Probability Theory: Events, Axioms, Rules & Applications

TAKEAWAYS:

• Probability quantifies uncertainty using axioms (Kolmogorov) and rules (addition, multiplication) to model real-world events.
• Conditional probability and independence help analyze dependent vs. independent scenarios (e.g., loan defaults, product recalls).
• Bayes’ theorem updates beliefs with new evidence, critical for medical testing, fraud detection, and business forecasting.
• Probability distributions (discrete/continuous) describe random variables, used in inventory management, insurance premiums, and quality control.
• Complementary events and mutually exclusive events simplify complex probability calculations in risk analysis.
• Real-world tools like eSewa’s fraud detection (Bayes’ theorem) and Ncell’s call routing (conditional probability) rely on these principles.

1. Introduction to Probability Theory

Probability is the branch of mathematics that studies randomness and uncertainty. It provides a framework to quantify the likelihood of events occurring, which is essential in business for decision-making under uncertainty.

1.1 Definitions

• Experiment: An action or process whose outcome is uncertain (e.g., drawing a ball from a bag, rolling a die).
• Sample Space (S): The set of all possible outcomes of an experiment.
• Event (E): A subset of the sample space (e.g., "drawing a multiple of 3").
• Probability of an Event (P(E)): A measure of how likely an event is, ranging from 0 (impossible) to 1 (certain).

1.2 Axioms of Probability (Kolmogorov)

Probability must satisfy three axioms:

1. Non-negativity: $P(E)\ge 0$ for any event $E$.
2. Normalization: $P(S)=1$ (the entire sample space is certain).
3. Additivity: For mutually exclusive events $E_1$ and $E_2$, $P(E_1\cup E_2)=P(E_1)+P(E_2)$.

Worked Example 1: Bag of Balls (Past Exam) A bag contains 20 balls numbered 1 to 20. Find the probability of drawing a ball that is a multiple of 3 or 7.

Solution:

1. Sample Space: $S=\{1,2,...,20\}$.
2. Event $E$: Multiples of 3 or 7.
   • Multiples of 3: $\{3,6,9,12,15,18\}$ → 6 outcomes.
   • Multiples of 7: $\{7,14\}$ → 2 outcomes.
   • Overlap (multiple of 21): None in 1–20.
3. Total favorable outcomes: $6+2=8$.
4. Probability: $$P(E)=\frac{\text{Favorable outcomes}}{\text{Total outcomes}}=\frac{8}{20}=0.4$$

Visualization:

2. Types of Events

2.1 Mutually Exclusive (Disjoint) Events

Two events cannot occur simultaneously.

• Example: Drawing a red or black ball (not both).
• Addition Rule: $P(A\cup B)=P(A)+P(B)$.

2.2 Independent Events

The occurrence of one event does not affect the other.

• Example: Rolling a die and flipping a coin.
• Multiplication Rule: $P(A\cap B)=P(A)\times P(B)$.

2.3 Dependent Events

The occurrence of one event affects the other.

• Example: Drawing two balls without replacement.
• Conditional Probability: $$P(A|B)=\frac{P(A\cap B)}{P(B)}$$

Worked Example 2: Committee Formation (Past Exam) A committee of 5 is formed from 8 boys and 7 girls. Find the probability that the committee has: (a) All boys. (b) All girls.

Solution:

1. Total ways to form a committee: $\left(\genfrac{}{}{0ex}{}{15}{5}\right)=3003$.
2. (a) All boys: $\left(\genfrac{}{}{0ex}{}{8}{5}\right)=56$. $P(\text{All boys})=\frac{56}{3003}\approx 0.0186$.
3. (b) All girls: $\left(\genfrac{}{}{0ex}{}{7}{5}\right)=21$. $P(\text{All girls})=\frac{21}{3003}\approx 0.00699$.

Visualization:

mindmap
  root((Committee Formation))
    Boys[8]
    Girls[7]
    Total[15]
    Committee[5]
      AllBoys[56/3003]
      AllGirls[21/3003]

3. Conditional Probability and Independence

3.1 Conditional Probability Formula

$$P(A|B)=\frac{P(A\cap B)}{P(B)}$$

• Used when the probability of an event depends on another event.

3.2 Independence

Two events $A$ and $B$ are independent if: $$P(A\cap B)=P(A)\times P(B)$$

• Example: Ncell’s call routing uses independence assumptions to distribute calls efficiently across towers.

Worked Example 3: Drawing Balls Without Replacement A bag has 3 red, 2 black, and 5 white balls. Two balls are drawn at random. Find the probability that both are red.

Solution:

1. First draw: $P({\text{Red}}_1)=\frac{3}{10}$.
2. Second draw (without replacement): $P({\text{Red}}_2|{\text{Red}}_1)=\frac{2}{9}$.
3. Combined probability: $$P(\text{Both red})=\frac{3}{10}\times \frac{2}{9}=\frac{6}{90}=\frac{1}{15}\approx 0.0667$$

Visualization:

4. Bayes’ Theorem

Bayes’ theorem updates probabilities based on new evidence: $$P(A|B)=\frac{P(B|A)\times P(A)}{P(B)}$$

• Used in fraud detection (eSewa), medical testing, and spam filtering.

Worked Example 4: Medical Testing (Hypothetical) Suppose a disease affects 1% of the population. A test for the disease is 99% accurate (true positive and true negative rates). If a person tests positive, what is the probability they actually have the disease?

Solution: Let:

• $D$: Disease present ($P(D)=0.01$).
• $\neg D$: Disease absent ($P(\neg D)=0.99$).
• $T^{+}$: Test positive ($P(T^{+}|D)=0.99$, $P(T^{+}|\neg D)=0.01$).

We want $P(D|T^{+})$: $$P(D|T^{+})=\frac{P(T^{+}|D)\times P(D)}{P(T^{+})}$$ where $P(T^{+})=P(T^{+}|D)P(D)+P(T^{+}|\neg D)P(\neg D)$.

Calculate: $$P(T^{+})=(0.99\times 0.01)+(0.01\times 0.99)=0.0099+0.0099=0.0198$$ Thus: P(D|T^+) = \frac{0.99 \times 0.01}{0.0198} \approx 0.5 \text{ (50%)} Surprising result: Even with a highly accurate test, prior probability matters!

Visualization:

mindmap
  root((Bayes' Theorem))
    Prior[P(D) = 1%]
    Likelihood[P(T+|D) = 99%]
    Evidence[P(T+) = 1.98%]
    Posterior[P(D|T+) ≈ 50%]

5. Probability Distributions

5.1 Discrete Probability Distributions

For countable random variables (e.g., number of defects in a batch).

• Expected Value (Mean): $$E(X)=\sum x\cdot P(x)$$
• Variance: $$\text{Var}(X)=E(X^2)-[E(X){]}^2$$

5.2 Continuous Probability Distributions

For unbounded random variables (e.g., height, time).

• Normal Distribution: Symmetric, bell-shaped curve. $$f(x)=\frac{1}{\sigma \sqrt{2\pi }}{e}^{-\frac{(x-\mu {)}^2}{2{\sigma }^2}}$$

Worked Example 5: Expected Value and Variance (Past Exam) A random variable $X$ has the following distribution:

Find $E(X)$ and $\text{Var}(X)$.

Solution:

1. Expected Value: $$E(X)=(0\times 0.22)+(1\times 0.18)+(2\times 0.35)+(3\times 0.15)+(4\times 0.10)$$ $$E(X)=0+0.18+0.70+0.45+0.40=1.73$$
2. Variance:
   • First, compute $E(X^2)$: $$E(X^2)=(0^2\times 0.22)+(1^2\times 0.18)+(2^2\times 0.35)+(3^2\times 0.15)+(4^2\times 0.10)$$ $$E(X^2)=0+0.18+1.40+1.35+1.60=4.53$$
   • Then: $$\text{Var}(X)=E(X^2)-[E(X){]}^2=4.53-(1.73{)}^2=4.53-2.99=1.54$$

Visualization:

6. Normal Distribution (Gaussian Distribution)

• Symmetry: Mean = Median = Mode.
• Empirical Rule (68-95-99.7):
  • 68% of data within $\mu \pm \sigma$.
  • 95% within $\mu \pm 2\sigma$.
  • 99.7% within $\mu \pm 3\sigma$.

Worked Example 6: Bank Staff Income (Past Exam) Daily income of bank staff is normally distributed with mean Rs 1200 and SD Rs 120. Find the probability that an employee earns: (a) Between Rs 1050 and Rs 1350.

Solution:

1. Convert to Z-scores: $$Z=\frac{X-\mu }{\sigma }$$
   • For Rs 1050: $Z=\frac{1050-1200}{120}=-1.25$.
   • For Rs 1350: $Z=\frac{1350-1200}{120}=1.25$.
2. Use standard normal table:
   • $P(Z<1.25)\approx 0.8944$.
   • $P(Z<-1.25)\approx 0.1056$.
3. Probability between: P(1050 < X < 1350) = 0.8944 - 0.1056 = 0.7888 \text{ (78.88%)}

Visualization:

In the Real World

1. eSewa’s Fraud Detection

   • Uses Bayes’ theorem to update fraud probabilities based on transaction patterns.
   • Example: If a user’s spending suddenly spikes, the system recalculates the probability they’re fraudulent using prior fraud rates and new transaction data.

2. Ncell’s Call Routing

   • Relies on conditional probability to distribute calls to the least congested towers.
   • Example: If tower A has a 70% success rate and tower B has 80%, but tower A is less busy, Ncell’s algorithm weighs these probabilities dynamically.

3. Daraz’s Inventory Management

   • Uses probability distributions (e.g., Poisson for demand) to predict stockouts.
   • Worked Example: If historical data shows 30% chance of selling 5+ items on a product, Daraz stocks extra to cover this probability.

4. NEPSE’s Stock Volatility

   • Analyzes normal distributions of stock returns to set risk thresholds.
   • Example: If a stock’s return is normally distributed with μ=5% and σ=2%, NEPSE calculates the probability of a 10% drop (Z = -2.5, P ≈ 0.6%).

Exam Tips

1. Memorize Key Formulas:

   • Addition rule, multiplication rule, conditional probability, Bayes’ theorem.
   • Normal distribution Z-scores (e.g., $P(Z<1.96)\approx 0.975$).

2. Diagrams > Words:

   • Always draw Venn diagrams for mutually exclusive/independent events.
   • For normal distributions, shade the area under the curve for probabilities.

3. Worked Examples > Theory:

   • Past exams love bag-and-ball, committee formation, and normal distribution problems.
   • Show all steps (e.g., Z-score calculations, binomial probabilities).

4. Real-World Tie-Ins:

   • Link answers to business scenarios (e.g., "Like eSewa’s fraud detection, Bayes’ theorem helps update probabilities with new evidence").
   • Use Ncell/Daraz examples to explain conditional probability.

5. Common Pitfalls:

   • Misapplying independence: Assume events are independent only if explicitly stated.
   • Ignoring replacement: Without replacement changes probabilities (e.g., drawing two red balls from a bag).
   • Normal approximation errors: For discrete data, use continuity correction (e.g., $P(X<5)\approx P(Z<4.5)$).

6. Time Management:

   • Spend 10–12 minutes per question in the exam.
   • For probability distributions, compute E(X) and Var(X) systematically (table method helps).

Final Note: Probability is not just about numbers—it’s about making sense of uncertainty. Master the rules, practice calculations, and always connect theory to real-world tools like eSewa or Ncell. Good luck!

In the real world

• eSewa’s Fraud Detection: Uses Bayes’ Theorem to update the probability of a transaction being fraudulent based on new evidence (e.g., unusual location, high amount). If prior fraud rate is 0.5% and the test flags 95% of frauds but also 1% of legitimate transactions, Bayes’ theorem calculates the actual risk after a flag.
• Ncell’s Call Routing: Applies conditional probability to route calls to the nearest available tower. If Tower A handles 60% of calls and Tower B 40%, but Tower A fails 10% of the time, the system recalculates routing probabilities dynamically.
• Nepal Rastra Bank’s Loan Approval: Uses probability distributions (e.g., Poisson for default events) to set risk-based interest rates. A borrower with a 3% annual default probability might pay a 12% interest rate to cover expected losses.