Applied Foundations & Computational Lab

Basic Probability & Python from Zero

An exact mathematical foundation: sample spaces, empirical distributions, multi-stage counting principles, and conditional theorems implemented with foundational Python 3.

Essential Syntax Toolkit

Python from Zero: Foundational Rules

NO.CORE COMPUTATIONAL CONCEPTSYNTAX EXAMPLE

01Variables store numeric values in memoryx = 5

02Basic arithmetic operators (+, -, *, /)(x + y) / 2

03Powers and square roots via exponent operatorx ** 0.5

04Parentheses enforce explicit order of operations(33 / 72)

05Print output to standard display consoleprint(...)

06Round decimal float values for clean presentationround(val, 4)

# 1. Assign variable values
x = 5
y = 3

# 2. Arithmetic with parentheses precedence
answer = (x + y) / 2

# 3. Output result to screen
print("Answer:", answer)

STANDARD OUTPUT CONSOLEEXIT CODE: 0

Answer: 4.0

PLOTTING NOTEProbability graphs later in this manual use matplotlib for visual distribution curves.

Probability Experiment Analysis

Sample Space & Complement Rule

CLASSICAL RULEP(E) = n(Favorable) / n(Total Outcomes)

COMPLEMENTP(Eᶜ) = 1 - P(E)

STEP 01

Total Possible Outcomes: Coin toss (2) × Die roll (6) equally likely results

2 × 6 = 12

STEP 02

Target Event E (Tail and Odd): Favorable outcomes are {T1, T3, T5} = 3

P(E) = 3/12 = 0.25

STEP 03

Complement Event P(Eᶜ): Probability of NOT getting (Tail and Odd)

1 - 0.25 = 0.75

SAMPLE SPACE MATRIX S = {H, T} × {1..6}TOTAL = 12

H1

H2

H3

H4

H5

H6

T1 ★

T2

T3 ★

T4

T5 ★

T6

# 1. Classical Probability: 3 favorable / 12 total
probability = 3 / 12

# 2. Complement Rule
complement = 1 - probability

print(probability)
print(complement)

CONSOLE OUTPUTP(E) & P(Eᶜ)

0.25
0.75

Statistical Relative Frequency

Empirical Probability & Frequency Ratios

EMPIRICAL FORMULAP(E) = Frequency of Event (f) / Total Frequency (n)

Used when probabilities are derived from observed real-world frequencies.

STEP 01

Total Survey Responses (n): Sum all observed counts: Serious + Moderate + Not a Problem

123 + 115 + 82 = 320

STEP 02

Probability of "Serious" Response: Divide observed serious frequency (123) by total (320)

P(Serious) = 123 / 320

RESULT

Exact Decimal Probability: Approximates 38.44% probability

0.384375

LECTURE SURVEY FREQUENCY TABLETOTAL N = 320

Response Category	Observed (f)	Relative Probability P(E)
Serious TARGET	123	123 / 320 = 0.384375
Moderate	115	115 / 320 = 0.359375
Not a Problem	82	82 / 320 = 0.256250
Total Frequency (n)	320	Σ P(E) = 1.000000

# Empirical Probability: frequency / total
probability = 123 / 320

print(probability)

CONSOLE OUTPUTP(SERIOUS)

0.384375

Multi-Stage Combinatorics

Fundamental Counting Principle

COUNTING RULETotal Outcomes = n₁ × n₂ × ⋯ × nₖ

When a process has k stages with nᵢ choices at each stage.

STEP 01

Count All 8-Digit ID Combinations: Each of 8 positions allows 10 digits (0–9) with repetition.

10⁸ = 100,000,000

STEP 02

Probability of One Specific ID: Only 1 favorable outcome out of 100,000,000 total IDs.

P = 1 / 10⁸

RESULT

Scientific & Decimal Format: P(Specific ID) = 0.00000001

1.0 × 10⁻⁸

8-DIGIT ID MULTI-STAGE SLOTS10 CHOICES PER SLOT

POS 110

POS 210

POS 310

POS 410

POS 510

POS 610

POS 710

POS 810

TOTAL PERMUTATIONS:10 × 10 × 10 × 10 × 10 × 10 × 10 × 10 = 10⁸

# 1. Total permutations using power operator
total = 10 ** 8

# 2. Probability of one specific ID
probability = 1 / total

print("Total:", total)
print("Probability:", probability)

CONSOLE OUTPUTSCIENTIFIC NOTATION

Total: 100000000
Probability: 1e-08

Prior Condition Evaluation

Conditional Probability & Subsets

CONDITIONAL FORMULAP(A | B) = P(A ∩ B) / P(B)

Calculates probability of event A given that event B has already occurred.

STEP 01

Identify Conditioned Sub-group B: Denominator restricted to children with the gene.

n(Gene Group) = 72

STEP 02

Count High IQ Within Gene Group: 33 of those 72 children have high IQ.

P(High IQ | Gene) = 33/72

RESULT

Rounded 4-Digit Decimal Result: Approximates 45.83% probability

0.4583

POPULATION SUBSET BREAKDOWNCONDITIONED ON GENE

Group Partition	Count	Conditional Ratio
High IQ & Gene A ∩ B	33	33 / 72 ≈ 0.4583
Normal IQ & Gene	39	39 / 72 ≈ 0.5417
Total Gene Group (B)	72	P(B\|B) = 1.0000

# Conditional Probability: P(High IQ | Gene)
probability = 33 / 72

# Round to 4 decimal places for clean display
print(round(probability, 4))

CONSOLE OUTPUTROUNDED(4)

0.4583

Sequential Joint Probability

Multiplication Rule & Sequences

JOINT RULEP(A and B) = P(A) × P(B | A)

Calculates sequential events when first draw affects the sample size of the second.

STEP 01

First Draw — King: 4 Kings in standard 52-card deck

P(K) = 4 / 52

STEP 02

Second Draw — Queen (Without Replacement): 4 Queens remain in 51 remaining cards

P(Q | K) = 4 / 51

STEP 03

Multiply Sequential Probabilities: (4/52) × (4/51) = 16 / 2652

≈ 0.006033

1st Draw (King)4 / 52

×

2nd Draw (Queen)4 / 51

=

Joint P(K & Q)0.006033

# Dependent draws without replacement
probability = (4/52) * (4/51)

# Round to 6 decimal places
print(round(probability, 6))

CONSOLE OUTPUTROUNDED(6)

0.006033

Repeated Trials Complement

At Least One Success Rule

COMPLEMENT THEOREMP(at least one) = 1 - P(none) = 1 - qⁿ

When evaluating "at least one" is easier by calculating the complement "none".

STEP 01

Find Failure Probability (q): Success p = 0.85; failure q = 1 - 0.85

q = 0.15

STEP 02

All Successes (p³) vs No Success (q³): Independent repeated 3 procedures

q³ = 0.15³ = 0.003375

STEP 03

Complement Calculation P(≥ 1): 1 - P(none) = 1 - 0.003375

0.996625

OUTCOME PROBABILITY DISTRIBUTION (N = 3)TOTAL = 100%

At Least One Success (99.66%)

P(All 3 Success) = 61.41%P(None) = 0.34%

p = 0.85
q = 1 - p

all_success = p ** 3
none = q ** 3
at_least_one = 1 - none

print(all_success)
print(none)
print(at_least_one)

CONSOLE OUTPUTALL / NONE / AT LEAST ONE

0.614125
0.003375
0.996625

STUDENT LAB TASKS // INDIVIDUAL WORK

Section 01 Assignment: Practice Problems

Write and submit a Python script solving the following 3 probability tasks. Ensure all formulas are coded from first principles with clear variable names and comments.

TASK 01

Complementary Defect Model (Ch 3.1 & 3.2): A production line has defective rate p = 0.08. For a sample of n = 4 items drawn independently, compute the probability of finding at least one defective item.

FIND: P(≥ 1 Defective)

TASK 02

Key Permutation Entropy (Ch 3.1.3): A security access key has 7 positions: 2 uppercase letters (A–Z, repeats allowed) followed by 5 digits (0–9, no repetition). Find the total unique keys and the probability of cracking it on the 1st attempt.

FIND: Total Keys & P(Crack)

TASK 03

Dependent Sequence Draw (Ch 3.2.2): From a standard shuffled 52-card deck, calculate the joint probability of drawing a King, then a Queen, then a Jack sequentially without replacement.

FIND: P(K → Q → J)

SUBMISSION SPECIFICATIONSWEIGHT: 10 PTS

FORMAT: .py Script

DUE: Before Next Lab

RUBRIC: 40% Math • 40% Code

DOCS: 20% Comments

# Task 1: Complementary Defect Model
p_defect = 0.08
n_items = 4
# TODO: Calculate P(at least 1 defective)
p_at_least_one = ...

# Task 2: Badge Key Permutations
# TODO: Apply Counting Principle for 2 letters + 5 digits
total_keys = ...
p_crack = ...

# Task 3: Dependent 3-Card Sequence (Without Replacement)
# TODO: Sequential multiplication (King -> Queen -> Jack)
p_sequence = ...

REQUIRED OUTPUT FORMATCONSOLE DISPLAY

Task 1 - P(At Least 1 Defect): <float>
Task 2 - Total Keys: <int> | P(Crack): <scientific>
Task 3 - P(K then Q then J): <float: 6 decimals>