Basic Probability Lecture Notes

Introduction to Uncertainty and Probability

  • Uncertainty is a fundamental aspect of making decisions with incomplete information, reflecting the way we generally operate in the world.

  • Everyday concepts used to describe the handling of uncertainty include chance, luck, and risk.

  • Probability is a branch of mathematics providing the language and tools necessary to quantify the uncertainty of events and reason in a principled manner.

Instances of Uncertainty

  • Common activities involving uncertain outcomes include flipping a coin, crossing the road without looking, or rolling a die.

  • These outcomes require weighing uncertainties against one another.

  • In a coin flip, there is no reason to expect heads appearing more often than tails.

  • If one crosses the road without looking, there is a risk of being squashed by a truck; looking before crossing significantly reduces the likelihood of this outcome.

Defining Probability and Experiments

  • Probability acts as the machinery used to describe and account for the fact that some outcomes occur more frequently than others.

  • It is a measure that quantifies the likelihood of an event occurring.

  • An Experiment is defined as any activity from which results are obtained.

  • A Random Experiment is one where results or outcomes cannot be predicted with total certainty. An example of a random experiment is flipping a coin and observing whether a head or tail appears.

Sample Space and Outcomes

  • An Outcome is the specific result of a single trial in a probability experiment.

  • The Sample Space, represented by the symbol SS, is the set of all possible outcomes for an experiment.

  • Each individual outcome within a sample space is referred to as an element, a member, or a sample point.

Examples of Sample Spaces
  • Rolling a Die: The sample space consists of six outcomes:     S={1,2,3,4,5,6}S = \{1, 2, 3, 4, 5, 6\}

  • Tossing a Coin Twice: There are four possible results based on the first and second flips:     S={HH,HT,TH,TT}S = \{HH, HT, TH, TT\}

  • Manufacturing Selection: If 3 items are randomly selected and classified as defective (DD) or nondefective (NN), a tree diagram can be used to list the elements of the sample space:     S={DDD,DDN,DND,DNN,NDD,NDN,NND,NNN}S = \{DDD, DDN, DND, DNN, NDD, NDN, NND, NNN\}

Types of Sample Spaces
  • Finite Sample Space: A space containing a fixed, finite number of points, such as {1,2,5,6}\{1, 2, 5, 6\}.

  • Countably Infinite Sample Space (Discrete): A space that has as many points as there are natural numbers (1,2,3,…1, 2, 3, \dots).

  • Noncountable Infinite Sample Space (Nondiscrete): A space containing as many points as there are in an interval on the xx axis, such as 0≤x≤10 \le x \le 1.

Events

  • An Event is a subset AA of the sample space SS, representing a set of possible outcomes.

  • A Simple Event is an event consisting of only a single outcome.

Examples of Events
  • In a die toss, the event that an outcome is divisible by 3 occurs if the outcome belongs to the subset A={3,6}A = \{3, 6\}.

  • In the manufacturing selection experiment of 3 items, the event that the number of defectives is greater than 1 occurs if the outcome is in the subset:     B={DDN,DND,NDD,DDD}B = \{DDN, DND, NDD, DDD\}

Classical and Empirical Probability

Classical Probability
  • Also known as theoretical probability, it is used when every outcome in a sample space is equally likely to occur.

  • The formula for the classical probability of event EE is:     P(E)=Number of outcomes in eventTotal number of outcomes in sample spaceP(E) = \frac{\text{Number of outcomes in event}}{\text{Total number of outcomes in sample space}}

  • Example: Rolling a die to find the probability of rolling a 5 (EventAEvent A). Since there is only one 5 in the set {5}\{5\}, the calculation is:     P(A)=16≈0.167P(A) = \frac{1}{6} \approx 0.167

Empirical Probability
  • Also known as statistical probability, it is based on actual observations from probability experiments.

  • The empirical probability of an event EE is its relative frequency:     P(E)=Frequency of Event (f)Total frequency (n)P(E) = \frac{\text{Frequency of Event (f)}}{\text{Total frequency (n)}}

  • Example: A travel agent finds that in every 50 reservations, 12 are for a cruise. The probability that the next reservation is a cruise is:     P(cruise)=1250=0.24P(\text{cruise}) = \frac{12}{50} = 0.24

Null Sets and Set Operations

  • A Null Set (ϕ\phi) contains no elements. For example, if B={x∣x is an even factor of 7},B = \{x \mid x \text{ is an even factor of 7}\}, BB is a null set because the factors of 7 (1 and 7) are odd.

Core Set Operations
  • Complement: The complement of event AA (A′A') consists of all elements in SS that are not in AA.

    • Example: If S={book, cell phone, mp3, paper, stationery, laptop}S = \{\text{book, cell phone, mp3, paper, stationery, laptop}\} and A={book, stationery, laptop, paper},A = \{\text{book, stationery, laptop, paper}\}, then A′={cell phone, mp3}A' = \{\text{cell phone, mp3}\}.

  • Intersection: Denoted by A∩BA \cap B, it is the event containing elements common to both AA and BB.

    • Example: If EE is the event of engineering majors and FF is the event of females, E∩FE \cap F is the set of all female engineering students.

  • Mutually Exclusive (Disjoint): Events AA and BB are mutually exclusive if they have no common elements, denoted as A∩B=ϕA \cap B = \phi.

  • Union: Denoted by A∪BA \cup B, it is the event containing elements belonging to AA, or BB, or both.

  • Difference: The difference of events AA and BB contains all outcomes included in AA but excluded from BB.

Venn Diagram Relationships
  • A∩BA \cap B: Regions common to both.

  • B∩CB \cap C: Regions common to BB and CC.

  • A∪CA \cup C: Regions in either AA or CC.

  • B′∩AB' \cap A: Regions in AA that are not in BB.

  • A∩B∩CA \cap B \cap C: The region common to all three.

  • (A∪B)∩C′(A \cup B) \cap C': Regions in the union of AA and BB that are not in CC.

Operational Rules and Axioms

Logical Rules for Sets
  1. A∩ϕ=ϕA \cap \phi = \phi

  2. A∪ϕ=AA \cup \phi = A

  3. A∩A′=ϕA \cap A' = \phi

  4. A∪A′=SA \cup A' = S

  5. S′=ϕS' = \phi

  6. ϕ′=S\phi' = S

  7. (A′)′=A(A')' = A

  8. (A∩B)′=A′∪B′(A \cap B)' = A' \cup B'

  9. (A∪B)′=A′∩B′(A \cup B)' = A' \cap B'

Axioms of Probability
  • Axiom 1: For every event AA, P(A)≥0P(A) \ge 0.

  • Axiom 2: For the certain event SS, P(S)=1P(S) = 1.

  • Axiom 3: For any number of mutually exclusive events A1,A2,…,AnA_1, A_2, \dots, A_n:     P(A1∪A2∪⋯∪An)=P(A1)+P(A2)+⋯+P(An)P(A_1 \cup A_2 \cup \dots \cup A_n) = P(A_1) + P(A_2) + \dots + P(A_n)

  • For two mutually exclusive events A1,A2A_1, A_2:     P(A1∪A2)=P(A1)+P(A2)P(A_1 \cup A_2) = P(A_1) + P(A_2)

Probability Theorems

  • Theorem 1: If A⊆BA \subseteq B, then P(A)≤P(B)P(A) \le P(B) and P(B−A)=P(B)−P(A)P(B - A) = P(B) - P(A).

  • Theorem 2: For every event AA, 0≤P(A)≤10 \le P(A) \le 1.

  • Theorem 3: The impossible event has zero probability: P(ϕ)=0P(\phi) = 0.

  • Theorem 4: P(A′)=1−P(A)P(A') = 1 - P(A).

  • Theorem 5 (Addition Rule): For any two events AA and BB:     P(A∪B)=P(A)+P(B)−P(A∩B)P(A \cup B) = P(A) + P(B) - P(A \cap B)

  • Theorem 6 (Triple Union Rule): For any three events A1,A2,A3A_1, A_2, A_3:     P(A1∪A2∪A3)=P(A1)+P(A2)+P(A3)−P(A1∩A2)−P(A2∩A3)−P(A3∩A1)+P(A1∩A2∩A3)P(A_1 \cup A_2 \cup A_3) = P(A_1) + P(A_2) + P(A_3) - P(A_1 \cap A_2) - P(A_2 \cap A_3) - P(A_3 \cap A_1) + P(A_1 \cap A_2 \cap A_3)

Worked Examples

Tossing a Coin Twice

Requirement: Find the probability of at least 1 head.

  • S={HH,HT,TH,TT}S = \{HH, HT, TH, TT\}

  • Assign a probability ω\omega to each equally likely point: 4ω=1  ⟹  ω=1/44\omega = 1 \implies \omega = 1/4.

  • Let AA be the event of at least 1 head: A={HH,HT,TH}A = \{HH, HT, TH\}.

  • P(A)=14+14+14=34P(A) = \frac{1}{4} + \frac{1}{4} + \frac{1}{4} = \frac{3}{4}

Loaded Die

Requirement: A die is loaded so an even number is twice as likely as an odd number. Find P(E)P(E) where EE is a number less than 44.

  • S={1,2,3,4,5,6}S = \{1, 2, 3, 4, 5, 6\}

  • Assign ω\omega to odds (1, 3, 5) and 2ω2\omega to evens (2, 4, 6).

  • Total probability: ω+2ω+ω+2ω+ω+2ω=9ω=1  ⟹  ω=1/9\omega + 2\omega + \omega + 2\omega + \omega + 2\omega = 9\omega = 1 \implies \omega = 1/9.

  • P(odd)=1/9P(\text{odd}) = 1/9; P(even)=2/9P(\text{even}) = 2/9.

  • E={1,2,3}  ⟹  P(E)=19+29+19=49E = \{1, 2, 3\} \implies P(E) = \frac{1}{9} + \frac{2}{9} + \frac{1}{9} = \frac{4}{9}.

  • Extended Example: For A={even}={2,4,6}A = \{\text{even}\} = \{2, 4, 6\} and B={divisible by 3}={3,6}B = \{\text{divisible by 3}\} = \{3, 6\}.

    • A∪B={2,3,4,6}  ⟹  P(A∪B)=29+19+29+29=79A \cup B = \{2, 3, 4, 6\} \implies P(A \cup B) = \frac{2}{9} + \frac{1}{9} + \frac{2}{9} + \frac{2}{9} = \frac{7}{9}.

    • A∩B={6}  ⟹  P(A∩B)=29A \cap B = \{6\} \implies P(A \cap B) = \frac{2}{9}.

Job Offers (John)

Requirement: Probability of an offer from Company A is 0.8, Company B is 0.6, and both is 0.5. Find probability of at least one offer.

  • P(A)=0.8P(A) = 0.8

  • P(B)=0.6P(B) = 0.6

  • P(A∩B)=0.5P(A \cap B) = 0.5

  • P(A∪B)=0.8+0.6−0.5=0.9P(A \cup B) = 0.8 + 0.6 - 0.5 = 0.9

Pairs of Dice

Requirement: Find the probability of a total of 7 or 11.

  • Total sample points = 36.

  • Event AA (Total 7): 6 points. P(A)=6/36=1/6P(A) = 6/36 = 1/6.

  • Event BB (Total 11): 2 points. P(B)=2/36=1/18P(B) = 2/36 = 1/18.

  • Since they are mutually exclusive: P(A∪B)=16+118=418=29P(A \cup B) = \frac{1}{6} + \frac{1}{18} = \frac{4}{18} = \frac{2}{9}.

  • Alternatively: n/N=8/36=2/9n/N = 8/36 = 2/9.

Car Color Preference

Requirement: Probabilities for green (0.09), white (0.15), red (0.21), and blue (0.23). Find probability of one of these colors.

  • Events are mutually exclusive.

  • P(G∪W∪R∪B)=0.09+0.15+0.21+0.23=0.68P(G \cup W \cup R \cup B) = 0.09 + 0.15 + 0.21 + 0.23 = 0.68

Mechanic Servicing Cars

Requirement: Probabilities for servicing 3, 4, 5, 6, 7, or 8+ cars are 0.12, 0.19, 0.28, 0.24, 0.10, and 0.07. Find the probability of servicing at least 5 cars.

  • Let EE be the event of servicing at least 5 cars.

  • E′E' is the event of servicing fewer than 5 cars (servicing 3 or 4 cars).

  • P(E′)=0.12+0.19=0.31P(E') = 0.12 + 0.19 = 0.31

  • P(E)=1−P(E′)=1−0.31=0.69P(E) = 1 - P(E') = 1 - 0.31 = 0.69