1
Conditioning restricts the space
P(A|B) counts the outcomes in A among those in B. The denominator is |B|, not |S|.
Lesson · Discrete Probability · Conditioning
Conditional Probability
P(A | B) = P(A ∩ B)/P(B) is not a new axiom: it is the original counting restricted to the outcomes where B holds. The traps are all about which of P(A|B) and P(B|A) you have computed.
P(A | B) = P(A ∩ B) / P(B) requires P(B) > 0
P(A ∩ B) = P(A | B) · P(B) the multiplication rule
P(A | B) ≠ P(B | A) in general
conditioning SHRINKS the sample space to B
A smaller sample space
Core ideas
What this lesson establishes
Conditioning is re-counting inside a subset. Everything difficult about it is about interpretation, not arithmetic.
2
The two conditionals are different numbers
P(test positive | disease) and P(disease | test positive) differ by the base rate, often by orders of magnitude.
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The multiplication rule chains
P(A ∩ B ∩ C) = P(A)P(B|A)P(C|A∩B) — useful for sequential draws, where each step conditions on the previous ones.
The material
Conditioning
The definition, why it is a definition rather than a theorem, and the standard traps.
Definition · Conditional probability
For events A, B with P(B) > 0, P(A | B) = P(A ∩ B) / P(B).
In the equally-likely model this is transparent: P(A|B) = (|A ∩ B|/|S|)/(|B|/|S|) = |A ∩ B|/|B|. The outcomes outside B have been discarded, and B is now the sample space.
Theorem · Conditional probability is a probability
For fixed B with P(B) > 0, the function A ↦ P(A|B) satisfies all three axioms. So every theorem about probability holds for conditional probability too.
Theorem · Multiplication rule
P(A ∩ B) = P(A|B)P(B) = P(B|A)P(A), and more generally P(A₁ ∩ ⋯ ∩ A_n) = P(A₁)P(A₂|A₁)P(A₃|A₁∩A₂)⋯.
Example · Sequential draws
Two cards from a deck without replacement. P(both aces) = P(first ace) · P(second ace | first ace) = (4/52)(3/51) = 1/221. Conditioning is exactly what handles "without replacement": the second probability changes because the first draw changed the deck.
The two conditionals are different
Example · Dice
Roll two dice. P(sum is 7 | first die is 4) = 1/6, since among the six outcomes with a 4 first, exactly one has sum 7. But P(first die is 4 | sum is 7) = 1/6 too — a coincidence arising from symmetry, not a rule.
Example · Where they differ wildly
P(is a mammal | is a dog) = 1, while P(is a dog | is a mammal) is small. The two conditionals share a numerator P(A ∩ B) and have different denominators, and the ratio of the denominators is exactly how much they differ.
The Monty Hall problem
Three doors, a prize behind one. You pick door 1. The host, who knows where the prize is, opens a different door with no prize behind it — say door 3 — and offers you a switch. Should you?
prize behind your choice host must open switching wins?
door 1 door 1 door 2 or 3 NO
door 2 door 1 door 3 YES
door 3 door 1 door 2 YES
Each row has probability 1/3, so switching wins with probability 2/3.
The step people miss is that the host's action is not random: he knows and is constrained to open a losing door. That constraint is information, and conditioning on it is what makes switching better. If the host opened a door at random and it happened to be empty, the answer would be 1/2 — a different problem with the same picture.
Interactive
Sample space, event, condition
Exact fractions, every outcome listed
Method
Conditioning correctly
Say which event is the condition, out loud.
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Identify the conditioning event
P(A|B) conditions on B. Writing the sentence "given that B happened" prevents the swap.
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Restrict the sample space
In the equally-likely model, count within B. The denominator changes and the numerator becomes |A ∩ B|.
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Use the multiplication rule for sequences
Each draw conditions on the previous ones. Without replacement, the denominators shrink.
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Ask whether you wanted the other conditional
The two are related by Bayes, which is “Bayes' Theorem”. Computing the wrong one is the most consequential error in this course.
Conditioning
Score 0 / 0Choose the answer you can justify.
The two-child problem
A family has two children. The sample space is {BB, BG, GB, GG}, equally likely.
"At least one is a boy." What is P(both boys)?
B = {BB, BG, GB} |B| = 3
A ∩ B = {BB} |A ∩ B| = 1
P(A | B) = 1/3
"The ELDER is a boy." What is P(both boys)?
B' = {BB, BG} |B'| = 2
A ∩ B' = {BB} |A ∩ B'| = 1
P(A | B') = 1/2
Same family, two different pieces of information, two answers.
The difference is real and not a trick: "at least one is a boy" and "the elder is a boy" restrict the sample space to sets of different sizes. Almost every disputed probability puzzle turns out to be an ambiguity about which restriction was intended.
Common mistakes
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Swapping the two conditionalsP(A|B) and P(B|A) are different numbers with the same numerator. “Bayes' Theorem” gives the exact relationship.
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Forgetting the denominator changedConditioning restricts the sample space. Dividing by |S| instead of |B| gives P(A ∩ B), not P(A|B).
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Ignoring how the information arrivedMonty Hall and the two-child problem both hinge on it. "At least one boy" and "the elder is a boy" are different conditions.
Check your understanding
✓
Finish when you can state the restricted sample space.
For each conditional probability you compute, write down explicitly which outcomes remain. That sentence catches the swap, the wrong denominator and the ambiguity all at once.
Note: Conditional probability is where probability stops being counting and starts being modelling. The arithmetic never gets harder than division; deciding what to condition on is where the difficulty lives.