Analysing Frequency Outcomes in Probability

The Frequency Approach to Probability

In probability experiments, we often deal with raw counts rather than theoretical ratios. To analyse these effectively, we use frequency distributions. A frequency distribution shows how a total number of trials or individuals is distributed across different possible outcomes. In the TMUA, you are expected to use two-way tables and frequency trees to organise this data. These tools are particularly useful because they allow you to visualize the relationships between different categorical variables without the need for complex algebraic probability formulas.

Two-Way Tables

A two-way table is a grid used to represent frequencies for two categorical variables. One variable is represented in the rows, and the other in the columns. The intersection of a row and a column shows the frequency of individuals that satisfy both criteria.

Construction and Properties

When constructing a two-way table, keep the following properties in mind:

1. Row Totals: The sum of all frequencies in a row must equal the total for that specific category.
2. Column Totals: The sum of all frequencies in a column must equal the total for that specific category.
3. Grand Total: Found in the bottom-right cell, this is the sum of all row totals, all column totals, and every individual cell in the grid. It represents the total population $N$.

Worked Example: Student Activities

Suppose we have a group of 100 students. 60 are in Year 12 and 40 are in Year 13. We also know that 45 students play a musical instrument, and of those, 25 are in Year 12.

To find how many Year 13 students do not play an instrument, we construct a table:

	Instrument	No Instrument	Total
Year 12	$25$	$35$	$60$
Year 13	$20$	$20$	$40$
Total	$45$	$55$	$100$

Step 1: Enter the known values: Year 12 Total ($60$), Year 13 Total ($40$), Instrument Total ($45$), and Year 12 Instrument ($25$). Step 2: Calculate Year 12 No Instrument by $60 - 25 = 35$. Step 3: Calculate Year 13 Instrument by $45 - 25 = 20$. Step 4: Calculate Year 13 No Instrument by $40 - 20 = 20$.

From this table, the probability that a randomly selected student is in Year 13 and does not play an instrument is $\frac{20}{100} = 0.2$.

Frequency Trees

Frequency trees are branching diagrams used to show how a total frequency is split into various categories. Unlike probability trees, which use fractions or decimals on the branches, frequency trees use actual counts at the nodes. This makes them highly intuitive for multi-stage experiments.

Rules for Frequency Trees

1. The Root: The first node represents the total frequency $N$.
2. Partitioning: At each node, the branches must represent mutually exclusive outcomes. This means an individual can only go down one branch.
3. Summation: The sum of the frequencies at the end of the branches must exactly equal the frequency of the node from which they originated.

Worked Example: Diagnostic Testing

In a population of 1000 people, 100 have a specific condition and 900 do not. A test for this condition is $90$ percent accurate for those who have it (positive result) and $80$ percent accurate for those who do not (negative result).

Node 1 (Total): $1000$ Branches from Root:

• Condition: $100$
• No Condition: $900$

Branches from Condition ($100$):

• Test Positive (Correct): $0.90 \times 100 = 90$
• Test Negative (Incorrect): $100 - 90 = 10$

Branches from No Condition ($900$):

• Test Negative (Correct): $0.80 \times 900 = 720$
• Test Positive (Incorrect): $900 - 720 = 180$

By following the branches, we can see that the total number of people who test positive is $90 + 180 = 270$. The probability that a person who tests positive actually has the condition is $\frac{90}{270} = \frac{1}{3}$.

Converting Frequency to Probability

Once a table or tree is complete, the probability of an event $A$ occurring is simply calculated as:

$P(A) = \frac{\text{Frequency of event } A}{\text{Total frequency } N}$

If the question asks for a conditional probability, such as the probability of $A$ given $B$ has already occurred, you change the denominator to the frequency of $B$:

$P(A|B) = \frac{\text{Frequency of } A \text{ and } B}{\text{Frequency of } B}$