Two Different Relationships Between Events
In Using Complements with Disjoint Events, you used the fact that mutually exclusive events cannot occur together. That describes whether events share outcomes. Independence describes a different relationship: whether knowing that one event occurred changes the probability of the other. Confusing these ideas can lead to an incorrect probability calculation, so it helps to test them separately.
Let \(A\) and \(B\) be events from the same chance process. They are mutually exclusive, or disjoint, if they have no outcomes in common, so \(P(A\cap B)=0\). They are independent if the probability that both occur equals the product of their separate probabilities.
The distinction is about what happens when the two events are considered together. Disjoint events cannot both occur. Independent events, on the other hand, have a joint probability equal to what their individual probabilities predict under independence. In particular, independent events with positive probabilities can occur together; their intersection then has positive probability.
The product rule also explains the common-language interpretation of independence. If \(B\) has positive probability, the probability of \(A\) among outcomes where \(B\) occurs is \(P(A\cap B)/P(B)\). When \(P(A\cap B)=P(A)P(B)\), that value is \(P(A)\): knowing that \(B\) occurred has not changed the probability of \(A\). You can use the product rule directly without calculating this conditional probability.
Why Disjoint Events with Positive Probabilities Are Dependent
Suppose \(A\) and \(B\) are disjoint and both have positive probabilities. Disjointness makes \(P(A\cap B)=0\). But the product \(P(A)P(B)\) is positive, because it multiplies two positive numbers. Therefore, the independence equation cannot be true: \(0\) is not equal to a positive number.
There is also a direct way to understand this result. If \(B\) occurs, and \(A\) and \(B\) are disjoint, then \(A\) cannot occur. The probability of \(A\) after learning that \(B\) occurred is zero, rather than its original positive probability. The occurrence of \(B\) has changed what we know about \(A\), so the events are dependent.
This argument also points out an exception worth remembering. If one event has probability zero, the product \(P(A)P(B)\) can be zero even when the events are disjoint. The product test may then show that they are independent. The rule is not “disjoint events are always dependent”; it is “disjoint events with positive probabilities are dependent.”
A Reliable Independence Check
When a question asks whether two events are independent, begin with their definitions and probabilities for the same chance process. Then find the intersection probability and compare it with the product of the marginal probabilities. Do not decide independence just because events sound unrelated or because their outcomes happen in separate parts of a story; use the probability relationship.
State exactly what \(A\) and \(B\) mean for one chance process.
Determine \(P(A)\) and \(P(B)\) from the specified model or information.
Calculate \(P(A\cap B)\) and \(P(A)P(B)\), using the same probability model for both.
If the two values are equal, the events are independent; otherwise, they are dependent. Explain what the result means in context.
If the events are disjoint, you already know that their intersection probability is zero. When both individual probabilities are positive, you can conclude they are dependent without doing any further counting. Still, showing the product makes the reason explicit.
Worked Examples
Worked Example: Even and Odd on One Die Roll
A fair six-sided die is rolled once. Let \(A\) be the event that the result is even, and let \(B\) be the event that the result is odd. Are \(A\) and \(B\) independent?
State: We are checking whether the event “the result is even” and the event “the result is odd” are independent on this one roll.
Plan: List the outcomes in each event. Since a single die result cannot be both even and odd, the events are disjoint and \(P(A\cap B)=0\). Then compare zero with \(P(A)P(B)\).
Do: There are three even results, \(2,4,6\), and three odd results, \(1,3,5\), out of six equally likely outcomes. Thus:
The product of the individual probabilities is:
Because \(P(A\cap B)=0\) is not equal to \(P(A)P(B)=1/4\), the product rule for independence is not satisfied.
Conclude: The events are dependent. In context, once the die result is known to be odd, it cannot be even; knowing that one event occurred changes the probability of the other from \(1/2\) to zero.
The same logic applies to any two categories that cannot occur together on one selection. For example, if one randomly selected student chooses exactly one workshop, “chooses robotics” and “chooses ceramics” are disjoint. If each has positive probability, they are dependent, even if the workshop topics seem unrelated.
Worked Example: Heads on Two Separate Coin Tosses
A fair coin is tossed twice. Let \(A\) be the event that the first toss is heads, and let \(B\) be the event that the second toss is heads. Are these events independent?
State: \(A\) concerns the first toss and \(B\) concerns the second toss. We will compare their intersection probability with the product of their individual probabilities.
Plan: The equally likely outcomes for the two tosses are \(HH, HT, TH,\) and \(TT\). Find \(P(A)\), \(P(B)\), and \(P(A\cap B)\) by counting the outcomes that meet each event definition. Then apply the product test.
Do: Two of the four outcomes have heads on the first toss, and two have heads on the second toss. Only \(HH\) has heads on both tosses. Therefore:
The product of the individual probabilities is:
The intersection probability and the product are both \(1/4\), so the product rule is satisfied.
Conclude: The events are independent. Getting heads on the first toss does not change the probability of heads on the second toss in this fair-coin model. Notice that these events are not disjoint: the outcome \(HH\) belongs to both.
Worked Example: Testing Independence from Counts
Imagine an invented check of 100 devices. Of these, 40 display a low-battery alert, 30 have a wireless connection issue, and 12 have both. Let \(A\) be the event that a randomly selected device displays a low-battery alert, and \(B\) the event that it has a wireless connection issue. Based on these counts, are \(A\) and \(B\) independent?
State: We are testing whether the two device events satisfy \(P(A\cap B)=P(A)P(B)\) for a random selection from these 100 devices.
Plan: Use the marginal counts for \(P(A)\) and \(P(B)\), and the count in both categories for \(P(A\cap B)\). Then compare the resulting probabilities.
Do: The individual and joint probabilities are:
Calculate the product of the individual probabilities:
Since the product \(0.12\) equals the intersection probability \(0.12\), the independence equation is satisfied.
Conclude: For a random selection from these devices, the events are independent according to the product test. They are not disjoint, since 12 devices have both characteristics. This example illustrates that independence does not mean that two events cannot occur together.
Common Mistakes and AP Exam Tips
- Using “mutually exclusive” and “independent” as synonyms. They are different properties. Disjoint events have an intersection probability of zero; independent events have an intersection probability equal to the product of the individual probabilities.
- Assuming disjoint events are independent because they do not overlap. If both disjoint events have positive probabilities, their product is positive while their intersection probability is zero. The events are dependent.
- Assuming independent events must be disjoint. Independent events can overlap. In the two-toss example, both events occur for \(HH\), and the intersection probability matches the product.
- Checking only whether the events sound unrelated. A story may make two events seem separate, but the probability model determines independence. Compare \(P(A\cap B)\) with \(P(A)P(B)\).
- Using probabilities from different chance processes or different populations. To use the product test, all probabilities must refer to the same model and the same definitions of \(A\) and \(B\).
- Giving only “yes” or “no.” A full-credit explanation states the product and intersection probabilities and compares them. For example: “Because \(P(A\cap B)=0\) but \(P(A)P(B)=1/4\), the events are dependent.”
For a quick check, first ask whether the events are disjoint. If they are disjoint and both have positive probability, dependence follows immediately. Otherwise, use the product test rather than relying on a verbal guess.
Check Your Understanding
For each situation, decide whether the events are disjoint, independent, both, or neither. Support your answer with the relevant probabilities or reasoning.
- A fair six-sided die is rolled once. \(A\) is rolling a 1, and \(B\) is rolling a 4. Are the events independent? Show the intersection probability and the product.
- A fair coin is tossed twice. \(A\) is getting tails on the first toss, and \(B\) is getting heads on the second toss. Find \(P(A)\), \(P(B)\), and \(P(A\cap B)\), then test independence.
- Explain why two mutually exclusive events with positive probabilities cannot be independent.
- Can two disjoint events ever satisfy the product rule for independence if one event has probability zero? Explain briefly.
- For events \(A\) and \(B\), suppose \(P(A)=0.6\), \(P(B)=0.2\), and \(P(A\cap B)=0.12\). Are the events independent? Show the comparison.