Conditional Probability
What happened already changes what's likely next -- a vertical line and a restricted sample space make that precise.
When one event changes another's probability
The probability of an event can change depending on the outcome of a previous event. For example, the probability of being late for work may change depending on whether you oversleep or not.
Situations like this can be modelled using conditional probability. A vertical line is used to indicate conditional probabilities.
The probability that $B$ occurs given that $A$ has already occurred is written as $\mathrm{P}(B\mid A)$. Similarly, $\mathrm{P}(B\mid A')$ describes the probability of $B$ occurring given that $A$ has not occurred.
Knowing that $B$ has happened shrinks the sample space down to just $B$ — $\mathrm{P}(A\mid B)$ asks what fraction of that smaller space is also inside $A$. This idea of shrinking the sample space is called a restricted sample space, and it's the key to solving most conditional probability problems without needing the formula at all.
For independent events, $\mathrm{P}(A\mid B)=\mathrm{P}(A\mid B')=\mathrm{P}(A)$, and $\mathrm{P}(B\mid A)=\mathrm{P}(B\mid A')=\mathrm{P}(B)$. You can use this condition to determine independence: if conditioning on $A$ ever changes $B$'s probability, the events cannot be independent.
Worked example 1
a Draw a two-way table to show this information.
| $C$ | $C'$ | Total | |
|---|---|---|---|
| $W$ | 16 | 34 | 50 |
| $W'$ | 20 | 20 | 40 |
| Total | 36 | 54 | 90 |
The values in bold above are found from the totals: $16+34=50$ (row total), $90-50-20=20$ (the remaining members attend neither), and the column/grand totals follow the same way.
b Find: i $\mathrm{P}(W'\cap C')$ ii $\mathrm{P}(W\mid C)$ iii $\mathrm{P}(C\mid W')$
For $\mathrm{P}(W\mid C)$, don't divide by 90 — divide by the row or column total for the given condition ($36$, the total number of cardio members), not the grand total. This is exactly the restricted sample space idea: $C$ is now the entire world under consideration.
Worked example 2
State one modelling assumption used in your calculations.
Each of the 36 mark-pairs is equally likely — that is, every mark from 1 to 6 is equally likely for each student.
Counting directly off the sample-space diagram is faster than computing $\mathrm{P}(\text{sum}=8\cap\text{one mark}=5)$ and $\mathrm{P}(\text{one mark}=5)$ separately and dividing — whenever every outcome is equally likely, a restricted sample space collapses to simple counting.
Check yourself
Of 40 sixth-formers, 22 have a part-time job and 15 volunteer regularly; 9 both work and volunteer. Draw a two-way table, then find $\mathrm{P}(\text{volunteers}\mid\text{has a job})$.