MISSION · Data and probability
Conditional Probability
How new information changes the odds
LEVEL 1 · EXPLORE
Make the idea move
Change the controls and watch what happens. Follow your curiosity and see what you notice.
Loading the activity…
LEVEL 2 · LEARN
Unlock the explanation
Open the chapters in any order. Try the examples as you go.
The idea in one sentence
\(P(A \mid B)\) is the probability of \(A\) given that \(B\) has happened — the same question asked in a smaller world.
\[P(A \mid B) = \frac{P(A \text{ and } B)}{P(B)}\]Shrinking the sample space
Roll a die. The chance of a 2 is \(\frac{1}{6}\). Now someone tells you the result is even. Three outcomes remain — 2, 4, 6 — and the chance of a 2 is now \(\frac{1}{3}\).
Nothing about the die changed. What changed is the set of outcomes still in play. That’s all conditioning is: discard the outcomes ruled out, then recount within what’s left. The formula says the same thing — divide by \(P(B)\) to rescale the reduced world back to a total of 1.
Independence
Two events are independent when conditioning changes nothing:
\[P(A \mid B) = P(A)\]Two coin flips are independent — the first tells you nothing about the second. Drawing two cards without replacement is not: removing an ace leaves 51 cards and 3 aces, so the second draw’s odds have genuinely moved.
For independent events the multiplication rule is simple, \(P(A \text{ and } B) = P(A) \times P(B)\). For dependent ones the second factor must be conditional:
\[P(A \text{ and } B) = P(A) \times P(B \mid A)\]Tree diagrams exist to keep this straight. Each branch after the first carries a conditional probability, and multiplying along a path gives that path’s chance.
The two directions are not the same
\(P(A \mid B)\) and \(P(B \mid A)\) are different numbers, and confusing them is one of the most consequential errors in applied statistics.
Nearly all footballers own boots. That doesn’t make boot-owners likely footballers.
The serious version shows up in medical testing. A test is 99% accurate for a disease affecting 1 person in 10,000. You test positive. The chance you have the disease is not 99% — it’s under 1%. Among a million people, 100 have it and test positive; roughly 10,000 healthy people also test positive. You are one of 10,100 positives, only 100 of whom are ill.
The test’s accuracy is \(P(\text{positive} \mid \text{ill})\). What you want is \(P(\text{ill} \mid \text{positive})\). The base rate — how rare the disease is — governs the gap between them, and ignoring it is so common it has a name: the base rate fallacy.
Common mix-ups
- Swapping the two directions. \(P(A \mid B) \ne P(B \mid A)\).
- Assuming independence when events clearly affect each other.
- Ignoring the base rate when interpreting a test result.
- Forgetting the world shrank. After conditioning, probabilities must still total 1 over the remaining outcomes.
Check yourself
A bag holds 4 red and 6 blue counters. Two are drawn without replacement. What’s the probability both are red?
\(\frac{4}{10} \times \frac{3}{9} = \frac{12}{90} = \frac{2}{15}\). The second factor is conditional — one red is already gone, so 3 reds remain among 9 counters.
LEVEL 3 · REFLECT
Add this quest to your trail
When you feel ready, mark this idea as explored. You can always come back and try it again.
EXPLORER STRATEGIES