Probability Theory: Rules, Events & Real-World Applications

TAKEAWAYS:

• Probability quantifies uncertainty using sample spaces and events, with rules to combine probabilities (addition for OR, multiplication for AND).
• The addition rule $P(A\cup B)=P(A)+P(B)-P(A\cap B)$ avoids double-counting overlapping events.
• The multiplication rule $P(A\cap B)=P(A)\cdot P(B|A)$ accounts for conditional dependencies (e.g., drawing two red balls without replacement).
• Independent events simplify calculations: $P(A\cap B)=P(A)\cdot P(B)$ when $P(B|A)=P(B)$.
• Real-world applications include fraud detection (eSewa’s transaction probability rules), supply chain logistics (Daraz’s order fulfillment probabilities), and risk assessment (bank loan approvals).
• Always visualize probabilities with Venn diagrams for events and tree diagrams for sequential dependencies.

1. Core Concepts: Sample Spaces and Events

Probability is the mathematical study of randomness, quantifying how likely an event is to occur. Every probability problem starts with two key ideas:

Sample Space (S)

• The set of all possible outcomes of an experiment.
• Example: Rolling a die → $S=\{1,2,3,4,5,6\}$.
• Example: Drawing a ball from a bag → $S=\{{\text{Red}}_1,{\text{Red}}_2,{\text{Black}}_1,{\text{White}}_1,\dots \}$.

Event (E)

• A subset of the sample space representing one or more outcomes of interest.
• Example: "Drawing a red ball" → $E=\{{\text{Red}}_1,{\text{Red}}_2,{\text{Red}}_3\}$.
• Example: "Rolling an even number" → $E=\{2,4,6\}$.

Probability of an Event

For a finite sample space with equally likely outcomes: $$P(E)=\frac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}}=\frac{n(E)}{n(S)}$$ Example 1 (Past Exam): A bag contains 20 balls numbered 1–20. Find the probability of drawing a ball that is:

1. A multiple of 3 or 7.
2. A multiple of 3 or 4.

Solution:

• Sample space $S$: $\{1,2,\dots ,20\}$.
• Multiples of 3: $\{3,6,9,12,15,18\}$ → $n(E_1)=6$.
• Multiples of 7: $\{7,14\}$ → $n(E_2)=2$.
• Multiples of both (3 and 7): $\{21\}$ → But 21 ∉ $S$, so $n(E_1\cap E_2)=0$.

Part (a): Multiples of 3 or 7 Use the addition rule: $$P(E_1\cup E_2)=P(E_1)+P(E_2)-P(E_1\cap E_2)=\frac{6}{20}+\frac{2}{20}-0=\frac{8}{20}=0.4$$

Part (b): Multiples of 3 or 4

• Multiples of 4: $\{4,8,12,16,20\}$ → $n(E_3)=5$.
• Multiples of both (3 and 4): $\{12\}$ → $n(E_1\cap E_3)=1$. $$P(E_1\cup E_3)=\frac{6}{20}+\frac{5}{20}-\frac{1}{20}=\frac{10}{20}=0.5$$

2. Addition Rule: Probability of "OR" Events

The addition rule combines probabilities of two events occurring together or separately: $$P(A\cup B)=P(A)+P(B)-P(A\cap B)$$ Why subtract $P(A\cap B)$? Without subtraction, the intersection $A\cap B$ is counted twice (once in $P(A)$ and once in $P(B)$).

Example 2 (Past Exam): Committee Formation

A committee of 5 is formed from 8 boys and 7 girls. Find the probability that:

1. All members are boys.
2. All members are girls.

Solution:

• Total ways to form a committee of 5 from 15 people: $$\left(\genfrac{}{}{0ex}{}{15}{5}\right)=\frac{15!}{5!\cdot 10!}=3003$$

1. All boys:

   • Favorable outcomes: $\left(\genfrac{}{}{0ex}{}{8}{5}\right)=56$.
   • Probability: $P(\text{All boys})=\frac{56}{3003}\approx 0.0186$.

2. All girls:

   • Favorable outcomes: $\left(\genfrac{}{}{0ex}{}{7}{5}\right)=21$.
   • Probability: $P(\text{All girls})=\frac{21}{3003}\approx 0.0070$.

Real-World Tie-In: This mirrors Nepal’s House of Representatives elections, where seats are allocated based on probabilistic combinations of candidates from different constituencies.

3. Multiplication Rule: Probability of "AND" Events

The multiplication rule calculates the probability of two events occurring together (sequentially or simultaneously): $$P(A\cap B)=P(A)\cdot P(B|A)$$

• $P(B|A)$: Conditional probability of $B$ given that $A$ has already occurred.

Key Cases:

1. With Replacement:
   • Probabilities remain unchanged (e.g., drawing two balls with replacement).
2. Without Replacement:
   • Probabilities change after the first draw (e.g., drawing two balls without replacement).

Example 3 (Past Exam): Drawing Balls Without Replacement

A bag contains 3 red, 2 black, and 5 white balls. Two balls are drawn at random. Find the probability both are red.

Solution:

• Total balls initially: $3+2+5=10$.
• First draw (Red): $P({\text{Red}}_1)=\frac{3}{10}$.
• Second draw (Red, without replacement): $P({\text{Red}}_2|{\text{Red}}_1)=\frac{2}{9}$.
• Combined probability: $$P({\text{Red}}_1\cap {\text{Red}}_2)=\frac{3}{10}\times \frac{2}{9}=\frac{6}{90}=\frac{1}{15}\approx 0.0667$$

Real-World Tie-In: This models eSewa’s transaction fraud detection. If a user attempts two transactions in quick succession with the same device, the probability of both being legitimate (or both being fraudulent) is calculated using conditional probabilities.

4. Independent vs. Dependent Events

Independent Events

Two events $A$ and $B$ are independent if: $$P(B|A)=P(B)\text{or}P(A\cap B)=P(A)\cdot P(B)$$ Example: Rolling a die twice. The outcome of the first roll does not affect the second.

Dependent Events

If $P(B|A)\ne P(B)$, the events are dependent. Example: Drawing two balls without replacement (as in Example 3).

5. Conditional Probability

Conditional probability answers: "What is the probability of $B$ given that $A$ has already occurred?" $$P(B|A)=\frac{P(A\cap B)}{P(A)}$$ Example 4 (Real-World): Daraz Order Fulfillment Daraz uses conditional probability to estimate the chance that an order will be delayed given that the warehouse is out of stock.

• Suppose:
  • $P(\text{Delay})=0.10$ (10% of orders are delayed).
  • $P(\text{Out of Stock})=0.05$.
  • $P(\text{Delay}\cap \text{Out of Stock})=0.04$.
• Probability of delay given out of stock: $$P(\text{Delay}|\text{Out of Stock})=\frac{0.04}{0.05}=0.8$$ (80% of out-of-stock orders are delayed.)

6. Probability Trees and Real-World Applications

Probability trees visualize sequential events and their probabilities.

Example 5: Ncell Network Connectivity

Ncell tracks the probability that a user’s call drops based on network congestion.

• First call:
  • $P(\text{Success})=0.95$.
  • $P(\text{Failure})=0.05$.
• Second call (if first fails):
  • $P(\text{Success}|\text{First Failure})=0.80$.
• Probability both calls fail: $$P({\text{Failure}}_1\cap {\text{Failure}}_2)=0.05\times 0.20=0.01$$

flowchart TD
    A["Start"] --> B["First Call: Success (0.95)"]
    A --> C["First Call: Failure (0.05)"]
    B --> D["End"]
    C --> E["Second Call: Success (0.80)"]
    C --> F["Second Call: Failure (0.20)"]
    E --> D
    F --> G["Both Fail (0.01)"]

In the Real World

1. eSewa Transaction Fraud Detection

   • Idea Used: Conditional probability and multiplication rule.
   • How? eSewa calculates the probability that a transaction is fraudulent given unusual activity (e.g., multiple transactions from the same device in 5 minutes).
   • Example: If $P(\text{Fraud})=0.01$ and $P(\text{Multiple Transactions}|\text{Fraud})=0.70$, then: $$P(\text{Fraud}\cap \text{Multiple Transactions})=0.01\times 0.70=0.007$$

2. Daraz Order Fulfillment Probabilities

   • Idea Used: Addition and multiplication rules for independent/dependent events.
   • How? Daraz estimates the probability that an order is delivered on time by combining:
     • Probability the warehouse has stock ($P(\text{Stock})=0.98$).
     • Probability the delivery partner is on time ($P(\text{On Time})=0.90$).
     • Combined: $P(\text{On Time}\cap \text{Stock})=0.98\times 0.90=0.882$.

3. Nepal Rastra Bank Loan Approvals

   • Idea Used: Conditional probability for risk assessment.
   • How? Banks calculate the probability that a loan will default given the applicant’s credit score.
   • Example: If $P(\text{Default}|\text{Poor Score})=0.30$ and $P(\text{Poor Score})=0.20$, then: $$P(\text{Default}\cap \text{Poor Score})=P(\text{Default}|\text{Poor Score})\times P(\text{Poor Score})=0.30\times 0.20=0.06$$

Exam Tip

1. Always define your sample space clearly. Examiners deduct marks for ambiguous $S$.
2. Use Venn diagrams for "OR" problems and tree diagrams for sequential events.
3. Watch for "with/without replacement." Forgetting this leads to incorrect conditional probabilities.
4. Memorize the two forms of the multiplication rule:
   • $P(A\cap B)=P(A)\cdot P(B|A)$.
   • For independent events: $P(A\cap B)=P(A)\cdot P(B)$.
5. Real-world questions often involve conditional probability. Look for keywords like:
   • "Given that..."
   • "Assuming..."
   • "After the first event..."
6. Practice combining rules. Many questions mix addition and multiplication (e.g., "probability of A or B, given C").

Worked Example Summary Table: