Conditional Probability

Learn to update probabilities on new information by restricting to a smaller sample space.

1. The idea

Conditioning on an event $B$ means restricting attention to the outcomes where $B$ happened, then asking what fraction of those also satisfy $A$ . This gives $P (A ∣ B) = \frac{P ( A \cap B )}{P ( B )}$ .

The key mental shift is that $B$ becomes the new universe: probabilities outside $B$ are irrelevant, and probabilities inside $B$ get rescaled so they sum to 1. Any outcome not compatible with $B$ simply gets probability 0.

Interview traps almost always involve subtly changing the sample space. "Given the sum is 8" is a very different question from "one die shows 5," and the right approach is to explicitly relist the outcomes consistent with the condition.

2. Why it matters

Interviewers use conditional probability to test whether you actually re-derive the restricted sample space or just recycle unconditional numbers. It's also the backbone of Bayes' theorem and of most sequential/updating questions.

3. Core mathematics

Definition

P (A ∣ B) = \frac{P ( A \cap B )}{P ( B )}, P (B) > 0

Multiplication rule

P (A \cap B) = P (A ∣ B) P (B)

Chain rule (general)

P (A_{1} \cap \dots \cap A_{n}) = P (A_{1}) P (A_{2} ∣ A_{1}) \dots P (A_{n} ∣ A_{1} \cap \dots \cap A_{n - 1})

4. How to recognize it

Think of this when…

The problem says "given that" or "it is known that" some event occurred.
You are asked to update a probability after observing partial information.
The question can be answered by relisting the outcomes consistent with the condition.
Sequential draws without replacement, where each draw changes the remaining sample space.

5. Worked example

Worked example

Two fair dice are rolled. Given that the sum is 8, what is the probability that one of the dice shows a 5?

1
List all ordered pairs summing to 8: (2,6), (3,5), (4,4), (5,3), (6,2).
$P (sum = 8) = 5/36$
2
Within this restricted space, find outcomes where a 5 appears: (3,5) and (5,3).
$∣ A \cap B ∣ = 2$
3
Apply the conditional probability definition using outcome counts within B.
$P (A ∣ B) = \frac{2}{5}$
Answer: $P (a 5 appears ∣ sum = 8) = 2/5$

6. Common mistakes

Using the unconditional probability of A instead of recomputing within the restricted sample space B.
Confusing $P (A ∣ B)$ with $P (B ∣ A)$ — these are generally different (see Bayes' theorem).
Forgetting that once you condition on B, outcomes outside B have probability zero and should be discarded, not merely down-weighted.
Applying independence-based shortcuts to events that are only independent unconditionally, not conditionally.

7. Quick check

Two fair dice are rolled given the sum is 8. What is $P (one die shows 5)$ ?

$P (A ∣ B)$ equals $P (B ∣ A)$ whenever...

8. Practice

39 QuantSolve questions use this concept.

Practice this concept

9. Related methods

Conditioning on the First Event

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Recommended next lessonBayes' Theorem →