Probability I
Interactive curriculum lessons, worked examples, and geometric problem-solving techniques designed to help Kenyan students master CBC, KPSEA, KCSE, and IGCSE mathematics.
First Principles
Objective: Master how to represent sample spaces, compute probabilities of combined events, and use Venn diagrams and addition rules to solve everyday probabilistic problems.
The Core Idea: Imagine surveying a class of 30 learners in Nakuru about whether they play Football (\(F\)), Volleyball (\(V\)), or both. Each learner is an elementary outcome in your sample space \(S\). A Venn diagram organizes these learners into distinct, non-overlapping regions so we never double-count anyone.
1. Deconstructing the Venn Diagram
When two events \(A\) and \(B\) occur in a sample space \(S\), the universe is partitioned into 4 mutually exclusive regions:
- Only \(A\) \((A \cap B^c)\): Outcomes in \(A\) that are not in \(B\).
- Only \(B\) \((B \cap A^c)\): Outcomes in \(B\) that are not in \(A\).
- Intersection \((A \cap B)\): Outcomes belonging to both \(A\) AND \(B\).
- Neither \((A \cup B)^c\): Outcomes outside both \(A\) and \(B\).
The General Addition Rule:
\[ P(A \cup B) = P(A) + P(B) - P(A \cap B) \]We subtract \(P(A \cap B)\) because adding the full circle \(P(A)\) and full circle \(P(B)\) counts the overlap twice.
Interactive Venn Diagram Explorer
Click the buttons below to distribute 40 students between two school clubs (Drama Club \(D\) and Science Club \(S\)) and observe how probabilities calculate live.
Key Formulas
Key probability definitions, rules of combination, and set relationships:
1. Classical Probability Definition
\[ P(E) = \frac{n(E)}{n(S)} = \frac{\text{Number of favourable outcomes}}{\text{Total number of outcomes in sample space}} \]Where \(0 \le P(E) \le 1\).
2. Complementary Event Rule
\[ P(E^c) = 1 - P(E) \]The probability that event \(E\) does not happen.
3. General Addition Rule (Any two events \(A\) and \(B\))
\[ P(A \cup B) = P(A) + P(B) - P(A \cap B) \]In terms of counts: \[ n(A \cup B) = n(A) + n(B) - n(A \cap B) \]
4. Mutually Exclusive (Disjoint) Events
\[ P(A \cap B) = 0 \implies P(A \cup B) = P(A) + P(B) \]Two events cannot happen simultaneously.
5. Independent Events (Multiplication Rule)
\[ P(A \cap B) = P(A) \times P(B) \]The occurrence of event \(A\) does not affect the probability of event \(B\).
6. Conditional Probability
\[ P(A \mid B) = \frac{P(A \cap B)}{P(B)}, \quad \text{where } P(B) > 0 \]The probability of \(A\) occurring given that \(B\) is known to have already occurred.
Worked Examples
Problem: A 14-seater matatu in Eldoret starts its journey with 4 seats already occupied. If a passenger randomly boards at the next stage, what is the probability that they pick an unoccupied seat?
- Identify the sample space: Total seats \(n(S) = 14\).
- Find favourable outcomes: Number of unoccupied seats \(= 14 - 4 = 10\).
- Apply the formula: \[ P(\text{Empty}) = \frac{10}{14} = \frac{5}{7} \]
Problem: In a cohort of 50 students at a technical institute, 30 take Computer Studies, 25 take Entrepreneurship, and 12 take both subjects. What is the probability that a randomly chosen student takes Computer Studies or Entrepreneurship?
- State the given data: \[ n(S) = 50, \quad n(C) = 30, \quad n(E) = 25, \quad n(C \cap E) = 12 \]
- Compute individual and joint probabilities: \[ P(C) = \frac{30}{50}, \quad P(E) = \frac{25}{50}, \quad P(C \cap E) = \frac{12}{50} \]
- Apply the addition rule: \[ P(C \cup E) = P(C) + P(E) - P(C \cap E) = \frac{30}{50} + \frac{25}{50} - \frac{12}{50} = \frac{43}{50} = 0.86 \]
Problem: A logistics firm in Mombasa tracks two independent delivery vans, Van A and Van B. The probability that Van A arrives on time is \(0.85\), and the probability that Van B arrives on time is \(0.70\). Find the probability that at least one of the two vans arrives on time.
- Identify given probabilities: \(P(A) = 0.85\) and \(P(B) = 0.70\).
- Determine intersection using independence: \[ P(A \cap B) = P(A) \times P(B) = 0.85 \times 0.70 = 0.595 \]
- Apply the addition rule: \[ P(A \cup B) = P(A) + P(B) - P(A \cap B) = 0.85 + 0.70 - 0.595 = 0.955 \]
- Alternative (Complement method check): \[ P(\text{At least one}) = 1 - P(A^c \cap B^c) = 1 - (1 - 0.85)(1 - 0.70) = 1 - (0.15 \times 0.30) = 1 - 0.045 = 0.955 \]
Common Mistakes
Mistake 2: Confusing 'Mutually Exclusive' with 'Independent'.
Correction:
- Mutually Exclusive: Cannot happen at the same time: \(P(A \cap B) = 0\).
- Independent: Occurrence of one does not affect the other: \(P(A \cap B) = P(A) \times P(B)\).
Why it feels right: Both imply that the events are "separated" in some verbal sense, but mutual exclusivity is about set overlap, while independence is about informational causality.
Mistake 3: Putting totals instead of "Only" counts into Venn diagram crescents.
Correction: If 20 students play Football and 8 play both Football and Rugby, the Football-only crescent has \(20 - 8 = 12\) students, NOT 20.
Why it feels right: The problem states "20 play Football", so students write 20 directly in the circle without subtracting the intersection first.
Real World
Probability of combined events governs risk management, technology, and public planning across East Africa:
1. Mobile Money & Server Redundancy
Safaricom's M-Pesa network uses redundant data servers in Nairobi and Eldoret. If Server 1 has a failure rate of \(0.001\) and Server 2 has \(0.002\), independent failure gives a total outage risk of just \(0.001 \times 0.002 = 0.000002\), guaranteeing 99.9998% uptime.
2. Agricultural Co-operatives (SACCOs)
A tea farming co-op in Kericho calculates the joint probability that a member applies for an emergency loan and a fertilizer subsidy to ensure cash liquidity during harvest season.
3. Epidemic Disease Surveillance
Health clinics screen patients for malaria and dengue. When symptoms overlap, doctors use conditional probabilities and Venn diagrams to assess true infection rates and administer targeted treatments.
Practice