AcademyProbability
Academy
Partitions
Level 1 - Math II (Physics) topic page in Probability.
Principle
A partition breaks the sample space into exhaustive, non-overlapping cases. Total probability computes an event by adding its probability across those cases.
Notation
\(S\)
sample space
\(E_1,E_2,\ldots,E_n\)
partition events
\(E_i\cap E_j=\varnothing\)
different partition cases do not overlap
\(E_1\cup\cdots\cup E_n=S\)
the cases are exhaustive
\(A\)
event whose probability is being computed
\(P(A|E_i)\)
probability of A within case E_i
Method
Step 1: Check the cases form a partition
Every outcome must be in exactly one case \(E_i\). This means the cases are exhaustive and non-overlapping.
Step 2: Split the event by cases
Intersect \(A\) with each partition case.
Start with exhaustive cases
\[S=E_1\cup E_2\cup\cdots\cup E_n\]
Intersect with A
\[A=A\cap S\]
Substitute the partition
\[A=A\cap(E_1\cup E_2\cup\cdots\cup E_n)\]
Distribute intersection
\[A=(A\cap E_1)\cup(A\cap E_2)\cup\cdots\cup(A\cap E_n)\]
Use disjoint additivity
\[P(A)=P(A\cap E_1)+P(A\cap E_2)+\cdots+P(A\cap E_n)\]
Use multiplication rule
\[P(A)=P(A|E_1)P(E_1)+\cdots+P(A|E_n)P(E_n)\]
Rules
Partition conditions
\[E_i\cap E_j=\varnothing\ (i\ne j),\qquad E_1\cup\cdots\cup E_n=S\]
Total probability
\[P(A)=\sum_{i=1}^n P(A\cap E_i)\]
Conditional total probability
\[P(A)=\sum_{i=1}^n P(A|E_i)P(E_i)\]
Two-case form
\[P(A)=P(A|E)P(E)+P(A|E^c)P(E^c)\]
Examples
Question
A disease affects 2 percent of a population. A test is positive with probability 0.95 for diseased people and 0.04 for non-diseased people. Find the probability of a positive test.
Answer
Use the partition diseased and not diseased:
\[P(T)=0.95\cdot0.02+0.04\cdot0.98=0.019+0.0392=0.0582.\]
Checks
- Partition cases must be exhaustive: together they cover all of \(S\).
- Partition cases must be non-overlapping: no outcome belongs to two different cases.
- Each conditional probability \(P(A|E_i)\) requires \(P(E_i)>0\).
- Total probability is a weighted sum of case probabilities.