Independence is an important concept in probability because it allows us to simplify calculations by assuming that events are not affected by each other. When two events are independent, the probability of one event occurring does not depend on whether the other event has occurred or not.
For example, if you flip two coins, the likelihood of one landing on heads is not affected by the other coin. Therefore, we would say that these two events are independent.
On the other hand, if the temperature is extremely low, the probability of it snowing will increase. Therefore, these two events are not independent, or dependent, since the temperature does affect the likelihood of snow.
DEFINITION: Events A and B are independent if, and only if, knowing whether event A has occurred (or will occur) does not change the probability that event B will occur.
Quantifying Independence
If two events are independent, then we can use the multiplication rule to find the probability of both events occurring. The multiplication rule states that the probability of two independent events occurring is the product of the probabilities of the individual events. This is written as P(A and B) = P(A) * P(B).
Another thing to keep in mind: if two events are independent, then the probability of one event occurring does not depend on whether the other event has occurred or not. This means that the probability of event A occurring (P(A)) is the same regardless of whether event B has occurred or not. Similarly, the probability of event B occurring (P(B)) is the same regardless of whether event A has occurred or not.
Therefore, if, and only if, events A and B are independent, then P(A | B) = P(A) and P(B | A) = P(B).
In other words, the probability of event A occurring (P(A)) is the same as the probability of event A occurring given that event B has occurred (P(A | B)). Similarly, the probability of event B occurring (P(B)) is the same as the probability of event B occurring given that event A has occurred (P(B | A)).
Unions & The Addition Rule
In a previous section, you were briefly introduced to unions, the probability that event A or event B (or both) will occur. Unions are often denoted as P(A ∪ B). From here, we can develop a new rule of finding probabilities for independent events.
The addition rule is a way to find the probability of two events occurring together. It is also known as the "union" rule because it helps us find the probability of the union of two events.
The addition rule states that the probability of event A or event B occurring (denoted as P(A or B)) is equal to the probability of event A occurring (P(A)) plus the probability of event B occurring (P(B)) minus the probability of both events A and B occurring (P(A and B)). This is written as P(A or B) = P(A) + P(B) - P(A and B).
The addition rule applies when two events are not mutually exclusive (in other words, independent). This means that the events can occur at the same time and have some outcomes in common. In this case, we need to subtract the probability of both events occurring to avoid overcounting the common outcomes.
To keep the confusion at bay, here's a glimpse of the rules we've encountered so far, neatly organized in a visual:
Example
Imagine you are studying the attendance at a local music festival. You want to find the probability that a person who attends the festival will visit at least one of the two main stages. The main stage has a capacity of 10,000 people and is expected to be 75% full on average. The second stage has a capacity of 5,000 people and is expected to be 50% full on average.
You know that the attendance at the two stages is independent, so you can use the union rule to find the probability that a person will attend at least one of the stages. The probability that a person will attend the main stage is 0.75 and the probability that a person will attend the second stage is 0.50. Using the union rule, you can calculate the probability that a person will attend at least one of the stages:
P(attend main stage or attend second stage) = P(attend main stage) + P(attend second stage) - P(attend main stage and attend second stage)
= 0.75 + 0.50 - (0.75 * 0.50)
= 0.75 + 0.50 - 0.375
= 1.25 - 0.375
= 0.875
The probability that a person will attend at least one of the stages is 0.875, or 87.5%.
Now, let's say you also want to find the probability that a person who attends the festival will attend both stages. In this case, the attendance at the two stages is not independent because a person who attends one stage is more likely to attend the other stage as well. Therefore, you cannot use the union rule to find the probability of both stages being attended. Instead, you can use the multiplication rule to find the probability of both events occurring:
P(attend main stage and attend second stage) = P(attend main stage) * P(attend second stage given that they attend main stage)
= 0.75 * 0.50
= 0.375
The probability that a person will attend both stages is 0.375, or 37.5%.
Overall, the probability that a person will attend at least one of the stages is higher than the probability that they will attend both stages, which makes sense because attending one stage does not necessarily mean that they will also attend the other stage. Understanding the union rule and the multiplication rule from previous sections can help you calculate the probability of events occurring together in different situations.
Practice Problem
Imagine you are studying the likelihood of getting a high score on a math exam. You know that the probability of getting a high score on the exam is 0.7 if you study for at least 20 hours, and 0.4 if you do not study for at least 20 hours. You also know that the probability of studying for at least 20 hours is 0.6.
Find the probability of getting a high score on the exam regardless of whether you study for at least 20 hours or not.
Answer
To do this, you can use the union rule to find the probability of either event occurring:
P(high score | study for at least 20 hours or do not study for at least 20 hours) = P(high score | study for at least 20 hours) + P(high score | do not study for at least 20 hours) - P(high score | study for at least 20 hours AND do not study for at least 20 hours)
= 0.7 + 0.4 - (0.7 * 0.4)
= 0.7 + 0.4 - 0.28
= 1.1 - 0.28
= 0.82
The probability of getting a high score on the exam is 0.82, or 82%, regardless of whether you study for at least 20 hours or not.
Vocabulary
The following words are mentioned explicitly in the College Board Course and Exam Description for this topic.
| Term | Definition |
|---|---|
| addition rule | A probability rule stating that P(A ∪ B) = P(A) + P(B) - P(A ∩ B), used to find the probability of the union of two events. |
| conditional probability | The probability that one event will occur given that another event has already occurred, denoted P(A | B). |
| independent events | Events A and B are independent if knowing whether event A has occurred does not change the probability that event B will occur. |
| intersection | The set of outcomes that belong to both event A and event B, denoted A ∩ B. |
| union of events | The event that either event A or event B or both will occur, denoted P(A ∪ B). |
Frequently Asked Questions
How do I know if two events are independent or not?
Two events A and B are independent if knowing one happened doesn’t change the probability of the other (CED VAR-4.E.1). You can test this three equivalent ways (VAR-4.E.2): - Check conditional probability: P(A | B) = P(A) (or P(B | A) = P(B)). - Check multiplication rule: P(A ∩ B) = P(A) · P(B). - Intuitively: if learning B happened gives no new information about A, they’re independent. Quick checks/examples: two fair coin flips are independent (P(H on 1st ∩ H on 2nd)=.5·.5=.25). Drawing two cards without replacing is not independent because the first draw changes the deck (P(second is ace | first was ace) ≠ P(second is ace)). Remember independence ≠ mutually exclusive. If A and B are mutually exclusive and both have positive probability, they’re dependent (since P(A ∩ B)=0 ≠ P(A)P(B)). For more practice and AP-aligned problems, see the Topic 4.6 study guide (https://library.fiveable.me/ap-statistics/unit-4/independent-events-unions-events/study-guide/aMOuOhtDQIJAtgZx3JNz) and AP practice sets (https://library.fiveable.me/practice/ap-statistics).
What's the formula for P(A and B) when events are independent?
If A and B are independent, then knowing one doesn’t change the chance of the other. On the AP CED (VAR-4.E.2) that means P(A and B) = P(A ∩ B) = P(A) · P(B). Equivalently, P(A | B) = P(A) and P(B | A) = P(B). Use the product rule whenever you can justify independence (not just when events are disjoint). This is tested under Skill 3.A (using probability rules) on the AP exam. If you want a quick review or example problems, see the Topic 4.6 study guide (https://library.fiveable.me/ap-statistics/unit-4/independent-events-unions-events/study-guide/aMOuOhtDQIJAtgZx3JNz) and try practice questions (https://library.fiveable.me/practice/ap-statistics).
When do I use the addition rule vs the multiplication rule for probability?
Use the multiplication rule when you want the probability that two (or more) events happen together—i.e., the intersection. In general: P(A ∩ B) = P(A)·P(B | A). If A and B are independent (knowing A doesn’t change B), this simplifies to P(A ∩ B) = P(A)·P(B) (VAR-4.E.2). That’s the rule for independent trials (like repeated coin flips or binomial problems). Use the addition rule when you want the probability that at least one of two events happens—the union. In general: P(A ∪ B) = P(A) + P(B) − P(A ∩ B) (VAR-4.E.4). If A and B are mutually exclusive (can’t both happen), then P(A ∩ B) = 0 and you just add P(A)+P(B) (Topic 4.4). For AP exam problems, first ask: are you finding “and” (use multiplication/intersection) or “or” (use addition/union)? Check independence vs conditional probability to decide which multiplication formula applies. For a quick review see the Topic 4.6 study guide (https://library.fiveable.me/ap-statistics/unit-4/independent-events-unions-events/study-guide/aMOuOhtDQIJAtgZx3JNz). For more practice, try the AP practice set (https://library.fiveable.me/practice/ap-statistics).
What's the difference between P(A or B) and P(A and B)?
P(A or B) means the probability that at least one of A or B happens (A ∪ B): A occurs, B occurs, or both. Use the addition rule: P(A ∪ B) = P(A) + P(B) − P(A ∩ B). You subtract P(A ∩ B) to avoid double-counting the overlap (CED VAR-4.E.3–4). P(A and B) means the probability both A and B happen (A ∩ B), sometimes called the joint probability. If A and B are independent (VAR-4.E.1–2), then P(A ∩ B) = P(A)·P(B). If they’re mutually exclusive, P(A ∩ B)=0, so P(A ∪ B)=P(A)+P(B). On the AP exam you’ll be asked to identify when to use the addition rule vs. the multiplication rule and to state independence or dependence (Skill 3.A). For extra practice, check the Topic 4.6 study guide (https://library.fiveable.me/ap-statistics/unit-4/independent-events-unions-events/study-guide/aMOuOhtDQIJAtgZx3JNz) and try problems at (https://library.fiveable.me/practice/ap-statistics).
I'm confused about when to subtract P(A ∩ B) in the addition rule - can someone explain?
You subtract P(A ∩ B) whenever you add P(A) + P(B) to find P(A ∪ B) because the overlap (the intersection) is counted twice. Think Venn diagram: the part where A and B both happen sits in both circles, so adding P(A) and P(B) double-counts that region. The clean formula from the CED is P(A ∪ B) = P(A) + P(B) − P(A ∩ B). Two quick special cases to remember (AP terms): - If A and B are mutually exclusive, P(A ∩ B) = 0, so you don’t subtract anything. - If A and B are independent (knowing A doesn’t change B), then P(A ∩ B) = P(A)·P(B), so you subtract that product. On the exam you’ll often show this with a Venn diagram or by explaining double-counting; if events are independent, use the multiplication rule (VAR-4.E.2). For more examples and practice, check the Topic 4.6 study guide (https://library.fiveable.me/ap-statistics/unit-4/independent-events-unions-events/study-guide/aMOuOhtDQIJAtgZx3JNz) and try practice problems at (https://library.fiveable.me/practice/ap-statistics).
How do I calculate P(A ∪ B) step by step?
Step-by-step: P(A ∪ B) means the chance A or B (or both) happens. 1. Write the addition (union) rule: P(A ∪ B) = P(A) + P(B) − P(A ∩ B). (CED VAR-4.E.4) 2. Find P(A) and P(B) from the problem. 3. Find P(A ∩ B)—the probability both happen. If A and B are independent (CED VAR-4.E.2), compute P(A ∩ B) = P(A)·P(B). If they’re mutually exclusive, P(A ∩ B)=0. Otherwise use given joint info or a Venn diagram. 4. Substitute into the formula and simplify. Quick example: P(A)=0.4, P(B)=0.3, independent → P(A ∩ B)=0.4·0.3=0.12. So P(A ∪ B)=0.4+0.3−0.12=0.58. On the AP exam you’ll be expected to state whether events are independent or mutually exclusive and justify how you got P(A ∩ B) (use VAR-4.E.1–4). For a focused review see the Topic 4.6 study guide (https://library.fiveable.me/ap-statistics/unit-4/independent-events-unions-events/study-guide/aMOuOhtDQIJAtgZx3JNz) and practice problems at (https://library.fiveable.me/practice/ap-statistics).
What does it mean when P(A
If P(A | B) = P(A), that means A and B are independent events—knowing B happened doesn’t change the probability that A happens (CED VAR-4.E.1). Equivalent ways to say this (CED VAR-4.E.2): P(B | A) = P(B), and P(A ∩ B) = P(A)·P(B). So to check independence you can either compare the conditional probability to the marginal probability or test whether the joint probability equals the product of the marginals. Quick example: if P(A)=0.3 and P(B)=0.5 and P(A ∩ B)=0.15, then 0.15 = 0.3·0.5 so A and B are independent (and P(A|B)=0.3). On the AP exam you may be asked to use either conditional probability or the multiplication rule; knowing both forms helps (see the Topic 4.6 study guide: https://library.fiveable.me/ap-statistics/unit-4/independent-events-unions-events/study-guide/aMOuOhtDQIJAtgZx3JNz). For extra practice, check Fiveable’s AP Stats practice problems (https://library.fiveable.me/practice/ap-statistics).
B) = P(A) and how do I use this?
When you see P(A | B) = P(A) it’s the formal test for independence: knowing B happened doesn’t change A’s probability (VAR-4.E.1–2). Use it two ways: 1) To check independence: compute P(A | B) = P(A ∩ B) / P(B). If that equals P(A), A and B are independent. 2) To compute joint probabilities when you already know independence: if A and B are independent, P(A ∩ B) = P(A)·P(B) (VAR-4.E.2). Quick example: P(A)=0.3, P(B)=0.5. If P(A ∩ B)=0.15 then P(A | B)=0.15/0.5=0.3 so P(A | B)=P(A) → independent, and indeed 0.3·0.5=0.15. If P(A ∩ B) ≠ P(A)·P(B) then they’re not independent. This idea appears on the AP exam in Topic 4.6 problems (you may need to check P(A|B) vs. P(A) or use the multiplication rule). For a quick topic review, see the Fiveable study guide (https://library.fiveable.me/ap-statistics/unit-4/independent-events-unions-events/study-guide/aMOuOhtDQIJAtgZx3JNz) and practice problems (https://library.fiveable.me/practice/ap-statistics).
If I know two events are independent, what's the easiest way to find their intersection probability?
If A and B are independent, the easiest way is the multiplication rule from the CED: P(A ∩ B) = P(A) · P(B). That comes from P(A | B) = P(A) (or P(B | A) = P(B)) for independent events (VAR-4.E.2). Quick example: if P(A) = 0.4 and P(B) = 0.3, then P(A ∩ B) = 0.4 × 0.3 = 0.12. If you only know complements, use them too: P(Aʹ ∩ B) = P(Aʹ)·P(B). And if you need P(A ∪ B), use the addition rule P(A ∪ B) = P(A)+P(B)−P(A ∩ B) (VAR-4.E.4). This is exactly what AP Stats expects you to use on the exam (Topic 4.6). For a quick refresher, see the Topic 4.6 study guide (https://library.fiveable.me/ap-statistics/unit-4/independent-events-unions-events/study-guide/aMOuOhtDQIJAtgZx3JNz) and try practice problems at (https://library.fiveable.me/practice/ap-statistics).
When solving union problems, how do I know if I need to subtract the overlap?
You subtract the overlap whenever A and B can both happen—i.e., whenever they’re not mutually exclusive. The addition rule from the CED is P(A ∪ B) = P(A) + P(B) − P(A ∩ B) (VAR-4.E.4). If A and B are mutually exclusive, P(A ∩ B) = 0 so you don’t subtract anything. If A and B are independent, they still can both happen, and P(A ∩ B) = P(A)·P(B) (VAR-4.E.2), so you do subtract that product. Quick example: P(A)=0.30, P(B)=0.40 and A, B independent → P(A ∩ B)=0.12, so P(A ∪ B)=0.30+0.40−0.12=0.58. Want more practice problems and step-by-step explanations for Topic 4.6? See the Fiveable study guide (https://library.fiveable.me/ap-statistics/unit-4/independent-events-unions-events/study-guide/aMOuOhtDQIJAtgZx3JNz) and the practice bank (https://library.fiveable.me/practice/ap-statistics).
I don't understand the difference between independent events and mutually exclusive events - are they the same thing?
They’re different—don’t mix them up. Independent events: knowing A happened doesn’t change the chance of B. Formally P(A|B)=P(A) (and P(A ∩ B)=P(A)·P(B)). Example: two fair coin flips, A = “first is heads” (0.5), B = “second is heads” (0.5). P(A ∩ B)=0.25. Mutually exclusive (disjoint) events: they cannot happen together, so P(A ∩ B)=0. Example: rolling a die, A = “roll 1”, B = “roll 2.” If P(A)>0 and P(B)>0, then they’re not independent, because P(A|B)=0 ≠ P(A). Key takeaway: if A and B are mutually exclusive and both have positive probability, they cannot be independent. Use the union/addition rule P(A ∪ B)=P(A)+P(B)−P(A ∩ B) for both cases (with P(A ∩ B)=0 for disjoint events, or P(A ∩ B)=P(A)P(B) if independent). For AP practice, review Topic 4.6 (study guide: https://library.fiveable.me/ap-statistics/unit-4/independent-events-unions-events/study-guide/aMOuOhtDQIJAtgZx3JNz) and try problems at (https://library.fiveable.me/practice/ap-statistics).
What's the formula for the addition rule and when exactly do I use it?
Addition rule (union formula): P(A ∪ B) = P(A) + P(B) − P(A ∩ B). When to use it: use this any time you want the probability that A or B (or both) occur—i.e., the union. You subtract P(A ∩ B) because A and B overlap and you’d otherwise double-count that overlap. Special cases: if A and B are independent, compute the intersection as P(A ∩ B) = P(A)·P(B) (CED VAR-4.E.2); if they’re mutually exclusive, P(A ∩ B) = 0 so the rule reduces to P(A ∪ B) = P(A) + P(B) (CED VAR-4.E.4). Tip for the exam: the AP formula sheet includes P(A ∪ B) = P(A)+P(B)−P(A∩B). For extra practice on independence and unions, check the Topic 4.6 study guide (https://library.fiveable.me/ap-statistics/unit-4/independent-events-unions-events/study-guide/aMOuOhtDQIJAtgZx3JNz) and the practice problems page (https://library.fiveable.me/practice/ap-statistics).
How do I solve word problems about independent events like rolling dice or flipping coins?
Start by asking: are the events independent? If knowing A doesn’t change P(B), they’re independent (CED VAR-4.E). For independent events use the multiplication rule: P(A and B) = P(A)·P(B). Example: two fair coin flips, P(both heads) = 1/2 · 1/2 = 1/4. For two independent dice rolls, P(roll a 4 then a 6) = (1/6)(1/6) = 1/36. If the problem asks “A or B (or both)”, use the union (addition) rule: P(A ∪ B) = P(A) + P(B) − P(A ∩ B) (CED VAR-4.E.3–4). Often it’s easier to use complements: P(at least one success) = 1 − P(none). Example: three coin flips, P(at least one head) = 1 − (1/2)^3 = 7/8. On the exam you’ll be judged on correct identification of independence and correct use of multiplication/addition rules (Skill 3.A, VAR-4.E). For practice and worked examples check the Topic 4.6 study guide (https://library.fiveable.me/ap-statistics/unit-4/independent-events-unions-events/study-guide/aMOuOhtDQIJAtgZx3JNz), the Unit 4 overview (https://library.fiveable.me/ap-statistics/unit-4), and many practice problems (https://library.fiveable.me/practice/ap-statistics).
Why does P(A ∩ B) = P(A) × P(B) only work for independent events?
Because P(A ∩ B) = P(A)·P(B) comes from the definition of conditional probability. By definition P(A ∩ B) = P(A | B)·P(B). If A and B are independent (CED VAR-4.E.1), knowing B happened doesn’t change the chance of A, so P(A | B) = P(A). Substitute that and you get P(A ∩ B) = P(A)·P(B) (CED VAR-4.E.2). If A and B aren’t independent, P(A | B) ≠ P(A), so you can’t replace P(A | B) with P(A) and the product rule fails. Example: drawing two cards without replacement—the probability the second is an ace depends on whether the first was an ace, so the joint probability is not the simple product. This is a core AP idea (use conditional probability and the multiplication rule). For a focused review, see the Topic 4.6 study guide (https://library.fiveable.me/ap-statistics/unit-4/independent-events-unions-events/study-guide/aMOuOhtDQIJAtgZx3JNz) and practice questions (https://library.fiveable.me/practice/ap-statistics).
Can someone walk me through how to check if events are independent using conditional probability?
You check independence by seeing whether knowing one event happened changes the probability of the other. Steps: 1) Find P(A|B) = P(A ∩ B) / P(B). 2) Compare that to P(A). If P(A|B) = P(A), then A and B are independent (same idea: P(A ∩ B) = P(A)·P(B)). Equivalently you can compute P(B|A) and compare to P(B). Example: if P(A)=0.4, P(B)=0.5, and P(A ∩ B)=0.20, then P(A|B)=0.20/0.50=0.4 so A and B are independent (0.20 = 0.4·0.5). On the AP, this is VAR-4.E (you may be asked to use P(A|B)=P(A) or P(A∩B)=P(A)P(B)); practice both forms (Skill 3.A). For a short review and extra examples see the Topic 4.6 study guide (https://library.fiveable.me/ap-statistics/unit-4/independent-events-unions-events/study-guide/aMOuOhtDQIJAtgZx3JNz) and try more problems at (https://library.fiveable.me/practice/ap-statistics).
I keep getting confused about whether to add or multiply probabilities - when do I do each one?
Short rule: multiply when you want the probability that BOTH (or all) things happen; add when you want the probability that AT LEAST ONE happens (A or B or both). Key details from the CED: - If A and B are independent, use the multiplication rule: P(A ∩ B) = P(A)·P(B). Example: two fair coin tosses → P(both heads) = 1/2 · 1/2 = 1/4 (VAR-4.E.2). - If not independent, use the general multiplication rule: P(A ∩ B) = P(A | B)·P(B) (use conditional probability). - For unions use the addition (union) rule: P(A ∪ B) = P(A) + P(B) − P(A ∩ B) (VAR-4.E.4). If A and B are mutually exclusive (disjoint), then P(A ∩ B)=0 so you simply add. On the AP exam, always check whether independence or mutual exclusivity is stated or implied before choosing multiply vs add (see Topic 4.6 in the study guide: https://library.fiveable.me/ap-statistics/unit-4/independent-events-unions-events/study-guide/aMOuOhtDQIJAtgZx3JNz). For extra practice, try problems at the unit page (https://library.fiveable.me/ap-statistics/unit-4) and the 1000+ practice questions (https://library.fiveable.me/practice/ap-statistics).