Test Independence by Comparing Probabilities
In What Independent Events Mean, you learned that independence means knowing one event occurred does not change the probability of the other. This tutorial turns that meaning into a direct test using probabilities that are given to you. Compare the probability of an event with its conditional probability after learning that another event occurred.
For events \(A\) and \(B\), compare \(P(A)\), the probability of \(A\) in the full chance process, with \(P(A\mid B)\), the probability of \(A\) among outcomes where \(B\) occurred. If these probabilities are equal and \(P(B)>0\), knowing that \(B\) occurred has not changed the probability of \(A\). That is the test for independence in this comparison.
The condition \(P(B)>0\) matters because \(P(A\mid B)\) describes outcomes where \(B\) occurred. If \(B\) has probability zero, there are no such outcomes in the model, so that conditional probability is undefined. For an ordinary comparison, first confirm that the condition event is possible.
This equality is consistent with the product-rule definition of independence from Mutually Exclusive Versus Independent Events. When \(P(B)>0\), the conditional probability formula gives \(P(A\mid B)=P(A\cap B)/P(B)\). Setting that equal to \(P(A)\) means the probability of both events matches the product of their individual probabilities. The comparison is often convenient because it directly expresses whether learning \(B\) changes the chance of \(A\).
A Direct Comparison Routine
Use this routine when a problem provides probabilities or enough information to find them. Keep the events in the same order: if you compare \(P(A\mid B)\) with \(P(A)\), the event whose probability you are tracking is \(A\), and \(B\) is the condition.
State what \(A\) and \(B\) mean in the situation.
Confirm that \(P(B)>0\), so \(P(A\mid B)\) is defined.
Compare \(P(A\mid B)\) with \(P(A)\). If they are equal, the events are independent; if they differ, the events are dependent.
Say whether knowing \(B\) occurred changes the chance of \(A\), and connect the comparison to the stated situation.
The order of the comparison is not a claim that \(B\) causes \(A\). It is a way to ask whether information about \(B\) changes the probability assigned to \(A\). If \(P(A)>0\), you could also make the reverse comparison, \(P(B\mid A)\) with \(P(B)\). Independence gives matching probabilities in either direction when both conditionals are defined. You do not need to calculate both directions if the given values already allow a valid comparison.
Worked Example: Compare Probabilities from a Table
Worked Example: Compare Probabilities from a Table
Suppose a hypothetical group of 200 people is classified by whether each person uses a tablet for reading and whether each person uses night mode. The table gives the counts. Let \(A\) be the event that a randomly selected person uses night mode, and \(B\) the event that the person uses a tablet. Are \(A\) and \(B\) independent?
| Uses night mode | Does not use night mode | Total | |
|---|---|---|---|
| Uses a tablet | 20 | 30 | 50 |
| Does not use a tablet | 60 | 90 | 150 |
| Total | 80 | 120 | 200 |
Find the unconditional probability: The probability of night-mode use in the full group is the total using night mode divided by the grand total.
Find the conditional probability: Given that a person uses a tablet, restrict attention to the tablet row. Of those 50 people, 20 use night mode.
The condition is possible because \(P(B)=50/200=0.25>0\). The two probabilities match: \(P(A\mid B)=P(A)=0.40\). In this chance model, knowing that a randomly selected person uses a tablet does not change the probability that the person uses night mode. Therefore, \(A\) and \(B\) are independent.
As a check using the earlier product-rule definition, \(P(A\cap B)=20/200=0.10\), while \(P(A)P(B)=(0.40)(0.25)=0.10\). The same result supports the conclusion.
Conclusion: In this hypothetical group, tablet use and night-mode use are independent according to the table. The proportion using night mode among tablet users equals the proportion using night mode in the full group.
Worked Example: A Conditional Probability That Is Larger
Worked Example: A Conditional Probability That Is Larger
In a hypothetical model of commute trips, let \(A\) mean a randomly selected trip is late and \(B\) mean the trip occurs on a rainy day. Suppose \(P(A)=0.20\), \(P(B)=0.30\), and \(P(A\cap B)=0.09\). Are \(A\) and \(B\) independent?
Check the condition: \(P(B)=0.30>0\), so the conditional probability of a late trip given a rainy day is defined.
Calculate the conditional probability: Use the conditional probability formula, dividing the probability of both events by the probability of the condition.
The unconditional probability of a late trip is \(P(A)=0.20\), while the conditional probability of a late trip on a rainy day is \(P(A\mid B)=0.30\). They differ, so the events are dependent in this model. Knowing a trip occurred on a rainy day raises the probability that it is late from 0.20 to 0.30, a difference of 0.10, or 10 percentage points.
Check the arithmetic another way: \(0.30\times 0.30=0.09\), which recovers the stated intersection probability from \(P(A\mid B)P(B)\). For independence, the product of the unconditional probabilities would instead be \((0.20)(0.30)=0.06\). This does not match the given intersection probability of 0.09, consistent with the conditional comparison.
Conclusion: The events “a commute trip is late” and “a commute trip occurs on a rainy day” are dependent in this model. Rainy-day trips have a higher probability of being late than trips considered across all days.
Worked Example: Equality Given Directly
Worked Example: Equality Given Directly
A hypothetical online learning model describes \(A\) as the event that a randomly selected learner completes a practice set and \(B\) as the event that the learner receives a reminder. The model gives \(P(A)=0.55\), \(P(B)=0.40\), and \(P(A\mid B)=0.55\). Are \(A\) and \(B\) independent?
The condition has positive probability, since \(P(B)=0.40>0\). Now compare the two stated probabilities:
The values are equal, so knowing that a learner received a reminder does not change the model’s probability that the learner completes the practice set. Therefore, \(A\) and \(B\) are independent in this model.
A product-rule check gives the same conclusion. The conditional probability formula implies \(P(A\cap B)=P(A\mid B)P(B)=(0.55)(0.40)=0.22\). The product of the unconditional probabilities is \(P(A)P(B)=(0.55)(0.40)=0.22\). Both calculations give 0.22.
Conclusion: In the stated model, receiving a reminder and completing the practice set are independent events. The conditional probability of completion among learners who received a reminder equals the unconditional probability of completion.
Equal Values, Rounded Values, and the Direction of Comparison
In a probability model, independence is based on equality, not on one probability being “close” to the other. If the stated values are exact and \(P(B)>0\), equality establishes independence, while a difference establishes dependence. For example, \(0.40\) and \(0.40\) match exactly, whereas \(0.30\) and \(0.20\) do not.
Sometimes probabilities are reported to a limited number of decimal places. Two displayed values may look equal because of rounding even if their underlying values differ slightly. If a problem explicitly gives rounded estimates, describe what the reported values indicate at that precision rather than claiming that the underlying probabilities are exactly equal. Follow any directions in the problem about treating displayed values as exact.
Be equally careful with the condition. To test whether \(A\) and \(B\) are independent by this comparison, you need \(P(A\mid B)\) and \(P(A)\), with \(P(B)>0\). Comparing \(P(A\mid B)\) with \(P(B)\) does not answer the test, because it compares probabilities of different events. Likewise, \(P(A\mid B)\) and \(P(A\mid B^c)\) compare two conditional groups, not the conditional probability with the unconditional probability named in the independence definition.
If a problem supplies \(P(A)\), \(P(B)\), and \(P(A\cap B)\) instead of \(P(A\mid B)\), calculate the conditional value first, as in the rainy-commute example. If it supplies counts in a two-way table, use the relevant event total for the unconditional probability and the condition-group total for the conditional probability, as in the tablet example. This follows the reference-group reasoning from Conditional Probability from a Two-Way Table.
Common Mistakes and AP Exam Tips
- Comparing the wrong quantities. For the test \(P(A\mid B)=P(A)\), compare the chance of \(A\) given \(B\) with the overall chance of \(A\). Do not compare the conditional probability with \(P(B)\).
- Using the grand total in a conditional probability. In a table, \(P(A\mid B)\) uses only the outcomes in \(B\) as its reference group. The denominator is the total for \(B\), not the grand total.
- Forgetting to check \(P(B)>0\). If \(P(B)=0\), \(P(A\mid B)\) is undefined. An AP response should not use an undefined conditional probability as evidence of independence.
- Calling a small difference equality. If exact values differ, the events are dependent, even if the difference seems minor. If values are rounded estimates, qualify the conclusion to match the precision provided.
- Giving only “independent” or “dependent.” Show both probabilities and state what their comparison means. Full-credit communication identifies whether the condition changes the chance of the event in context.
- Treating the result as cause and effect. Independence is a probability relationship. A difference does not prove that one event causes the other.
Key Takeaway
The independence test in this tutorial is a comparison: ask whether learning that \(B\) occurred changes the probability of \(A\). Use \(P(A\mid B)=P(A)\) when \(P(B)>0\). Equality means the events are independent; unequal values mean they are dependent. Then explain that result using the situation’s events and reference group.
Check Your Understanding
For each situation, decide whether the events are independent or dependent using the given probabilities. State the comparison that supports your decision.
- Let \(A\) mean a customer chooses curbside pickup and \(B\) mean the customer shops on a weekend. Suppose \(P(A)=0.35\), \(P(B)=0.40\), and \(P(A\mid B)=0.35\). Are the events independent?
- Let \(A\) mean a community garden plot is watered on a given day and \(B\) mean it is a hot day. Suppose \(P(A)=0.60\), \(P(B)=0.25\), and \(P(A\cap B)=0.20\). Find \(P(A\mid B)\) and decide whether the events are independent.
- In a two-way table, 45 of 150 people are in event \(A\), and 12 of the 30 people in event \(B\) are also in \(A\). Find \(P(A)\) and \(P(A\mid B)\). What does the comparison show?
- Suppose \(P(A\mid B)=0.42\), \(P(A)=0.42\), and \(P(B)=0\). Can you use the conditional-probability comparison to conclude that the events are independent? Explain.
- A report rounds both \(P(A)\) and \(P(A\mid B)\) to 0.37. What should you consider before claiming that the underlying probabilities are exactly equal?