Events where P(A and B) = P(A) * P(B). No influence on each other.
Deep-dive lesson - accessible entry point but dense material. Use worked examples and spaced repetition.
Sometimes learning that one event happened should change your expectations about another event. Independence is the special case where it doesn’t.
Two events A and B are independent if knowing one occurred gives you no information about the other. Formally: P(A ∩ B) = P(A)P(B). Equivalently (when probabilities are nonzero): P(A|B) = P(A) and P(B|A) = P(B).
In probability, we constantly update beliefs. If you hear “B happened,” you typically revise how likely A is. Independence names the situation where that update does nothing.
Independence is important because it lets you break complex joint questions into simpler pieces. Many probability calculations become easy once you can multiply marginals instead of reasoning about complicated “overlap.”
Events A and B are independent if:
So independence is about no probabilistic influence.
Write “A is independent of B” as A ⟂ B.
The defining equation is:
P(A ∩ B) = P(A) · P(B)
This says: the probability that both happen equals the product of the probabilities of each happening alone.
From conditional probability:
P(A|B) = P(A ∩ B) / P(B) (when P(B) ≠ 0)
If A and B are independent, substitute P(A ∩ B) = P(A)P(B):
P(A|B)
= P(A ∩ B) / P(B)
= [P(A)P(B)] / P(B)
= P(A)
So independence means:
This is the “no information gained” interpretation.
Independence is exactly a statement about how the joint relates to the marginals: the joint factorizes into a product.
Independence is not something an event has “by itself.” It is a relationship between two events A and B.
Also, independence is not the same as “A and B don’t overlap.” That’s a different idea (mutual exclusivity). You can have independent events that overlap a lot, and mutually exclusive events that are almost never independent (except in trivial cases).
Depending on what information you’re given, one form of the definition is easier to use than another. Sometimes you know a joint probability; other times you know conditionals.
Below are the common equivalent ways to check independence.
Assume P(A) > 0 and P(B) > 0.
1) Product (joint = product of marginals)
P(A ∩ B) = P(A)P(B)
2) Conditional equals marginal
P(A|B) = P(A)
3) The other conditional equals marginal
P(B|A) = P(B)
These are all equivalent when the conditionals are defined.
Start from the definition of conditional probability:
P(A|B) = P(A ∩ B) / P(B)
If P(A ∩ B) = P(A)P(B), then:
P(A|B)
= P(A ∩ B) / P(B)
= [P(A)P(B)] / P(B)
= P(A)
Conversely, if P(A|B) = P(A), multiply both sides by P(B):
P(A|B) = P(A)
⇒ P(A ∩ B) / P(B) = P(A)
⇒ P(A ∩ B) = P(A)P(B)
So either form can serve as a test.
| What you know | Most convenient independence check | Notes | ||
|---|---|---|---|---|
| P(A), P(B), P(A ∩ B) | Check if P(A ∩ B) = P(A)P(B) | Direct “joint vs marginals” test | ||
| P(A), P(A | B) | Check if P(A | B) = P(A) | Requires P(B) ≠ 0 |
| P(B), P(B | A) | Check if P(B | A) = P(B) | Requires P(A) ≠ 0 |
Two events are mutually exclusive if they cannot happen together:
A ∩ B = ∅ ⇒ P(A ∩ B) = 0
If A and B were both mutually exclusive and independent, then:
P(A ∩ B) = P(A)P(B)
0 = P(A)P(B)
This forces P(A) = 0 or P(B) = 0 (a trivial event). So for nontrivial events, mutual exclusivity and independence usually contradict each other.
If A ⟂ B, then independence extends to complements:
Why? Here is one piece of the reasoning:
P(A ∩ Bᶜ)
= P(A) − P(A ∩ B)
= P(A) − P(A)P(B)
= P(A)[1 − P(B)]
= P(A)P(Bᶜ)
So A and Bᶜ also satisfy the product rule.
Independence is not directly about “separate causes” or “separate stories.” Two events may feel unrelated but still be dependent due to hidden structure. The safest intuition comes from looking at how probability mass is distributed.
In a uniform sample space, probabilities are proportional to counts.
If all outcomes are equally likely:
P(A) = |A| / |Ω|
P(B) = |B| / |Ω|
P(A ∩ B) = |A ∩ B| / |Ω|
Independence becomes:
|A ∩ B| / |Ω| = (|A|/|Ω|)(|B|/|Ω|)
⇒ |A ∩ B| = |A||B| / |Ω|
So in uniform spaces, independence means the overlap size matches what you would expect from “random mixing.”
When A and B are yes/no events, it helps to arrange probabilities like this:
| B | Bᶜ | Total | |
|---|---|---|---|
| A | P(A ∩ B) | P(A ∩ Bᶜ) | P(A) |
| Aᶜ | P(Aᶜ ∩ B) | P(Aᶜ ∩ Bᶜ) | P(Aᶜ) |
| Total | P(B) | P(Bᶜ) | 1 |
Independence says the top-left cell should equal the product of its row and column totals:
P(A ∩ B) = P(A)P(B)
And if that holds, the other cells automatically align too (because the table must sum correctly).
If P(A ∩ B) > P(A)P(B), then A and B are positively associated (A makes B more likely and vice versa).
If P(A ∩ B) < P(A)P(B), then they are negatively associated (A makes B less likely).
Independence is exactly the “neutral” middle case.
Independence does not claim:
It only claims a statement about the probability model:
P(A|B) = P(A)
A classic way this goes wrong: two events can be dependent because of a common cause, even if neither causes the other. (You don’t need to model causality to see dependence; it shows up in the numbers.)
Independence is what justifies multiplying probabilities when combining events.
If A ⟂ B:
P(A and B) = P(A)P(B)
This appears everywhere:
Many “simple” probability models assume independence because it makes calculations tractable.
But independence is an assumption you should challenge:
A useful checklist:
| Situation | Independence likely? | Why |
|---|---|---|
| Coin flips with a fair coin | Often yes | Physical resets approximate independence |
| Drawing cards without replacement | No | First draw changes composition |
| Drawing with replacement | Yes (for the same event definition per draw) | Composition resets |
| Two sensors reading same environment | Maybe not | Common noise source can induce dependence |
Even with just two events, independence is already powerful. It’s also the stepping stone to more advanced ideas:
For this node, the key habit is: When you see a joint probability, ask whether you can factor it. If you can justify independence, many problems collapse into simple arithmetic.
Without independence, you always have:
P(A ∩ B) = P(A|B)P(B)
Independence is the special case where P(A|B) becomes P(A), so:
P(A ∩ B)
= P(A|B)P(B)
= P(A)P(B)
So independence is exactly when “conditioning on B doesn’t change A.”
Suppose P(A) = 0.4, P(B) = 0.5, and P(A ∩ B) = 0.2. Are A and B independent?
Use the product test: independence requires P(A ∩ B) = P(A)P(B).
Compute the product:
P(A)P(B) = 0.4 · 0.5 = 0.20
Compare with the given joint probability:
P(A ∩ B) = 0.20
Since P(A ∩ B) = P(A)P(B), conclude A ⟂ B.
Insight: Independence is a precise numerical condition. If the joint equals the product of marginals, the events behave like “probabilistically unrelated” in this model.
You are told P(B) = 0.25 and P(A|B) = 0.60. Also P(A) = 0.60. Are A and B independent? What is P(A ∩ B)?
Independence would require P(A|B) = P(A) (assuming P(B) ≠ 0).
Check the condition:
P(A|B) = 0.60 and P(A) = 0.60
So P(A|B) = P(A). This supports independence.
Compute the joint using the conditional probability identity:
P(A ∩ B) = P(A|B)P(B)
Substitute values:
P(A ∩ B) = 0.60 · 0.25 = 0.15
Optionally verify the product form:
P(A)P(B) = 0.60 · 0.25 = 0.15
Matches P(A ∩ B), so A ⟂ B.
Insight: Often you don’t need the joint probability upfront. If conditioning on B doesn’t change A, the joint automatically becomes P(A)P(B).
Let P(A) = 0.3 and P(B) = 0.4, and suppose A and B are mutually exclusive (cannot both happen). Are they independent?
Mutual exclusivity implies:
P(A ∩ B) = 0
If A and B were independent, we would need:
P(A ∩ B) = P(A)P(B)
Compute the product:
P(A)P(B) = 0.3 · 0.4 = 0.12
Compare:
P(A ∩ B) = 0 but P(A)P(B) = 0.12
Not equal ⇒ not independent.
Insight: If two nontrivial events cannot occur together, learning one happened strongly changes the probability of the other (it becomes 0), so they cannot be independent.
Independence is a relationship between two events: A ⟂ B.
Definition: A ⟂ B ⇔ P(A ∩ B) = P(A)P(B).
Equivalent view (when defined): A ⟂ B ⇔ P(A|B) = P(A) and ⇔ P(B|A) = P(B).
Independence is about joint vs marginal probabilities: the joint “factorizes.”
Mutual exclusivity (A ∩ B = ∅) is not independence except in trivial cases (P(A)=0 or P(B)=0).
If A ⟂ B then complements are also independent: A ⟂ Bᶜ, Aᶜ ⟂ B, Aᶜ ⟂ Bᶜ.
Independence is a modeling property of the probability distribution, not a direct claim about causality.
Confusing independence with mutual exclusivity (thinking “they don’t affect each other” means “they can’t both happen”).
Forgetting the condition P(B) ≠ 0 when using P(A|B) = P(A) as a test.
Assuming events are independent because they sound unrelated in words, without checking the probabilities or the sampling process.
Mixing up joint and conditional probabilities, e.g., treating P(A ∩ B) as if it were P(A|B).
Given P(A) = 0.2, P(B) = 0.7, and P(A ∩ B) = 0.18, are A and B independent?
Hint: Compute P(A)P(B) and compare it to P(A ∩ B).
Compute the product:
P(A)P(B) = 0.2 · 0.7 = 0.14
Given P(A ∩ B) = 0.18.
Since 0.18 ≠ 0.14, A and B are not independent.
Suppose P(A) = 0.5 and P(B) = 0.4. If A ⟂ B, compute (i) P(A ∩ B), (ii) P(A ∩ Bᶜ), and (iii) P(Aᶜ ∩ B).
Hint: Use P(A ∩ B) = P(A)P(B) and complements: P(Bᶜ) = 1 − P(B).
(i) P(A ∩ B) = P(A)P(B) = 0.5 · 0.4 = 0.20
(ii) Since A ⟂ Bᶜ, P(A ∩ Bᶜ) = P(A)P(Bᶜ) = 0.5 · (1 − 0.4) = 0.5 · 0.6 = 0.30
(iii) Since Aᶜ ⟂ B, P(Aᶜ ∩ B) = P(Aᶜ)P(B) = (1 − 0.5) · 0.4 = 0.5 · 0.4 = 0.20
You know P(B) = 0.3, P(A|B) = 0.5, and P(A|Bᶜ) = 0.5. Are A and B independent? Also compute P(A).
Hint: Use the law of total probability: P(A) = P(A|B)P(B) + P(A|Bᶜ)P(Bᶜ). If P(A|B) equals P(A), then A ⟂ B.
First compute P(Bᶜ) = 1 − 0.3 = 0.7.
Now compute P(A) by total probability:
P(A) = P(A|B)P(B) + P(A|Bᶜ)P(Bᶜ)
= 0.5 · 0.3 + 0.5 · 0.7
= 0.15 + 0.35
= 0.50
Now compare P(A|B) with P(A):
P(A|B) = 0.5 and P(A) = 0.5
Since P(A|B) = P(A) (and P(B) ≠ 0), A ⟂ B.
(Equivalently, because P(A|B) and P(A|Bᶜ) are equal, knowing B occurred does not change A.)
Prerequisite refresh: Conditional Probability
Next concepts you can build from here: