Conditional Probability Notation

Reading the bar

The chance of A, given that B has happened, is written P(A | B). The vertical bar is read "given". The event whose chance is wanted goes on the left of the bar, and the event that is known to have happened goes on the right.

In the club of 24 members, the chance that a member plays sport, given that they play music, was $\frac{5}{12}$5/12. In the notation that is $P\left(\text{sport}|\text{music}\right)=\frac{5}{12}$P(sport | music) = 5/12. The other way round, $P\left(\text{music}|\text{sport}\right)=\frac{5}{13}$P(music | sport) = 5/13: the same two events with the roles swapped, and a different answer.

Given B

Call the music players A and the sport players B, so the regions hold 7 in A only, 5 in both, 8 in B only and 4 in neither. To find P(A | B), start from what is given. Only the outcomes inside B are still possible: 5 shared with A and 8 in B alone, so 13 outcomes.

Given B, only the 5 + 8 = 13 outcomes in the shaded circle count. The 7 in A only and the 4 in neither are ruled out.

The overlap out of B

Of those 13 outcomes, the 5 in the overlap are in A as well. So $P\left(A|B\right)=\frac{5}{13}$P(A | B) = 5/13.

The overlap holds the 5 outcomes in A and B. Out of the 13 in B, that is $\frac{5}{13}$5/13.

The formula

In chances, the overlap has probability $\frac{5}{24}$5/24 and the condition has $P\left(B\right)=\frac{13}{24}$P(B) = 13/24. Dividing one by the other gives the same answer as the counts, because the 24s cancel: $\left(\frac{5}{24}\right)÷\left(\frac{13}{24}\right)=\frac{5}{13}$(5/24) ÷ (13/24) = 5/13.

So P(A | B) is the chance of both, P(A and B), divided by the chance of the condition, P(B). Multiplying both sides by P(B) turns it round: P(A and B) = P(B) × P(A | B).

Tree branches are conditional

The second branches of a tree are conditional probabilities. For the bag of 3 red and 2 blue counters drawn without replacement, the branch $\frac{2}{4}$2/4 after a red is P(second red | first red). Multiplying along the path is the formula turned round: the chance of red then red is the chance of a first red, $\frac{3}{5}$3/5, times the chance of a second red given a first red, $\frac{2}{4}$2/4, so $\frac{3}{5}\times \frac{2}{4}=\frac{3}{10}$3/5 × 2/4 = 3/10.

For independent events, knowing B changes nothing, so P(A | B) = P(A). A dice and a coin are independent, so $P\left(\text{head}|\text{six}\right)=\frac{1}{2}=P\left(\text{head}\right)$P(head | six) = 1/2 = P(head). In the club, $P\left(A\right)=\frac{12}{24}=\frac{1}{2}$P(A) = 12/24 = 1/2 but $P\left(A|B\right)=\frac{5}{13}$P(A | B) = 5/13, so playing music and playing sport are not independent.

R for a red counter and B for a blue one. The second branch on the colored path, $\frac{2}{4}$2/4, is the chance of a second red given a first red. The first branch, $\frac{3}{5}$3/5, is the chance of the condition.

From the chances alone

Out of 30 equally likely outcomes, the chance of both A and B is $\frac{6}{30}$6/30 and $P\left(B\right)=\frac{15}{30}$P(B) = 15/30. Then $P\left(A|B\right)=\frac{6}{15}=\frac{2}{5}$P(A | B) = 6/15 = 2/5: given B, 15 outcomes are left, and 6 of them are in A.

The usual mistakes

Reading the bar the wrong way round. P(A | B) divides by B; P(B | A) divides by A. With the club, $\frac{5}{13}$5/13 and $\frac{5}{12}$5/12.

Writing P(A and B) for P(A | B). The chance of both, $\frac{6}{30}$6/30, is out of every outcome, and no condition has narrowed them yet; P(A | B) is out of the 15 in B.

Counting the part of B that is not in A: $\frac{9}{15}$9/15. The question asks for the outcomes in B that are in A as well, the overlap.

Cyclists and helmets

In the application below, a two-way table of 160 cyclists sorts them by age and by helmet. The condition after the bar chooses the row or the column whose total goes underneath.