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Conditional Probability

P(A∣B)=P(A∩B)P(B)P(A \mid B) = \frac{P(A \cap B)}{P(B)}

Worked example: P(A|B) 0.4 times P(B) 0.25 → joint 0.1 — press Try an example to run it live, then adjust anything.

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Conditional Probability explained

P(B)P(A∩B)P(A|B)

Conditioning throws away every outcome in which B did not happen and re-normalises what is left, which is why you divide by P(B). Roll one die: P(it is a 2) = 1/6, but if you are told the roll is even, only three outcomes survive and P(2 | even) = (1/6)/(1/2) = 1/3. Probabilities go in as decimals from 0 to 1 (the % unit is available too).

In September 1990 Marilyn vos Savant told Parade readers that a contestant on Let's Make a Deal should switch doors after Monty Hall opens a losing one, because switching wins 2/3 of the time. Some 10,000 letters arrived, roughly a thousand from people with PhDs, insisting she was wrong; Paul Erdős stayed unconvinced until he saw a simulation. The host's choice is not independent of where the car is, and conditioning on which door he opened is the whole game. The standard trap is confusing P(A|B) with P(B|A): almost every fatal cycling accident involves a bicycle, yet almost no bicycle ride is fatal.

Conditional Probability formula

P(A∣B)=P(A∩B)P(B)P(A \mid B) = \frac{P(A \cap B)}{P(B)}
Where
  • P(A∣B)P(A \mid B)= Probability of A given B
  • P(A∩B)P(A \cap B)= Probability both occur
  • P(B)P(B)= Probability of B

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