This problem illustrates the Special Rule of Multiplication.

• Delta Airlines On-Time Arrival Example:

  • Probability any flight arrives within $15$ minutes of scheduled time = $0.90$.
  • Four flights selected on different days (independent events).
  • Part A - Probability all $4$ flights arrive within $15$ minutes:         $0.90 \times 0.90 \times 0.90 \times 0.90 = 0.90^4 = 0.6561$
  • Part B - Probability none of the $4$ flights arrive within $15$ minutes:
    • Probability a single flight is late = $1 - 0.90 = 0.10$
    • Probability all $4$ flights are late = $0.10 \times 0.10 \times 0.10 \times 0.10 = 0.10^4 = 0.0001$
  • Part C - Probability at least one of the selected flights did not arrive within $15$ minutes:
    • Using the complement rule: $P(\text{at least 1 late}) = 1 - P(\text{all 4 on time})$
    • $1 - 0.6561 = 0.3439$

General Rule of Multiplication and Dependent Events

• Dependent Events Definition:

  • Occurs when the outcome or occurrence of the first event alters the probability of the second event.

• Conditional Probability:

  • The probability of event $B$ occurring given that event $A$ has already occurred.
  • Notation: $P(B \mid A)$

• General Rule of Multiplication:

  • Applied when events are dependent.
  • Formula:         $P(A \text{ and } B) = P(A) \times P(B \mid A)$
  • If events are independent, $P(B \mid A) = P(B)$, returning to the Special Rule of Multiplication.

• Marble Box Demonstration:

  • A box contains $10$ marbles total: $7$ green, $3$ red.
  • Initial probabilities: $P(\text{Green}) = \frac{7}{10}$, $P(\text{Red}) = \frac{3}{10}$
  • First event: Pick $1$ green marble and do not replace it.
  • Remaining marbles = $9$ total ($6$ green, $3$ red).
  • Second event: Probability of picking a red marble given the first was green:         $P(\text{Red}_2 \mid \text{Green}_1) = \frac{3}{9}$
  • Because the first event altered the sample space for the second event, these events are dependent.

• Golfer Shirts Example:

  • A closet contains $12$ golf shirts: $9$ white, $3$ blue.
  • The golfer picks shirts in the dark on $2$ consecutive days without replacing them.
  • Probability of wearing white shirts on both days:
    • Probability of picking a white shirt on Day 1: $P(W_1) = \frac{9}{12}$
    • Probability of picking a white shirt on Day 2 given Day 1 was white: $P(W_2 \mid W_1) = \frac{8}{11}$
    • Combined probability: $P(W_1 \text{ and } W_2) = \frac{9}{12} \times \frac{8}{11} = \frac{72}{132} = \frac{6}{11} \approx 0.5455$

• Clean Brush Products Toothbrush Example:

  • A shipment of $20$ toothbrushes contains $3$ defective and $17$ non-defective toothbrushes.
  • Part A - Probability the first two toothbrushes sold are defective:
    • First toothbrush defective: $P(D_1) = \frac{3}{20}$
    • Second toothbrush defective: $P(D_2 \mid D_1) = \frac{2}{19}$
    • $P(D_1 \text{ and } D_2) = \frac{3}{20} \times \frac{2}{19} = \frac{6}{380} \approx 0.0158$
  • Part B - Probability the first two toothbrushes sold are non-defective:
    • First toothbrush non-defective: $P(ND_1) = \frac{17}{20}$
    • Second toothbrush non-defective: $P(ND_2 \mid ND_1) = \frac{16}{19}$
    • $P(ND_1 \text{ and } ND_2) = \frac{17}{20} \times \frac{16}{19} = \frac{272}{380} \approx 0.7158$
  • Note: Both the numerator and denominator decrease sequentially because sampling is conducted without replacement.

• Blackie Investment Real Estate Example:

  • Total land tracts purchased = $10$ ($4$ tracts in Holly Farms, $6$ tracts in Newburgh Woods).
  • Part A - Probability the next two tracts sold are in Newburgh Woods:
    • First tract in Newburgh Woods: $P(N_1) = \frac{6}{10}$
    • Second tract in Newburgh Woods: $P(N_2 \mid N_1) = \frac{5}{9}$
    • $P(N_1 \text{ and } N_2) = \frac{6}{10} \times \frac{5}{9} = \frac{30}{90} = \frac{1}{3} \approx 0.3333$
  • Part B - Probability that of the next four tracts sold, at least one will be in Holly Farms:
    • Using complement rule: $P(\text{at least 1 Holly Farms}) = 1 - P(\text{all 4 in Newburgh Woods})$
    • $P(\text{1st Newburgh}) = \frac{6}{10}$
    • $P(\text{2nd Newburgh}) = \frac{5}{9}$
    • $P(\text{3rd Newburgh}) = \frac{4}{8}$
    • $P(\text{4th Newburgh}) = \frac{3}{7}$
    • $P(\text{all 4 Newburgh}) = \frac{6}{10} \times \frac{5}{9} \times \frac{4}{8} \times \frac{3}{7} = \frac{360}{5040} = \frac{1}{14}$
    • $P(\text{at least 1 Holly Farms}) = 1 - \frac{1}{14} = \frac{13}{14} \approx 0.9286$

Summary of Probability Rules

• Addition Rules (Calculating $P(A \text{ or } B)$):

  • General Rule (Any events):$P(A \text{ or } B) = P(A) + P(B) - P(A \text{ and } B)$
  • Special Rule (Mutually Exclusive events where $P(A \text{ and } B) = 0$):$P(A \text{ or } B) = P(A) + P(B)$

• Multiplication Rules (Calculating $P(A \text{ and } B)$):

  • General Rule (Dependent events):$P(A \text{ and } B) = P(A) \times P(B \mid A)$
  • Special Rule (Independent events where $P(B \mid A) = P(B)$):$P(A \text{ and } B) = P(A) \times P(B)$

• Complement Rule:

  • $P(A) = 1 - P(\text{not } A)$

Contingency Tables and Conditional Calculations

• Contingency Table Definition:

  • A tabular representation used to classify sample observations according to two or more identifiable categories or classes.
  • Row totals and column totals are critical for finding marginal and joint probabilities.

• Facebook Account and Age Survey Example:

  • Sample size = $150$ adults.
  • Categories: Age ($> 50$ years vs. $< 50$ years) and Facebook Accounts used.
  • $60$ total adults have $0$ Facebook accounts ($20$ aged $> 50$, $40$ aged $< 50$).
  • $70$ total adults are aged $> 50$ ($10$ of whom have $2$ or more Facebook accounts).

• Movie Theater Survey Example:

  • Sample size = $500$ moviegoers.
  • Age Categories:
    • $B_1$: Less than $30$ years old ($100$ total people)
    • $B_2$: $30$ up to $60$ years old ($225$ total people)
    • $B_3$: $60$ years or older ($175$ total people)
  • Monthly Movie Attendance Categories:
    • $A_1$: $0$ movies ($75$ total people)
    • $A_2$: $1$ or $2$ movies ($200$ total people)
    • $A_3$: $3$, $4$, or $5$ movies
    • $A_4$: $6$ or more movies ($50$ total people)

• Calculations based on Movie Theater Table:

  • 1. Probability of selecting a person who attends $6$ or more movies per month ($A_4$):         $P(A_4) = \frac{50}{500} = 0.10$
  • 2. Probability of selecting a person who attends $2$ or fewer movies per month ($A_1 \text{ or } A_2$):         $P(A_1 \text{ or } A_2) = P(A_1) + P(A_2) = \frac{75}{500} + \frac{200}{500} = \frac{275}{500} = 0.55$
  • 3. Probability of selecting a person who attends $6$ or more movies per month OR is $60$ years of age or older ($A_4 \text{ or } B_3$):
    • Uses General Rule of Addition due to overlap ($30$ people fit both conditions):         $P(A_4 \text{ or } B_3) = P(A_4) + P(B_3) - P(A_4 \text{ and } B_3) = \frac{50}{500} + \frac{175}{500} - \frac{30}{500} = \frac{195}{500} = 0.39$
  • 4. Conditional Probability of attending $6$ or more movies given the person is $60$ years of age or older ($P(A_4 \mid B_3)$):
    • Restrict the focus entirely to row $B_3$ (total $175$ people).
    • Count of people in $A_4$ within row $B_3 = 30$.         $P(A_4 \mid B_3) = \frac{30}{175} \approx 0.1714$
  • 5. Joint Probability of attending $6$ or more movies AND being $60$ years of age or older ($P(A_4 \text{ and } B_3)$):
    • Evaluates the intersection relative to the overall sample size ($500$).         $P(A_4 \text{ and } B_3) = \frac{30}{500} = 0.06$
  • 6. Probability of selecting an adult who is $30$ up to $60$ years old ($B_2$):         $P(B_2) = \frac{225}{500} = 0.45$
  • 7. Probability of selecting an adult who is under $60$ years of age ($B_1 \text{ or } B_2$):         $P(B_1 \text{ or } B_2) = P(B_1) + P(B_2) = \frac{100}{500} + \frac{225}{500} = \frac{325}{500} = 0.65$
  • 8. Probability of selecting an adult who is less than $30$ years old OR went to $0$ movies ($B_1 \text{ or } A_1$):
    • Overlap ($B_1 \text{ and } A_1$) = $15$ people.         $P(B_1 \text{ or } A_1) = P(B_1) + P(A_1) - P(B_1 \text{ and } A_1) = \frac{100}{500} + \frac{75}{500} - \frac{15}{500} = \frac{160}{500} = 0.32$
  • 9. Probability of selecting an adult who is less than $30$ years old AND went to $0$ movies ($B_1 \text{ and } A_1$):         $P(B_1 \text{ and } A_1) = \frac{15}{500} = 0.03$

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