Classify the Chance Process, Not Just the Story
In Checking Independence in a Two-Way Table, you compared conditional percentages with overall percentages. The same central question applies to events in a sequence: does knowing that one event occurred change the probability of the other? To answer, pay attention to how the chance process works, not simply to whether the events happen one after another.
For instance, making one free throw does not remove a ball or change the hoop. That suggests the next attempt could have the same chance of success. Drawing a card and keeping it out of the deck does change what cards remain, so the next draw can have a different chance of meeting a description. These features help us classify scenarios, but we must state any assumptions about the process.
As in The Definition \(P(A\mid B)=P(A)\) and The Multiplication Rule for Independent Events, the comparison is about probabilities in a chance process. In a written scenario, you may not be given exact probabilities. You can still explain whether the process changes between events and what that implies, while making clear when your conclusion depends on a model assumption.
A Practical Way to Reason About Independence
Start by naming the two events precisely. Then ask what is different about the chance process after the first event. Has an item been removed? Has the population or set of possible outcomes changed? Is the same trial repeated under the same conditions? Could learning the first result alter a later action or outcome? These questions help identify whether the probability of the second event stays the same.
For example, \(A\) might mean “the first free throw is made,” and \(B\) might mean “the second free throw is made.”
Check whether the first outcome changes the objects, conditions, or decisions involved in the next chance event.
If probabilities are available, compare \(P(A\mid B)\) with \(P(A)\), or compare the second-event probability with and without knowledge of the first outcome. Otherwise, explain why the process does or does not change, and state relevant assumptions.
Say whether the events are independent or dependent in the stated chance model, and explain the reason. Do not claim more than the scenario supports.
A sequence of events is not automatically dependent just because it happens in order. Nor is it automatically independent because the events seem unrelated. For example, two events might be connected through a shared condition, even if neither event physically changes an object. The probability comparison, or a clear account of the chance process, is what supports the classification.
Independence is also different from mutual exclusivity. As explained in Mutually Exclusive Versus Independent Events, mutually exclusive events cannot happen together. If both have positive probabilities, they are dependent: knowing that one happened makes the other impossible. Independence, by contrast, means that knowing one event occurred does not change the probability of the other.
Worked Example: Two Free Throws
Worked Example: Two Free Throws
Suppose a player’s free-throw routine is modeled so that each attempt has a \(0.70\) probability of being made, regardless of the result of the previous attempt. Let \(A\) be the event that the first free throw is made, and \(B\) the event that the second free throw is made. Under this stated model, are \(A\) and \(B\) independent?
State. We will classify \(A\) and \(B\) in the model. The key assumption is that the chance of making the second shot remains \(0.70\), even if we know the result of the first shot.
Plan. Compare \(P(B\mid A)\) with \(P(B)\). If these probabilities are equal, knowing that the first shot was made does not change the chance that the second is made.
Do. The model gives \(P(B)=0.70\) and \(P(B\mid A)=0.70\). They match. As a check using the multiplication rule for independent events, the probability of making both is \(P(A)P(B)=0.70(0.70)=0.49\). The model also gives \(P(A\cap B)=0.70(0.70)=0.49\).
Conclude. In this model, making the first free throw does not change the probability of making the second, so \(A\) and \(B\) are independent. This conclusion follows from the stated model; it does not prove that every player’s real attempts are independent. Fatigue, a change in technique, or a reaction to the first result could make a real situation more complicated.
This example illustrates an important qualification: “repeated free throws” is a description, not a proof of independence. Independence is justified when the probability model says the chance of success stays the same regardless of earlier results. If the problem explicitly supplies that assumption, use it. If it describes a possible effect of the first result on the second, account for that effect instead.
Worked Example: Two Cards Drawn Without Replacement
Worked Example: Two Cards Drawn Without Replacement
A standard deck has 52 cards, including 26 red cards. Two cards are drawn one after the other, and the first card is kept out of the deck. Let \(A\) be the event that the first card is red, and \(B\) the event that the second card is red. Classify the events.
State. We want to determine whether \(A\) and \(B\) are independent for this two-draw process.
Plan. Compare the probability that the second card is red overall with the probability that it is red given that the first card was red. Because the first card is not returned, check how the number of red cards and the total number of cards change under the condition \(A\).
Do. Before any card is drawn, the second position is equally likely to contain any card in the deck, so \(P(B)=26/52=0.50\). Given that the first card is red, 25 red cards remain among 51 cards. Therefore, \(P(B\mid A)=25/51\approx0.4902\), rounded to four decimal places. Since \(0.4902\ne0.50\), the conditional and overall probabilities differ.
The multiplication rule gives a second check. The probability that both cards are red is
If the events were independent, their intersection probability would instead be \(P(A)P(B)=0.50(0.50)=0.25\). The values \(0.2451\) and \(0.25\) differ, confirming the classification.
Conclude. The events are dependent. Given that the first card is red, there is one fewer red card among the cards available for the second draw, so the probability that the second card is red decreases. This is a specific consequence of drawing without replacement.
Worked Example: Two Cards Drawn With Replacement
Worked Example: Two Cards Drawn With Replacement
Now suppose a red card is drawn from a standard deck, returned to the deck, and the deck is thoroughly shuffled before the second draw. Define \(A\) and \(B\) as in the previous example: the first card is red and the second card is red. Are the events independent in this process?
State. We will compare the probability that the second card is red with and without knowing that the first card was red.
Plan. Returning and shuffling the first card restores the original deck composition for the second draw. Check whether that makes the conditional and overall probabilities equal.
Do. The probability of a red card on the second draw is \(P(B)=26/52=0.50\). After the first card is returned and the deck is shuffled, the probability of red on the second draw remains \(P(B\mid A)=26/52=0.50\). Thus the probabilities match. The probability of two red cards is \(P(A\cap B)=0.50(0.50)=0.25\), which also equals \(P(A)P(B)\).
Conclude. In this model, the events are independent. Returning and shuffling restores the same deck composition before the second draw, so knowing that the first card was red does not change the probability that the second is red. The replacement step is the key difference from the previous example.
Worked Example: A First Result Changes a Later Chance
Worked Example: A First Result Changes a Later Chance
Imagine a practice model in which a player makes the second free throw with probability \(0.78\) after making the first, but with probability \(0.52\) after missing the first. Let \(A\) be the event that the first shot is made and \(B\) the event that the second shot is made. Assume \(P(A)=0.70\). Are the events independent?
State. We will check whether the chance of the second shot being made depends on the first result.
Plan. Compare \(P(B\mid A)\) with \(P(B\mid A^c)\). If these conditional probabilities differ, the second-shot probability changes with the first outcome, so the events are dependent.
Do. The model gives \(P(B\mid A)=0.78\) and \(P(B\mid A^c)=0.52\). The second-shot chance is different after a make and after a miss. We can also find the overall second-shot probability by adding the two disjoint cases:
Substituting the model values gives \(P(B)=0.70(0.78)+0.30(0.52)=0.546+0.156=0.702\). Since \(P(B\mid A)=0.78\ne0.702=P(B)\), knowing that the first shot was made changes the probability of making the second.
Conclude. In this model, \(A\) and \(B\) are dependent. The model assigns different second-shot probabilities depending on the first result. This example does not claim that such probabilities describe a real player; it shows how a stated relationship between attempts justifies classifying events as dependent.
Common Mistakes and AP Exam Tips
- Calling every sequence dependent. Events can happen at different times and still be independent. Explain whether the first outcome changes the probability of the later event.
- Assuming repeated attempts are automatically independent. A repeated free-throw model may assume a stable chance, but a real player’s later attempt could be affected by the earlier result. State the assumption or use the probabilities provided.
- Ignoring what happens to the objects. In card draws, replacement restores the deck composition; keeping the first card out changes it. Describe that change and connect it to the conditional probability.
- Confusing dependence with mutual exclusivity. Mutually exclusive events cannot occur together, while dependent events are events for which knowing one occurred changes the probability of the other. They are not synonyms.
- Giving only a label. “Dependent” by itself is not a justification. A strong response names the events and explains how the conditional probability differs, or identifies the feature of the process that changes the later chance.
- Overstating what a model establishes. If a scenario says the attempts have the same probability regardless of previous outcomes, conclude independence in that model. Do not turn that assumption into a claim about every real-world repetition.
Key Takeaway
To classify events in context, identify what could change the probability of one event after learning that the other occurred. Repeated free throws can be independent under a stable-probability model; card draws without replacement are dependent when the first draw changes the remaining deck; and replacing and shuffling a card can restore the original probability for the next draw.
Check Your Understanding
For each scenario, classify the events when the information allows it and give a reason tied to the chance process.
- A player’s model assigns the same probability of making each free throw regardless of previous results. Are the first- and second-shot make events independent in this model? Explain what assumption matters.
- A card is drawn from a deck and kept out before a second card is drawn. Explain why the events “first card is an ace” and “second card is an ace” are dependent or independent.
- A card is drawn, returned to the deck, and the deck is shuffled before the next draw. What feature of the process supports independence for the events that each draw is red?
- A model gives \(P(B)=0.40\) and \(P(B\mid A)=0.40\), with \(P(A)>0\). What does this comparison say about \(A\) and \(B\)?
- Explain why saying “the events happen at different times” is not enough to establish that they are dependent.