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Learning Resources

Probability & Grouped Data

Interactive curriculum lessons, worked examples, and geometric problem-solving techniques designed to help Kenyan students master CBC, KPSEA, KCSE, and IGCSE mathematics.

Form 3 Pathway: N/A

First Principles

Objective

Calculate probabilities of combined events and estimate statistical measures (mean and variance) from grouped frequency distributions.

Concrete Scenario: A teacher in Nakuru records test scores for 30 students grouped into class intervals (e.g., 40–49, 50–59). Since exact individual marks are unknown, we use class midpoints \( m_i \) to estimate the mean mark. In probability, drawing beads from a jar without replacement changes the total count, making combined events dependent.

Statistical & Probability Insights:

  • Probability Tree Diagrams: For sequential combined events, multiply probabilities along branches to find combined outcomes.
  • Addition Rule: \( P(A \cup B) = P(A) + P(B) - P(A \cap B) \) (subtract overlap so it is not double-counted).
  • Grouped Data Midpoint: \( m_i = \frac{\text{lower bound} + \text{upper bound}}{2} \).
  • Grouped Mean: \( \bar{x} = \frac{\sum f_i m_i}{\sum f_i} \).

Probability Tree Diagram Structure

P(A) = 0.6P(A') = 0.4P(B|A)P(B'|A)Event ANot A

Key Formulas

\[ P(A) = \frac{\text{Favourable Outcomes}}{\text{Total Outcomes}} \] — Fundamental probability ratio.
\[ P(A') = 1 - P(A) \] — Complement law: Probability of an event not occurring.
\[ P(A \cup B) = P(A) + P(B) - P(A \cap B) \] — General Addition Rule for combined events.
\[ P(A \cap B) = P(A) \times P(B) \] — Multiplication Rule for independent events.
\[ m_i = \frac{\text{Lower Bound} + \text{Upper Bound}}{2} \] — Class midpoint representative value.
\[ \bar{x} = \frac{\sum f_i m_i}{\sum f_i} \] — Estimated mean for grouped frequency data.
\[ \sigma^2 = \frac{\sum f_i (m_i - \bar{x})^2}{\sum f_i} = \frac{\sum f_i m_i^2}{\sum f_i} - \bar{x}^2 \] — Estimated variance for grouped data.

Worked Examples

Example 1 (Easy) — Addition Rule for Probability:

Problem: A student has a probability of 0.6 of passing Maths and 0.5 of passing English. The probability of passing both is 0.3. Find the probability of passing at least one subject.

  1. Apply General Addition Rule: \[ P(M \cup E) = P(M) + P(E) - P(M \cap E) \]
  2. Substitute values: \[ P(M \cup E) = 0.6 + 0.5 - 0.3 = 0.8 \]
Example 2 (Medium) — Dependent Probability Without Replacement:

Problem: A bag contains 5 red and 5 blue marbles. Two marbles are drawn successively without replacement. What is the probability that both are red?

  1. Probability of 1st marble red: \( P(R_1) = \frac{5}{10} \).
  2. Probability of 2nd marble red (given 1st was red): \( P(R_2|R_1) = \frac{4}{9} \).
  3. Multiply probabilities: \[ P(R_1 \cap R_2) = \frac{5}{10} \times \frac{4}{9} = \frac{20}{90} = \frac{2}{9} \approx 0.222 \]
Example 3 (Hard) — Grouped Mean & Variance:

Problem: Calculate the estimated mean for intervals: 10–14 (f=5), 15–19 (f=8), 20–24 (f=7).

  1. Find midpoints \( m_i \): 12, 17, and 22.
  2. Calculate \( f_i m_i \): \( 5(12)=60 \), \( 8(17)=136 \), \( 7(22)=154 \).
  3. Sum \( f_i m_i \): \( 60 + 136 + 154 = 350 \).
  4. Sum frequencies: \( 5 + 8 + 7 = 20 \).
  5. Estimated mean: \[ \bar{x} = \frac{350}{20} = 17.5 \]

Common Mistakes

Mistake Using class boundary limits instead of midpoints when calculating the estimated mean of grouped data.
Correction Always calculate the central representative value \( m_i = \frac{\text{lower} + \text{upper}}{2} \).
Why it feels right Picking class upper limits directly from the table skips the midpoint step.
Mistake Forgetting to subtract \( P(A \cap B) \) when finding \( P(A \cup B) \) for non-mutually exclusive events.
Correction \( P(A \cup B) = P(A) + P(B) - P(A \cap B) \). Without subtraction, double-counting occurs.
Why it feels right Assuming simple addition applies to all probability combinations.

Real World

Quality Control in Manufacturing: Testing combined defect probabilities across sequential production lines in Nairobi industrial parks.
Census & Demographic Analysis: Estimating average family household incomes from grouped census interval reports.
Insurance Risk Assessment: Computing probability distributions for driver accident claims across age group intervals.

Practice

A Form 3 student has a probability of 0.6 of passing Mathematics and 0.5 of passing English. The probability of passing both subjects is 0.3. What is the probability that the student passes at least one subject? (Type only the number, e.g., 0.8)
Review the concepts above.
A bag contains 5 red, 3 blue, and 2 green marbles. Two marbles are drawn successively without replacement. What is the probability that both marbles drawn are red? (Type only the fraction e.g. 2/9)
Review the concepts above.
A single card is drawn at random from a standard deck of 52 cards. What is the probability that the card is either a heart or a king? (Type only the fraction e.g. 4/13)
Review the concepts above.
The test scores of 30 Form 3 students are grouped as: 40-49 (5), 50-59 (8), 60-69 (10), 70-79 (4), 80-89 (2), 90-100 (1). Estimate the mean score using class midpoints rounded to 1 decimal place. (Type only the number, e.g., 62.2)
Review the concepts above.
The grouped data shows books read: 0-9 (4), 10-19 (6), 20-29 (10), 30-39 (8), 40-49 (2). What is the estimated median position class midpoint? (Type only the number, e.g., 25)
Review the concepts above.
Grouped age data for 20 pupils: 0-9 (4), 10-19 (6), 20-29 (5), 30-39 (5). Estimate the variance of ages using midpoints (rounded to 1 decimal place). (Type only the number, e.g., 114.8)
Review the concepts above.