STATGEN CLASS 2023 #8 Metode Statistika I (SATS4121) Pertemuan 4

Main ideas, concepts, and lessons (Module 3: Opportunity & Probability formulas)

0) Class flow / learning goals (framing)

• The instructor (Kak Rio) emphasizes that they will learn together, not just lecture.
• Students are encouraged to ask questions or write clarifications during explanations.
• Today’s material is Module 3 (BMP: SATS4121 / “Metode Statistika I”), which contains three topics:
  1. Basic concept of opportunity/probability
  2. Probability formulas using set algebra (complements, unions, intersections, mutually exclusive vs not, etc.)
  3. Conditional probability & Bayes’ theorem (not fully reached due to time)

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1) Basic concept of opportunity (probability)

1.1 Meaning of “opportunity” in statistics

• “Opportunity” is treated as probability.
• Probability measures the extent to which an event/outcome is possible.
• Probability is typically expressed as a number between 0 and 1:
  • 0 ⇒ the event cannot occur
  • 1 ⇒ the event will definitely occur

1.2 Key objects: sample space and sample points

• Sample space (S): the set of all possible outcomes in an experiment/event.
• Sample points: each individual possible outcome in the sample space.
• Examples:
  • Throw one balanced die:
    • (S = {1,2,3,4,5,6})
    • sample points: 6
  • Select ball color from a box: sample space depends on the available colors (e.g., red, blue, green, yellow, white).
  • Throw one coin: sample space includes heads/tails.

1.3 Counting sample points using the multiplication principle

• If one operation can be done in (N_1) ways and another in (N_2) ways, then both together occur in (N_1 \times N_2) ways.
• Example:
  • Throw 1 coin and 1 die:
    • coin outcomes: 2
    • die outcomes: 6
    • total sample points: (2 \times 6 = 12)

1.4 Practice: sample space and sample points

• Q1: One balanced die is thrown once
  • Answer: (S={1,2,3,4,5,6}), sample points = 6
• Q2: Two balanced dice are thrown once each
  • Answer: sample points = (6 \times 6 = 36)

1.5 Types of events: simple vs compound

• Event: a subset of the sample space.
• Simple event: event set contains exactly one sample point.
• Compound event: event set contains multiple sample points.
• Examples with a die:
  • Simple event: “even prime number” (only 2 in 1–6) ⇒ one outcome
  • Compound event: “even number” (2, 4, 6) ⇒ multiple outcomes

1.6 Probability of an event (basic formula)

• Probability of event (A) (written as (P(A)), or (P_a) in some subtitles): [ P(A) = \frac{N_a}{N_s} ] where:

  • (N_a) = number of outcomes favorable to event (A)
  • (N_s) = total number of outcomes
  • Constraint:
  • (0 \le P(A) \le 1)

1.7 Worked example: weighted die probabilities (probability via weights)

• Scenario: a non-uniform (weighted) die
  • Odd outcomes (1,3,5) have weight (B)
  • Even outcomes (2,4,6) have weight (2B)

• Normalization (total probability equals 1): [ 3B + 3(2B) = 1 \Rightarrow 9B = 1 \Rightarrow B=\frac{1}{9} ]

• Therefore:

  • Each odd number: (\frac{1}{9})
  • Each even number: (\frac{2}{9})
• Probability of “(K<4)” (outcomes 1,2,3): [ P = \frac{1}{9}+\frac{2}{9}+\frac{1}{9}=\frac{4}{9} ]

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2) Probability formulas (set algebra)

2.1 Probability of complements

• Complement of (A) (written as (A^c) or “not (A)”) contains all outcomes not in (A): [ P(A^c)=1-P(A) ]

• Examples:

  • Bag with 10 balls (4 red, 6 blue):
    • (P(\text{red})=4/10=2/5)
    • (P(\text{not red})=P(\text{blue})=1-2/5=3/5)
  • Two dice:
    • (P(\text{total}=12)=1/36)
    • (P(\text{not 12})=1-1/36=35/36)

2.2 Union / “combination” of events (A or B)

• Union corresponds to: [ A \cup B = \text{all elements in A or in B (or both)} ]

• Lesson from the session:

  • If overlap information (intersection) is not given, you determine whether the problem implies a simple “add” situation (union without overlap), depending on the stated structure.

2.3 Intersection of events (A and B)

• Intersection: [ A \cap B = \text{elements that are in both A and B} ]

• Example:

  • (A={1,2,3,4,5}), (B={2,3,5,7})
  • (A \cap B={2,3,5})
• Practice sets:
  • (A={2,3,5,7,11,13,17,19}) (primes < 20)
  • (B={1,3,5,7,9}) (odd numbers < 10)
  • Intersection: ({3,5,7})

2.4 Mutually exclusive vs non-mutually exclusive events

• Mutually exclusive:

  • Cannot happen at the same time
  • (A \cap B = \varnothing)

  • Then: [ P(A\cup B)=P(A)+P(B) ]

  • Cue: questions using “or” with implied exclusivity.

  • Non-mutually exclusive:
  • Can occur simultaneously
  • (A \cap B \neq \varnothing)
  • Then: [ P(A\cup B)=P(A)+P(B)-P(A\cap B) ]

2.5 Worked example: sum of dice totals (7 or 11)

• Two dice total outcomes: (6 \times 6 = 36)
• Event definitions:
  • Event A: total = 7
  • Event B: total = 11

• Key takeaway:

  • If “7 or 11” corresponds to mutually exclusive outcomes, use: [ P(A\cup B)=P(A)+P(B) ]

• Note: subtitles included a correction during explanation, but the main method remains.

2.6 Practice example: odd on first die OR even on second die

• Treated as mutually exclusive in the session’s framing.
• Method shown:
  • Compute separately:
    • Odd on one die: (3/6 = 1/2)
    • Even on one die: (3/6 = 1/2)
  • Then apply the mutually exclusive union approach.
• Teaching point:
  • Identify event type first (mutually exclusive vs not) before selecting the formula.

2.7 Independent events (final probability formula section reached)

• Independent events:
  • The outcome of one event does not affect the probability of the other.

• For independent events (A) and (B): [ P(A\cap B)=P(A)\times P(B) ]

• Cue: question language indicating “independent/no influence.”

2.8 Worked example: fuses drawn sequentially without replacement

• Problem:
  • Box: 20 fuses, 5 defective
  • Draw 2 fuses sequentially without replacement
  • Find probability both are defective
• Let:
  • (A): first fuse is defective
  • (B): second fuse is defective after the first draw
• Computation:
  • (P(A)=5/20=1/4)
  • After one defective is taken:
    • remaining defective = 4
    • remaining total = 19
    • (P(B)=4/19)
  • Both defective: [ P(A\cap B)=P(A)\cdot P(B)=\frac{1}{4}\cdot\frac{4}{19}=\frac{1}{19} ]

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3) Conditional probability and Bayes’ theorem

• The instructor states this topic exists in Module 3:
  • Conditional probabilities and Bayes’ theorem
• The session indicates it was not fully covered due to time (e.g., “Only 15 minutes left…”) and then wrapped up with questions.

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Methodology / instruction-like steps explicitly used in the video

• Step 1: Identify the sample space
  • List all possible outcomes (coin, die, sets, etc.).
• Step 2: Identify the event
  • Determine which outcomes satisfy the question.
• Step 3: Count favorable outcomes
  • Compute (N_a) (favorable) and (N_s) (total).
• Step 4: Compute probability
  • Use (P(A)=N_a/N_s) for equally likely outcomes.
• Step 5: For complements
  • Use (P(A^c)=1-P(A)).

• Step 6: For unions (A or B)

  • If mutually exclusive: [ P(A\cup B)=P(A)+P(B) ]

  • If not mutually exclusive: [ P(A\cup B)=P(A)+P(B)-P(A\cap B) ]

• Step 7: For intersections (A and B)

  • If independent: [ P(A\cap B)=P(A)\times P(B) ]

  • For sequential draws without replacement:

    • compute (P(A)), update counts, compute (P(B)), then multiply (P(A)\cdot P(B)).
  • Step 8: Use set logic vocabulary
  • Simple vs compound events
  • Intersection vs union vs complement
  • Mutually exclusive vs non-mutually exclusive

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Speakers / sources featured (as indicated by subtitles)

• Kak Rio (main instructor/host)
• Kak Hidayat (MC/closing role)
• MC today (“MC”, unnamed explicitly, and later referenced as returning to Kak Hidayat)
• Students/participants named in subtitles:
  • Sis Nisa
  • Sis Khairunnisa
  • Sis Rizieq Noval
  • Sis Anju
  • Aurelia
  • Siti / Sis Siti Alhamdulillah
  • Sis Saiful / Muhammad Saiful
  • Sis Muhammad Syaifullah
  • Sis Jubaedah
  • Zoom participants / classmates (group participation; additional individual names appear where shown)