AcademyProbability
Academy
Conditional Probability
Level 1 - Math II (Physics) topic page in Probability.
Principle
Conditional probability updates the sample space after information is known. The probability \(P(A|B)\) is the probability of \(A\) after restricting attention to outcomes in \(B\), and it is defined only when \(P(B)>0\).
Notation
\(A,B\)
events in the same sample space
\(P(A|B)\)
probability of A given that B occurred
\(P(B)\gt 0\)
condition required before conditioning on B
\(A\cap B\)
event that both A and B occur
Method
Step 1: Identify the given event
The event after the vertical bar is the new reference event. For \(P(A|B)\), restrict attention to \(B\).
Step 2: Find the overlap
Only outcomes in \(A\cap B\) count as favourable after conditioning on \(B\).
Step 3: Divide by the probability of the given event
The new total probability is \(P(B)\), so divide the overlap probability by \(P(B)\).
New sample space
\[B\]
Favourable part inside B
\[A\cap B\]
Conditional probability
\[P(A|B)=\frac{P(A\cap B)}{P(B)}\]
Multiply by P(B)
\[P(A\cap B)=P(A|B)P(B)\]
Rules
Conditional probability
\[P(A|B)=\frac{P(A\cap B)}{P(B)}\quad(P(B)\gt 0)\]
Multiplication rule
\[P(A\cap B)=P(A|B)P(B)\]
Reverse multiplication
\[P(A\cap B)=P(B|A)P(A)\quad(P(A)\gt 0)\]
Chain form
\[P(A\cap B)=P(A)P(B|A)\]
Examples
Question
A fair die is rolled. Given that the result is even, find the probability that it is greater than 3.
Answer
Let
\[B=\{2,4,6\}\]
and \[A=\{4,5,6\}\]
Then \[A\cap B=\{4,6\}\]
so \[P(A|B)=\frac{P(A\cap B)}{P(B)}=\frac{2/6}{3/6}=\frac{2}{3}.\]
Checks
- The conditioning event must have nonzero probability.
- \(P(A|B)\) need not equal \(P(B|A)\).
- The event after the vertical bar is the information being assumed.
- Conditional probabilities are still probabilities, so they must lie between \(0\) and \(1\).