Check Independence by Comparing a Joint Probability with a Product
In The Multiplication Rule for Independent Events, you learned that independent events satisfy \(P(A\cap B)=P(A)P(B)\). A two-way table lets us check that relationship using counts: calculate the observed proportion in the “both” cell, then compare it with the product of the two overall event proportions.
This is a check of whether the events are independent for the group described by the table. If the joint proportion equals the product of the marginal proportions, the events satisfy the product rule in that group. If the values differ, they do not satisfy it exactly in that group. When a table comes from a sample, its proportions describe the sampled individuals; a small difference does not by itself establish that the events are dependent in a larger population.
The joint probability \(P(A\cap B)\) is the proportion of the full group that belongs to both events. The marginal probabilities \(P(A)\) and \(P(B)\) are the proportions of the full group that belong to each event separately. The row and column totals are called marginal totals because they appear at the edges of the table.
A useful way to organize the comparison is to first find the joint proportion in the table, then find the product of the marginal proportions. Keep the probabilities unrounded until the comparison is complete. If the counts allow it, also compare the observed number in the “both” cell with the number predicted by the product.
The comparison has a simple interpretation. Equality means that knowing whether an individual belongs to one event does not change the probability of belonging to the other, for the distribution represented by the table. Inequality means the observed proportions do not meet the product-rule definition of independence. The direction and size of the difference can help describe the pattern, but the basic check is whether the two values are equal.
A Reliable Table-Checking Process
Before calculating, identify exactly what the events mean and locate the table entries that represent them. Then use the full table total consistently. Mixing a row total, a column total, and the grand total as denominators can make a calculation look plausible while answering a different probability question.
State what \(A\) and \(B\) mean, and locate the cell where both occur.
Divide the “both” count by the grand total to calculate \(P(A\cap B)\).
Divide each event’s marginal total by the grand total, then multiply the two probabilities.
Say whether the values are equal, and describe the conclusion for the group represented by the table.
As a check on the arithmetic, the product can also be calculated by multiplying the two marginal counts and dividing by the square of the grand total. Or, multiply the product probability by the grand total to find the count that would match it. These are equivalent ways to check the same comparison.
Worked Example: Club Membership and Bringing Lunch
Worked Example: Club Membership and Bringing Lunch
A school records whether 200 students belong to an after-school club and whether they bring lunch from home. Let \(A\) mean “the student belongs to the club,” and \(B\) mean “the student brings lunch from home.” The table summarizes the students.
| Brings lunch | Does not bring lunch | Total | |
|---|---|---|---|
| Club member | 30 | 30 | 60 |
| Not a club member | 70 | 70 | 140 |
| Total | 100 | 100 | 200 |
State. We will check whether \(A\) and \(B\) satisfy \(P(A\cap B)=P(A)P(B)\) among these 200 students.
Plan. The joint probability comes from the club-member-and-brings-lunch cell. The two marginal probabilities come from the club-member row total and brings-lunch column total. All three use the grand total of 200.
Do. The observed joint probability is \(30/200=0.15\). The marginal probabilities are \(P(A)=60/200=0.30\) and \(P(B)=100/200=0.50\). Their product is \((0.30)(0.50)=0.15\). As a count check, the product predicts \(200(0.15)=30\) students in both events, matching the table’s 30. The same product can be checked from the counts: \((60)(100)/(200^2)=6000/40000=0.15\).
Conclude. The observed joint probability equals the product of the marginal probabilities. Therefore, club membership and bringing lunch satisfy the product rule for independence in this group of 200 students.
Worked Example: Garden Volunteers and Bus Travel
Worked Example: Garden Volunteers and Bus Travel
A community center records whether 100 participants volunteer in a garden and whether they travel to the center by bus. Let \(A\) mean “volunteers in the garden,” and let \(B\) mean “travels by bus.”
| Travels by bus | Does not travel by bus | Total | |
|---|---|---|---|
| Garden volunteer | 20 | 20 | 40 |
| Not a garden volunteer | 10 | 50 | 60 |
| Total | 30 | 70 | 100 |
The “both” count is 20, so the observed joint probability is \(P(A\cap B)=20/100=0.20\). The marginal probabilities are \(P(A)=40/100=0.40\) and \(P(B)=30/100=0.30\). Their product is \((0.40)(0.30)=0.12\). As a second check, multiply the marginal counts and divide by the squared grand total: \((40)(30)/(100^2)=1200/10000=0.12\). The product predicts \(100(0.12)=12\) people in both events, but the table has 20.
Because \(0.20\ne0.12\), garden volunteering and bus travel do not satisfy the product rule for independence in these 100 participants. The observed joint proportion is greater than the product of the marginal proportions. This describes the table’s pattern; it does not, on its own, establish a cause or prove that the same difference holds for all community-center participants.
Worked Example: A Small Difference in a Larger Table
Worked Example: A Small Difference in a Larger Table
A fictional survey records whether 160 people use a neighborhood recreation center and whether they own a bicycle. Let \(A\) mean “uses the recreation center,” and \(B\) mean “owns a bicycle.” The table gives 31 people in both groups, 33 who use the center but do not own a bicycle, 49 who own a bicycle but do not use the center, and 47 in neither group.
| Owns a bicycle | Does not own a bicycle | Total | |
|---|---|---|---|
| Uses recreation center | 31 | 33 | 64 |
| Does not use recreation center | 49 | 47 | 96 |
| Total | 80 | 80 | 160 |
The observed joint probability is \(31/160=0.19375\). The marginal probabilities are \(P(A)=64/160=0.40\) and \(P(B)=80/160=0.50\), so their product is \((0.40)(0.50)=0.20\). As a count check, independence by the product rule would correspond to \(160(0.20)=32\) people in both groups, while the observed count is 31. Calculating with counts gives the same product: \((64)(80)/(160^2)=5120/25600=0.20\).
The values are close, but they are not equal: \(0.19375\ne0.20\). Thus, the events do not satisfy the product rule exactly for these 160 people. Do not round \(0.19375\) to \(0.19\) and \(0.20\) to \(0.2\), then claim equality based on the similar-looking rounded values. The table describes a small difference in this group; if the people are a sample from a broader population, this check alone does not determine whether the population events are independent.
Common Mistakes and AP Exam Tips
- Using the wrong denominator. For this product-rule check, calculate each probability from the full table using the grand total. A within-row percentage is a conditional probability and is not the marginal probability \(P(A)\) or \(P(B)\).
- Comparing the wrong quantities. Compare the both-cell proportion with the product of the two marginal proportions. Do not compare the joint count directly with a probability; convert them to the same scale, or compare the joint count with the product-predicted count.
- Assuming a table must show independence. The table provides the values to check. Do not multiply the margins and report that product as the observed joint probability unless the comparison supports it.
- Rounding too early. Keep exact fractions or unrounded decimals through the comparison. Two values that look equal after rounding may differ before rounding.
- Overstating what a sample shows. If the table summarizes a sample, say whether the events satisfy the product rule in the sample. A sample comparison is not, by itself, proof about a larger population.
- Leaving the conclusion unstated. A calculation alone is incomplete. State whether the values are equal and identify the events and group in context.
Key Takeaway
A two-way table supplies the joint and marginal proportions needed to check independence with the product rule. Calculate \(P(A\cap B)\) from the both-cell count and compare it with \(P(A)P(B)\), using the same grand total throughout. Equality means the events satisfy the rule for the group represented; inequality means they do not satisfy it exactly in that group. Keep conclusions about a sample distinct from claims about a broader population.
Check Your Understanding
Use the grand total for each probability. Show the joint probability, the product of the marginal probabilities, and a conclusion in context.
- A table has a grand total of 120. Event \(A\) has a marginal total of 48, event \(B\) has a marginal total of 60, and the both-event cell has count 24. Do the events satisfy the product rule in this group?
- In a group of 200 people, 80 belong to event \(A\), 50 belong to event \(B\), and 30 belong to both. Find the two quantities to compare and state whether the events satisfy the product rule.
- A table has 150 individuals, with 45 in event \(A\), 60 in event \(B\), and 18 in both. How many in both would match the product rule? Compare that count with the observed count.
- A student calculates \(P(A\cap B)\) using a row total as the denominator, then compares it with \(P(A)P(B)\) calculated from the grand total. Explain why this is not a valid product-rule comparison.
- A sample table has a joint proportion close to, but not equal to, the product of its marginal proportions. What conclusion can be made about the sample, and what conclusion cannot be established from this comparison alone?