When One Event Rules Out the Other
In Independence When Sampling Without Replacement, you saw how learning the result of one draw can change the probability of a later event. This tutorial looks at a different, especially clear case: when two events are mutually exclusive, knowing that one occurred makes the other impossible.
The key comparison is the one from The Definition \(P(A\mid B)=P(A)\). If \(P(B)>0\), then \(A\) and \(B\) are independent when learning that \(B\) occurred does not change the probability of \(A\): \(P(A\mid B)=P(A)\). But if the events are mutually exclusive and \(B\) occurred, \(A\) cannot have occurred. So \(P(A\mid B)=0\). When \(P(A)>0\), zero is not equal to \(P(A)\).
The nonzero-probability qualification matters. If \(P(A)=0\), then the comparison \(P(A\mid B)=P(A)\) could be \(0=0\); that is not the situation this tutorial focuses on. Here, both events are possible on their own, but their overlap is empty.
The Conditional-Probability Reason
As in The Conditional Probability Formula, calculate \(P(A\mid B)\) by dividing the probability of both events by the probability of the condition. The condition is \(B\), so \(P(B)\) must be greater than zero. For mutually exclusive events, the numerator \(P(A\cap B)\) is zero. Therefore, the conditional probability is zero.
Independence would require this conditional probability to equal \(P(A)\). If \(P(A)>0\), then \(P(A\mid B)=0\ne P(A)\), so the events are dependent. In context, knowing that \(B\) occurred changes the chance of \(A\) from a positive probability to zero.
The product rule gives the same conclusion from another direction. As covered in The Multiplication Rule for Independent Events, independent events satisfy \(P(A\cap B)=P(A)P(B)\). For disjoint events, the left side is zero. If both individual probabilities are positive, their product is positive. The two sides cannot be equal.
These are two ways to express the same contradiction: one compares a conditional probability with an unconditional probability, and the other compares the intersection probability with the product of the individual probabilities. You only need one valid check to show that the events are not independent.
A Reliable Way to Check
Start by being precise about the chance process and the event definitions. In a single roll of a die, for example, “roll a 1” and “roll a 6” cannot both happen. But “roll an even number” and “roll a number greater than 3” can both happen, since a 4 or 6 meets both descriptions. Disjointness depends on what outcomes belong to each event; it is not established just because the event names sound different.
State what \(A\) and \(B\) mean in the same chance process.
Determine whether any outcome can be in both events. If none can, \(P(A\cap B)=0\).
For this conclusion, verify that \(P(A)>0\) and \(P(B)>0\). The positive value of \(P(B)\) also makes \(P(A\mid B)\) defined.
For disjoint events, \(P(A\mid B)=0\). If \(P(A)>0\), they are dependent, not independent.
A useful mental picture is the reduced sample space from Checking Answers by Reduced Sample Space. Once you are told that \(B\) occurred, consider only outcomes in \(B\). If \(A\) and \(B\) are disjoint, none of those remaining outcomes belong to \(A\). That leaves zero outcomes favorable to \(A\), so the conditional probability is zero.
Worked Examples
Worked Example: Two Colors on a Spinner
A fair spinner has eight equal sections: three are red, two are blue, and three are green. On one spin, let \(A\) mean “the spinner lands on red” and \(B\) mean “the spinner lands on blue.” Decide whether \(A\) and \(B\) are independent.
State. We will check whether learning that the spinner landed on blue changes the probability that it landed on red.
Plan. The color on a single spin can be only one of red, blue, or green. Thus, red and blue are mutually exclusive. Find \(P(A)\), \(P(B)\), and \(P(A\mid B)\), then use the conditional-probability definition of independence.
Do. There are eight equally likely sections, of which three are red and two are blue. Therefore, \(P(A)=3/8=0.375\) and \(P(B)=2/8=0.25\). Since a section cannot be both red and blue, \(P(A\cap B)=0\). The condition \(B\) has positive probability, so the conditional probability is defined:
If the events were independent, \(P(A\mid B)\) would equal \(P(A)=0.375\). Instead, it is zero. The product-rule check agrees: \(P(A)P(B)=(3/8)(2/8)=6/64=0.09375\), while \(P(A\cap B)=0\).
Conclude. Landing on red and landing on blue on the same spin are mutually exclusive and not independent. Given that the spinner landed on blue, the probability that it landed on red is zero, rather than the unconditional probability of \(0.375\).
Worked Example: Two Events in One Die Roll
A fair six-sided die is rolled once. Let \(A\) mean “the result is at most 2,” and let \(B\) mean “the result is at least 5.” Are the events independent?
State. We want to determine whether knowing that the result is at least 5 changes the probability that it is at most 2.
Plan. List the possible results in each event. If the lists do not overlap, use the conditional probability formula and compare the result with \(P(A)\).
Do. Event \(A\) contains results \(\{1,2\}\), and event \(B\) contains results \(\{5,6\}\). These sets have no result in common, so \(P(A\cap B)=0\). Each event contains two of the six equally likely outcomes, giving \(P(A)=2/6=1/3\) and \(P(B)=2/6=1/3\). Thus,
The unconditional probability \(P(A)\) is \(1/3\), not zero, so \(P(A\mid B)\ne P(A)\). The product-rule comparison is \(P(A)P(B)=(1/3)(1/3)=1/9\), which is not the actual intersection probability of zero.
Conclude. The events are mutually exclusive and dependent. If the die shows 5 or 6, it cannot also show 1 or 2; knowing \(B\) makes \(A\) impossible.
Worked Example: Primary Travel Mode in a Hypothetical Survey
Imagine a hypothetical survey of 200 students, where each student reports exactly one primary way of getting to school. Suppose 70 report bus and 30 report bicycle. For one student selected at random from these 200, let \(A\) mean “the student’s primary mode is bus” and \(B\) mean “the student’s primary mode is bicycle.” Find \(P(A\mid B)\) and decide whether the events are independent.
State. We are comparing the probability that a randomly selected student’s primary mode is bus with the probability of bus among students whose primary mode is bicycle.
Plan. Because each student gives exactly one primary mode, the same student cannot belong to both event groups. Use the overlap count and the total count in group \(B\) to calculate the conditional probability, then compare it with the overall bus proportion.
Do. The overall probability of bus is \(P(A)=70/200=0.35\), and the probability of bicycle is \(P(B)=30/200=0.15\). No student in this setup has both primary modes, so the overlap count is zero. Among the 30 students in \(B\), zero have primary mode bus:
This also follows from the probability formula: \(P(A\cap B)=0/200=0\), so \(P(A\mid B)=0/0.15=0\). Since \(P(A)=0.35\), learning that the selected student’s primary mode is bicycle changes the chance of bus from 0.35 to zero.
Conclude. In this hypothetical setting, primary mode of bus and primary mode of bicycle are mutually exclusive and not independent. Among students whose primary mode is bicycle, none have bus as their primary mode. This conclusion relies on the stated “exactly one primary mode” rule; if students could name multiple modes, the events might overlap.
Common Mistakes and AP Exam Tips
- Calling disjoint events independent because they are “separate.” Separate event names do not establish independence. If both events have positive probabilities and cannot occur together, conditioning on one makes the other impossible.
- Using \(P(A)\) as the conditional probability. That is the value independence would require, not the value for disjoint events. Calculate \(P(A\mid B)\) using the overlap divided by \(P(B)\).
- Forgetting the condition must have positive probability. The formula for \(P(A\mid B)\) requires \(P(B)>0\). If \(B\) cannot occur, the conditional probability is not defined.
- Confusing “cannot happen together” with “one affects the other.” For disjoint events, the logical restriction itself changes the conditional probability: once \(B\) has occurred, \(A\) is ruled out. That is dependence when \(P(A)>0\).
- Leaving out the comparison in a conclusion. Saying only “the events are disjoint” does not explain why they are not independent. State the values: \(P(A\mid B)=0\), while \(P(A)>0\).
- Assuming categories are automatically disjoint. Check how the chance process defines them. In a select-all survey, for instance, a person might fit both categories. The one-primary-mode rule in the example is what makes those categories disjoint.
Key Takeaway
Independence means that knowing one event occurred does not change the probability of the other. Disjoint events with positive probabilities do the opposite: knowing that \(B\) occurred makes \(A\) impossible. The conditional probability is zero, while \(P(A)\) is positive.
Check Your Understanding
For each question, explain how the events can or cannot occur together and use probabilities to support your conclusion.
- A fair eight-section spinner has three yellow and five purple sections. On one spin, let \(A\) be landing on yellow and \(B\) be landing on purple. Find \(P(A\mid B)\) and decide whether the events are independent.
- A fair die is rolled once. Let \(A\) be rolling an odd number and \(B\) be rolling a number greater than 3. Are these events disjoint? Give an outcome that supports your answer.
- Suppose \(P(A)=0.4\), \(P(B)=0.2\), and \(A\) and \(B\) are mutually exclusive. Find \(P(A\mid B)\). Are they independent?
- Why must \(P(B)>0\) to calculate \(P(A\mid B)\)?
- Write one sentence that uses \(P(A\mid B)\) and \(P(A)\) to justify why mutually exclusive events with positive probabilities are not independent.