Organizing Two Events in a Venn Diagram
In Independence of Three or More Events, you used the product rule when events are mutually independent. For two events, the same check can be organized visually: a Venn diagram separates the outcomes in \(A\) only, \(B\) only, both events, and neither event. Filling those four regions helps you see which probability is the intersection and which probabilities describe the whole events.
The diagram is a map of the sample space, not necessarily a picture drawn to scale. The rectangles or circles do not need to have areas proportional to their probabilities. What matters is that every outcome belongs to exactly one of the four regions.
The probability of \(A\) includes both the \(A\)-only region and the overlap. Likewise, \(P(B)\) includes the \(B\)-only region and the overlap. Therefore, if the overlap is known, subtract it from each marginal probability to fill the two non-overlapping parts.
The earlier tutorial Common Errors with the Addition Rule established that \(P(A\cup B)=P(A)+P(B)-P(A\cap B)\). The union consists of the \(A\)-only, \(B\)-only, and both regions. If you know the union and both marginal probabilities, rearrange the addition rule to find the overlap:
Once you have the overlap and the marginal probabilities, you can fill all four regions. For example, the probability of neither event is the complement of the union. This is the part of the sample space outside both circles.
Use the Regions to Check Independence
Independence is not determined by whether two circles appear to overlap, or by whether the overlap seems large. Use the probabilities. As established in The Multiplication Rule for Independent Events, events \(A\) and \(B\) are independent exactly when the probability that both occur equals the product of their individual probabilities.
The Venn diagram helps keep these quantities straight. \(P(A)\) is the total probability inside circle \(A\), including its overlap with \(B\). \(P(B)\) is the total inside circle \(B\). \(P(A\cap B)\) is only the shared region. Do not compare the product to the \(A\)-only region or to the union.
A useful way to proceed is to fill and check the diagram before making the independence comparison. If one region has a negative probability, the supplied values cannot all be correct probabilities for the same two events. A probability must be between 0 and 1, and none of the four disjoint regions can have a negative probability.
Identify the overlap, the two “only” regions, and the outside region representing neither event.
Use the intersection to find the “only” regions, or use the addition rule to recover the intersection from a union.
Confirm that every region is nonnegative and that the four probabilities add to 1.
Compare \(P(A\cap B)\) with \(P(A)P(B)\), using unrounded values when available.
This is an arithmetic check of independence in a probability model when the probabilities are given. If the values come from observed data and are rounded proportions, the equality may only be approximate. Do not treat a small rounding difference as proof of dependence; instead, compare the values at the precision supplied and describe the evidence appropriately.
Worked Examples
Worked Example: Filling Regions and Checking Independence
A randomly selected package from a production process has probability \(0.48\) of having a label error, event \(A\), and probability \(0.35\) of having a seal error, event \(B\). The probability that a package has both errors is \(0.168\). Fill the four Venn diagram regions and determine whether the events are independent.
State. Let \(A\) be the event that a package has a label error and \(B\) the event that it has a seal error. We need to fill the \(A\)-only, \(B\)-only, both, and neither regions, then check the product rule.
Plan. The overlap is given as \(P(A\cap B)=0.168\). Subtract it from each marginal probability to get the two “only” regions. Add the three regions inside the circles to find the union, then subtract from 1 to find neither. Finally, compare the overlap with \(P(A)P(B)\).
Do. The probability of a label error without a seal error is
The probability of a seal error without a label error is
The union is the sum of the three disjoint regions inside the circles:
So the probability of neither error is \(1-0.662=0.338\). The regions add to \(0.312+0.182+0.168+0.338=1.000\), and all are nonnegative. Now check independence:
The product calculation checks because \(48\times35=1680\), with four decimal places in the factors combined giving \(0.1680\).
Conclude. The four region probabilities are \(0.312\) for \(A\) only, \(0.182\) for \(B\) only, \(0.168\) for both, and \(0.338\) for neither. Because the intersection probability equals the product of the marginal probabilities, the label-error and seal-error events are independent in this probability model.
Worked Example: Recovering the Overlap from a Union
In a community garden, let \(A\) mean that a randomly selected plot grows tomatoes and \(B\) mean that it grows peppers. Suppose \(P(A)=0.60\), \(P(B)=0.40\), and \(P(A\cup B)=0.76\). Fill the four Venn diagram regions and check whether the events are independent.
State. The union probability is known, but the overlap is not. We will use the general addition rule to find \(P(A\cap B)\), then fill the other regions and test independence.
Plan. Rearrange the addition rule to find the intersection: add the two marginal probabilities and subtract the union. Then subtract the overlap from each marginal, and find the neither region by subtracting the union from 1. Compare the overlap with the product of the marginals.
Do. The overlap probability is
Thus, \(A\) only has probability \(0.60-0.24=0.36\), and \(B\) only has probability \(0.40-0.24=0.16\). Neither event has probability \(1-0.76=0.24\). As a check, the four regions sum to \(0.36+0.16+0.24+0.24=1.00\).
For independence, compare the intersection with the product:
The multiplication can also be checked as \(60\times40=2400\), then placing four decimal places gives \(0.2400=0.24\).
Conclude. The regions are \(0.36\) for tomato plots only, \(0.16\) for pepper plots only, \(0.24\) for plots growing both, and \(0.24\) for plots growing neither. Since the probability of growing both crops equals the product of the probabilities of growing each crop, these events are independent in the stated model.
Worked Example: The Overlap Does Not Match the Product
A school club survey uses a randomly selected student as its outcome. Let \(A\) mean the student participates in the robotics club and \(B\) mean the student participates in the art club. Suppose \(P(A)=0.55\), \(P(B)=0.30\), and \(P(A\cap B)=0.20\). Fill the diagram and decide whether the events are independent.
State. We are given both marginal probabilities and the overlap. The goal is to fill the other regions and compare the overlap with the product of the marginals.
Plan. Subtract the intersection from each marginal to find the “only” regions. Use the addition rule to find the union and its complement to find neither. Then test the product rule. The calculation tells us whether the events are independent; a mismatch means they are dependent.
Do. Robotics only has probability \(0.55-0.20=0.35\). Art only has probability \(0.30-0.20=0.10\). The union is
Therefore, neither club has probability \(1-0.65=0.35\). The region check gives \(0.35+0.10+0.20+0.35=1.00\), so the values form a possible probability assignment.
Now compare the intersection with the product of the marginals:
The product is \(55\times30=1650\), with four decimal places, so it is \(0.1650\). The given overlap is \(0.20\), which is not equal to \(0.165\). The difference is \(0.20-0.165=0.035\).
Conclude. The events are dependent in this probability model: the probability that a student participates in both clubs, \(0.20\), does not equal the product of the marginal probabilities, \(0.165\).
Common Mistakes and AP Exam Tips
- Using the wrong region for an event probability. \(P(A)\) includes both \(A\) only and the overlap. It is not just the \(A\)-only region. Similarly, \(P(A\cup B)\) includes the overlap as well as both “only” regions.
- Comparing the product with the union. The independence rule compares \(P(A\cap B)\) with \(P(A)P(B)\). The union \(P(A\cup B)\) is a different event.
- Forgetting to subtract the overlap. Adding \(P(A)\) and \(P(B)\) counts outcomes in both events twice. Use the general addition rule when finding the union, or rearrange it when finding the intersection from a union.
- Ignoring the outside region. “Neither” is \(1-P(A\cup B)\), not \(1-P(A)-P(B)\) unless the events are disjoint. The outside region and the circles together must account for the whole sample space.
- Calling events independent because their probabilities look similar. Independence is checked with the product rule, not by comparing the two marginal probabilities to each other.
- Failing to check the regions. A negative “only” region signals inconsistent supplied values: an intersection cannot be larger than either event’s marginal probability. Also confirm that the four regions total 1.
Key Takeaway
A two-event Venn diagram turns probability information into four non-overlapping regions. Fill the overlap and “only” regions carefully, use the complement for neither, and check that the regions sum to 1. Then test independence by comparing the overlap probability with the product of the two marginal probabilities.
Check Your Understanding
Show the region calculations and state the independence conclusion in context.
- Suppose \(P(A)=0.50\), \(P(B)=0.30\), and \(P(A\cap B)=0.15\). Find the probabilities of \(A\) only, \(B\) only, and neither. Are \(A\) and \(B\) independent?
- Suppose \(P(A)=0.70\), \(P(B)=0.20\), and \(P(A\cup B)=0.80\). Find \(P(A\cap B)\), then test independence.
- In your own words, explain why \(P(A)\) includes the overlap region but \(P(A\text{ only})\) does not.
- A proposed diagram has \(P(A)=0.25\) and \(P(A\cap B)=0.31\). What problem would you notice when filling the \(A\)-only region?
- What two probabilities must be equal for events \(A\) and \(B\) to be independent?