Probability (Grade 11 NSC Matric Mathematics): Revision Notes

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Contingency Tables

What are contingency tables?

A contingency table is a useful tool for organizing and recording data in probability problems. These tables are particularly helpful when you need to determine whether two events are dependent or independent.

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Contingency tables show the frequency counts or percentages for different combinations of outcomes. They make it easier to calculate probabilities and analyze relationships between different variables in your data.

Understanding two-way contingency tables

A two-way contingency table displays data for two different events and their complements, creating four possible combinations in total. This type of table shows:

  • The counts for each possible combination of events
  • The totals for each individual event and its complement
  • The overall total for all observations

Two-way contingency tables are especially useful for computing probabilities of various events and determining whether events are dependent or independent.

Table

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The table structure includes rows and columns representing the different categories, with marginal totals showing the sum of each row and column.

How to work with contingency tables

When working with contingency tables, always follow this systematic approach:

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Step 1: Complete the contingency table

The most important first step is to fill in all missing values in the table. Since each column and row must sum to its respective total, you can calculate missing values using simple arithmetic.

Table

Use the fact that:

  • Each row total equals the sum of all entries in that row
  • Each column total equals the sum of all entries in that column
  • The grand total can be found by summing either all row totals or all column totals
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Step 2: Identify what probability you need to calculate

Read the question carefully to determine exactly which probability is being asked for. Pay attention to conditional statements like "given that" or "if we know that."

Calculating probabilities from contingency tables

Conditional probabilities

When calculating conditional probabilities, use the appropriate subset as your denominator:

  • P(positive if female) = number of females who tested positivetotal number of females\frac{\text{number of females who tested positive}}{\text{total number of females}}

  • P(negative if male) = number of males who tested negativetotal number of males\frac{\text{number of males who tested negative}}{\text{total number of males}}

Joint probabilities

For joint probabilities, use the total sample size as your denominator:

  • P(female and positive) = number of females who tested positivetotal number of participants\frac{\text{number of females who tested positive}}{\text{total number of participants}}

Testing for independence

Two events A and B are independent if and only if:

P(A and B)=P(A)×P(B)P(A \text{ and } B) = P(A) \times P(B)

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Testing for Independence Procedure:

  1. Calculate P(A), P(B), and P(A and B) from your contingency table
  2. Multiply P(A) × P(B)
  3. Compare this product with P(A and B)
  4. If they are equal, the events are independent
  5. If they are not equal, the events are dependent

Worked example 1: Medical trial data

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Worked Example: Medical Trial Analysis

Question: A medical trial tested a new medication on 210 participants (120 females and 90 males). Out of these, 50 females and 30 males responded positively.

Table

Solution:

  1. Complete the table:

    • Negative females: 120 - 50 = 70
    • Negative males: 90 - 30 = 60
    • Total positive: 50 + 30 = 80
    • Total negative: 70 + 60 = 130

  2. Calculate probabilities:

    P(positive if female)=50120=512P(\text{positive if female}) = \frac{50}{120} = \frac{5}{12}

    P(negative if male)=6090=23P(\text{negative if male}) = \frac{60}{90} = \frac{2}{3}

  3. Test for independence:

    P(female)=120210=47P(\text{female}) = \frac{120}{210} = \frac{4}{7}

    P(positive)=80210=821P(\text{positive}) = \frac{80}{210} = \frac{8}{21}

    P(female and positive)=50210=521P(\text{female and positive}) = \frac{50}{210} = \frac{5}{21}

    Since 47×821=32147≠521\frac{4}{7} \times \frac{8}{21} = \frac{32}{147} \neq \frac{5}{21}, the events are dependent.

Worked example 2: Cellphone ownership by grade

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Worked Example: Cellphone Ownership Survey

Question: A survey of 118 high school students examined cellphone ownership by grade level.

Table

Solution:

  1. P(Grade 11 learner has cellphone) = 5965\frac{59}{65}

  2. P(learner without cellphone is from Grade 11) = 69=23\frac{6}{9} = \frac{2}{3}

  3. Test for independence:

    P(Grade 11)=65118P(\text{Grade 11}) = \frac{65}{118}

    P(has cellphone)=109118P(\text{has cellphone}) = \frac{109}{118}

    P(Grade 11 and has cellphone)=59118=12P(\text{Grade 11 and has cellphone}) = \frac{59}{118} = \frac{1}{2}

    Since 12≠65118×109118\frac{1}{2} \neq \frac{65}{118} \times \frac{109}{118}, grade level and cellphone ownership are dependent.

Worked example 3: Hair and eye colour

Table

This table shows the relationship between hair colour and eye colour in a sample of 230 people. You can use this data to:

  • Calculate conditional probabilities (e.g., probability of brown eyes given black hair)
  • Test whether hair colour and eye colour are independent characteristics
  • Find joint probabilities for specific combinations
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Key Points to Remember:

  • Always complete the contingency table first before attempting any probability calculations

  • Pay careful attention to the wording of probability questions - conditional probabilities have different denominators than joint probabilities

  • For independence testing, calculate P(A) × P(B) and compare it with P(A and B) - if they're equal, the events are independent

  • Use appropriate denominators - for conditional probabilities, use the given condition as your sample space; for overall probabilities, use the total sample size

  • Check your arithmetic - row totals and column totals should always add up correctly in your completed table

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