Conditional probability
A conditional probability restricts attention to outcomes where the conditioning event is known. The event after the vertical bar supplies the denominator, and the intersection supplies the numerator
\[\begin{gathered}\Pr(A\mid B)=\frac{\Pr(A\cap B)}{\Pr(B)}\\\Pr(B)>0\end{gathered}\]
- Read after the bar The event after the vertical bar is the known condition and becomes the reduced sample space \(\displaystyle \Pr(A\mid B)\text{ uses }B\)
- Use the intersection The numerator contains outcomes that satisfy both the required event and the condition \(\displaystyle \Pr(A\cap B)\)
- Use the condition denominator The conditioning event must have positive probability. Divide by its probability, not the whole sample space. If outcomes are equally likely, divide the intersection count by the condition count \(\displaystyle \begin{gathered}\Pr(A\mid B)=\frac{\Pr(A\cap B)}{\Pr(B)}\\\Pr(B)>0\end{gathered}\)
- Keep directions separate Reversing the events usually changes the reduced sample space and the answer \(\displaystyle \Pr(A\mid B)\ne\Pr(B\mid A)\text{ in general}\)
Worked example
All outcomes are equally likely. The complete sample-space count and the two event counts are shown. Find the displayed conditional probability
\[\begin{gathered}n(S)=30,\quad n(A)=12\\n(B)=15,\quad n(A\cap B)=6\\\Pr(A\mid B)=?\end{gathered}\]
- \(\displaystyle \Pr(A\mid B)\text{ has condition }B\)The event after the bar is the reduced sample space
- \(\displaystyle n(A\cap B)=6,\quad n(B)=15\)Use the intersection for the numerator and the known event for the denominator
- \(\displaystyle \Pr(A\mid B)=\frac{6}{15}=\frac{2}{5}=0.4\)Divide the intersection count by the count in the conditioning event and reduce
Common mistakes
Reversing the condition
✕\(\displaystyle \Pr(A\mid B)=\frac{6}{12}=\frac{1}{2}\)✓\(\displaystyle \Pr(A\mid B)=\frac{6}{15}=\frac{2}{5}\)
The event after the bar supplies the denominator, not the event before itUsing the whole space as denominator
✕\(\displaystyle \Pr(A\mid B)=\frac{6}{30}=\frac{1}{5}\)✓\(\displaystyle \Pr(A\mid B)=\frac{6}{15}=\frac{2}{5}\)
A conditional probability uses the known event as the reduced sample space, not all outcomesUsing a marginal numerator
✕\(\displaystyle \Pr(A\mid B)=\frac{12}{15}=\frac{4}{5}\)✓\(\displaystyle \Pr(A\mid B)=\frac{6}{15}=\frac{2}{5}\)
Count only outcomes in both events for the numerator, not every outcome in the required eventDifficulty levels
introReduce an equiprobable listed space to the outcomes in the conditioning event
standardCalculate from a completed count or probability table
challengeCalculate both directions or recover an intersection from a stated relation
examChoose the correct formula orientation and calculate an exact conditional probability
Questions about this topic
What does the vertical bar in a conditional probability mean?+
It means that the event after the bar is known to occur. That event must have positive probability and supplies the denominator
Why is the numerator an intersection?+
A favourable outcome must satisfy both the required event and the known condition
Why does reversing the two events usually change the answer?+
The two calculations generally use different reduced sample spaces and therefore different denominators
Can I use a calculator?+
A calculator is allowed unless a question is marked no calculator
Which year level and chapter is this?+
Year 12, VCE Units 3 & 4, Mathematical Methods, chapter 12: Continuous probability distributions
In this chapter
12A Continuous random variables and density functions: Validate and normalise a continuous probability density function12A Continuous random variables and density functions: Estimate probabilities from grouped continuous frequency dataGroup numerical data into class intervalsConstruct a histogram from a frequency table12A Continuous random variables and density functions: Finite events, probability assignments and complements12A Continuous random variables and density functions: Conditional probability from exact representationsEvaluate and determine function rules12B Probability density validity and normalisation: Validate and normalise a continuous probability density function12B Probability density validity and normalisation: Estimate probabilities from grouped continuous frequency dataSolve a linear equation12B Probability density validity and normalisation: Evaluate a polynomial or recover a coefficient parameterWork with sets and interval notation12B Probability density validity and normalisation: Round to decimal places and significant figuresInterval and conditional probabilities from a densityCompound events, the addition rule and mutual exclusivity12C Interval and conditional probabilities: Finite events, probability assignments and complements12C Interval and conditional probabilities: Conditional probability from exact representationsMean, variance and standard deviation of a continuous random variableMedian, quartiles and percentiles of a continuous random variableExpand and factor polynomial expressions12D Mean, variance, standard deviation and percentiles: Evaluate a polynomial or recover a coefficient parameterMove between roots, factors, polynomial rules and equationsSolve a quadratic exactly with the quadratic formula12D Mean, variance, standard deviation and percentiles: Round to decimal places and significant figuresLinear transformations and totals of independent continuous random variables