(Lecture 10) Probability in Our Daily Lives: Fundamental Concepts and Applications

Chapter 5: Probability in Our Daily Lives

Section 5.1: How Probability Quantifies Randomness

Random Phenomena

Random phenomena are observable occurrences whose outcomes cannot be predicted with certainty. The study of probability helps quantify the inherent randomness in such phenomena.

• Definition: Random phenomena are any observable occurrences where the outcome is uncertain.

• With a small number of observations, outcomes may appear unpredictable or vary widely from expectations.

• As the number of observations increases, summary statistics (such as proportions) tend to stabilize and approach specific values.

Long-Run Proportion and Probability

Probability is fundamentally linked to the long-run behavior of random phenomena. As more observations are made, the proportion of times a particular outcome occurs approaches a predictable value.

• The long-run proportion forms the basis for the definition of probability.

• For example, in repeated coin tosses, the proportion of heads approaches 0.5 as the number of tosses increases.

Relative Frequency Definition of Probability

The relative frequency of an event is the number of times the event occurs divided by the total number of trials.

• Formula: $\text{Relative Frequency} = \frac{\text{Number of times event occurs}}{\text{Total number of trials}}$

• Example: Tossing a coin 30 times and recording the number of heads.

Table: Coin Toss Results and Relative Frequency of Heads

The following table shows the outcome and relative frequency of heads for each trial:

Trial	Outcome	Relative Frequency of Heads
1	H	1.00
2	T	0.50
3	H	0.67
...	...	...
30	T	0.43

Additional info: Table entries inferred for illustration; actual outcomes and frequencies may vary.

Relative Frequency Graph

The graph of relative frequency versus number of trials typically shows high variability in the short run, but stabilizes as the number of trials increases.

• Initial trials may show large fluctuations.

• With more trials, the relative frequency approaches the theoretical probability (e.g., 0.5 for heads in a fair coin toss).

Long-Run Behavior of Random Outcomes

In random phenomena, the proportion of times an outcome occurs is unpredictable in the short run but predictable in the long run.

• Example: Rolling a die repeatedly; the proportion of times a 6 appears approaches $\frac{1}{6}$ as the number of rolls increases.

• Probability is interpreted as the long-run proportion of an outcome.

Probability Quantifies Long-Run Randomness

Probability is the proportion of times a particular outcome would occur in a long run of observations.

• Probability values range between 0 and 1.

• Example: The probability of rolling a 6 on a fair die is $\frac{1}{6}$.

Independent Trials

Trials are independent if the outcome of one trial does not affect the outcome of another.

• Example: Each roll of a fair die is independent of previous rolls.

• The law of large numbers states that long-run averages stabilize, but short-run results may vary significantly.

Finding Probabilities

Probabilities can be determined by making reasonable assumptions about the random phenomenon, such as symmetry or equal likelihood of outcomes.

• Empirical probabilities are estimated from observed relative frequencies.

• Sample proportions estimate actual probabilities, but accuracy improves with more trials.

Types of Probability: Relative Frequency and Subjective Probability

• Relative frequency probability: Based on long-run proportions from repeated trials.

• Subjective probability: Based on personal judgment or degree of belief, used when long-run data is unavailable.

Section 5.2: Finding Probabilities

Counting Techniques: Multiplication Principle

The Multiplication Principle is used to count the total number of possible outcomes in multi-stage experiments.

• If one experiment has $m$ outcomes and another has $n$ outcomes, there are $m \times n$ total outcomes.

• For $k$ experiments with $n_1, n_2, ..., n_k$ outcomes, total outcomes are $n_1 \times n_2 \times ... \times n_k$.

• Example: Tossing a coin three times yields $2 \times 2 \times 2 = 8$ outcomes.

Permutations and Combinations

• Permutation: An ordered arrangement of objects. Formula: $n!$ (where $n$ is the number of distinct objects)

• Combination: An unordered selection of $r$ objects from $n$ distinct objects. Formula: $\binom{n}{r} = \frac{n!}{r!(n-r)!}$

• Example: Number of ways to select 3 students from 10: $\binom{10}{3} = 120$

Sample Space

The sample space is the set of all possible outcomes of a random phenomenon.

• Example: For a three-question quiz, possible outcomes are: CCC, CCI, CIC, CII, ICC, ICI, IIC, III.

Events

An event is a subset of the sample space, representing one or more outcomes.

• Example: Event A: student answers all questions correctly (CCC). Event B: student passes (at least 2 correct answers).

Finding Probabilities of Events

• Each outcome in the sample space has a probability between 0 and 1.

• The sum of all individual probabilities equals 1.

• Formula: $P(A) = \text{sum of probabilities of outcomes in event A}$

Equally Likely Outcomes

• Example: Coin flip: S = {Head, Tail}

• Two coins: S = {(H, H), (H, T), (T, H), (T, T)}

• Six-sided die: S = {1, 2, 3, 4, 5, 6}

• Formula: $P(A) = \frac{\text{number of outcomes in A}}{\text{number of outcomes in sample space}}$

Example: Probability of Sum of Two Dice Equal to 7

• Enumerate all possible outcomes: (1,1), (1,2), ..., (6,6)

• There are 6 outcomes where the sum is 7: (1,6), (2,5), (3,4), (4,3), (5,2), (6,1)

• Probability: $P(\text{sum} = 7) = \frac{6}{36} = \frac{1}{6}$

Example: Selecting Coaches

• Number of ways to select 2 coaches from 4: $\binom{4}{2} = 6$

• Probability of selecting a particular pair: $\frac{1}{6}$

• Probability that coach C is chosen: $\frac{3}{6} = 0.5$

Example: Relative Frequency and Contingency Table

Consider a contingency table for tax audits by income level:

Income Level	Audited (Yes)	Audited (No)	Total
Under $200,000	1,233	139,305	140,538
$200,000 - $1,000,000	133	4,747	4,880
More than $1,000,000	39	324	363
Total	1,405	144,376	145,781

• Probability of audit: $\frac{1,405}{145,781} \approx 0.0096$

• Probability of income > $1,000,000$: $\frac{363}{145,781} \approx 0.0025$

Section 5.3: Basic Rules for Finding Probabilities

Complement of an Event

• The complement of event A, denoted A', consists of all outcomes not in A.

• Formula: $P(A') = 1 - P(A)$

Disjoint (Mutually Exclusive) Events

• Events A and B are disjoint if they have no outcomes in common.

• Example: Getting exactly one correct answer and exactly two correct answers on a quiz are disjoint events.

Intersection and Union of Events

• Intersection (A and B): Outcomes in both A and B. Formula: $P(A \cap B)$

• Union (A or B): Outcomes in A, B, or both. Formula: $P(A \cup B)$

Addition Rule: Probability of Union of Two Events

• General Rule: $P(A \cup B) = P(A) + P(B) - P(A \cap B)$

• If A and B are disjoint: $P(A \cup B) = P(A) + P(B)$

Example: Probability of Union in Family with Two Children

• Sample space: {FF, FM, MF, MM}

• Event A: first child is a girl; Event B: second child is a girl

• $P(A) = 0.5$, $P(B) = 0.5$, $P(A \cap B) = 0.25$

• $P(A \cup B) = 0.5 + 0.5 - 0.25 = 0.75$

Multiplication Rule: Probability of Intersection of Independent Events

• If A and B are independent: $P(A \cap B) = P(A) \times P(B)$

• Example: Probability of rolling a 6 on two dice: $\frac{1}{6} \times \frac{1}{6} = \frac{1}{36}$

Example: Guessing on a Multiple-Choice Quiz

• Probability of guessing all 3 questions correctly (each with 5 options): $0.2 \times 0.2 \times 0.2 = 0.008$

• Probability of guessing at least 2 correct: $0.008 + 3(0.032) = 0.104$

Events Often Are Not Independent

• Do not assume independence unless justified by context.

• Example: Actual student responses to quiz questions may show dependence between answers.

• Compare $P(A \cap B)$ to $P(A) \times P(B)$ to test independence.

Table: Student Quiz Responses

Outcome	Probability
I I	0.26
I C	0.11
C I	0.05
C C	0.58

• $P(A) = 0.05 + 0.58 = 0.63$ (first question correct)

• $P(B) = 0.11 + 0.58 = 0.69$ (second question correct)

• $P(A \cap B) = 0.58$

• If independent, $P(A \cap B) = 0.63 \times 0.69 = 0.43$

• Since $0.58 \neq 0.43$, events are not independent.

Additional info: All formulas and tables expanded for clarity and completeness.