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
By the end of this lesson, you will be able to calculate the theoretical probability of simple events, find the probability of complementary events, and determine probabilities of combined mutually exclusive and independent events using fundamental ratios.
Step 1 — Concrete Scenario
Imagine a farmer in Nakuru inspecting a certified seed packet containing \(12\) red-treated maize seeds, \(5\) white seeds, and \(3\) yellow seeds. If she picks a seed blindly from the bag without looking, what is the chance she selects a red seed? Intuitively, we compare the count of the desired colour to the total number of seeds available in the bag.
Step 2 — Sample Space and Probability Measure
The set of all possible outcomes of a random experiment is called the sample space, denoted by \(S\). An event \(E\) is any subset of outcomes in \(S\). If every single outcome in \(S\) is equally likely to occur, probability is defined geometrically as the fraction of the total space occupied by the event:
\[ P(E) = \frac{\text{Number of favourable outcomes in } E}{\text{Total number of possible outcomes in } S} = \frac{n(E)}{n(S)} \]Because the event \(E\) cannot contain fewer than \(0\) outcomes or more than the full set \(S\), probability is strictly bounded:
\[ 0 \le P(E) \le 1 \]Step 3 — Complement and Disjoint Events
Complementary Events: If \(E\) is an event, \(E^c\) (or \(E'\)) is the event that \(E\) does not occur. Since the total probability of all outcomes is \(1\):
\[ P(E^c) = 1 - P(E) \]Mutually Exclusive (Disjoint) Events: If two events \(A\) and \(B\) cannot happen at the same time (i.e. \(A \cap B = \emptyset\)), the probability that either \(A\) or \(B\) occurs is:
\[ P(A \cup B) = P(A) + P(B) \]Core Analogy
Think of the sample space as a full \(1\,\text{litre}\) jug of fresh milk. Each possible event is poured into a separate cup. You cannot pour out more than \(1\,\text{litre}\) in total (\(P \le 1\)), nor can any cup have negative milk (\(P \ge 0\)). The fraction of the jug in a cup represents its probability.
Interactive Seed Bag Experiment
Adjust the counts of Red, White, and Green seeds in the bag. Draw seeds at random to compare the experimental frequency against the theoretical probability \(\frac{n(E)}{n(S)}\).
Red Draws: 0 (0.0%) | Theoretical: 50.0%
White Draws: 0 (0.0%) | Theoretical: 30.0%
Green Draws: 0 (0.0%) | Theoretical: 20.0%
Key Formulas
1. Classical Definition of Probability
\[ P(E) = \frac{n(E)}{n(S)} = \frac{\text{Number of favourable outcomes}}{\text{Total number of equally likely outcomes}} \]2. Axiom of Probability Range
\[ 0 \le P(E) \le 1 \]For an impossible event \(\emptyset\), \(P(\emptyset) = 0\). For a certain event \(S\), \(P(S) = 1\).
3. Complementary Rule
\[ P(E^c) = 1 - P(E) \iff P(E) + P(E^c) = 1 \]4. Addition Rule for Mutually Exclusive Events
\[ P(A \cup B) = P(A) + P(B) \quad (\text{when } A \cap B = \emptyset) \]5. Multiplication Rule for Independent Events
\[ P(A \cap B) = P(A) \times P(B) \]For dependent events with conditional probability: \(P(A \cap B) = P(A) \times P(B \mid A)\).
Worked Examples
Problem: A basket at an open-air market in Machakos contains \(6\) mangoes, \(4\) passion fruits, and \(10\) avocados. If a customer picks one fruit at random, what is the probability that it is a mango?
- Find the total outcomes \(n(S)\): \(n(S) = 6 + 4 + 10 = 20\).
- Identify favourable outcomes \(n(M)\): \(n(M) = 6\).
- Calculate \(P(M)\): \[ P(M) = \frac{n(M)}{n(S)} = \frac{6}{20} = \frac{3}{10} = 0.30 \]
Answer: \(0.30\) (or \(\frac{3}{10}\)).
Problem: In a high school athletics meet in Eldoret, the probability that athlete Kiprono wins the \(800\,\text{m}\) race is \(0.45\), and the probability that athlete Cheruiyot wins is \(0.35\). Only one runner can win the race. What is the probability that neither Kiprono nor Cheruiyot wins?
- Identify the events: Let \(K\) be the event Kiprono wins and \(C\) be the event Cheruiyot wins. Since they cannot both win, the events are mutually exclusive.
- Find the combined probability of either winning: \[ P(K \cup C) = P(K) + P(C) = 0.45 + 0.35 = 0.80 \]
- Use the complement rule for neither winning: \[ P(\text{Neither}) = 1 - P(K \cup C) = 1 - 0.80 = 0.20 \]
Answer: \(0.20\).
Problem: A quality assurance officer in Nairobi tests electrical components from two assembly lines, \(A\) and \(B\). Line \(A\) produces \(70\%\) of components, with a defective rate of \(4\%\) (\(0.04\)). Line \(B\) produces \(30\%\) of components, with a defective rate of \(10\%\) (\(0.10\)). What is the overall probability that a randomly chosen component is defective?
- Identify the mutually exclusive pathways to a defective component:
- Pathway 1: Chosen from Line \(A\) and Defective \((A \cap D)\)
- Pathway 2: Chosen from Line \(B\) and Defective \((B \cap D)\)
- Calculate each pathway probability: \[ P(A \cap D) = P(A) \times P(D \mid A) = 0.70 \times 0.04 = 0.028 \] \[ P(B \cap D) = P(B) \times P(D \mid B) = 0.30 \times 0.10 = 0.030 \]
- Sum the mutually exclusive probabilities: \[ P(D) = P(A \cap D) + P(B \cap D) = 0.028 + 0.030 = 0.058 \]
Answer: \(0.058\).
Common Mistakes
Example: A box has \(3\) red pens and \(7\) blue pens. A student writes \(P(\text{red}) = \frac{3}{7} \approx 0.43\).
Why it feels right: Students confuse odds ratios (favourable to unfavourable, \(3:7\)) with probability (favourable to total, \(3:10\)).
Example: Adding percentages incorrectly or writing \(P(A) = \frac{15}{12} = 1.25\).
Example: When finding \(P(\text{roll a 2 OR a 4})\) on a fair 6-sided die, multiplying \(\frac{1}{6} \times \frac{1}{6} = \frac{1}{36}\).
Real World
Practice