Search topics

Search

Conditional probability from exact representations

Start a set of 8
Step 2 of 5

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

\[\Pr(A\mid B)=\frac{\Pr(A\cap B)}{\Pr(B)},\quad\Pr(B)>0\]
  1. 1Read after the bar In Pr(A|B), event B is the known condition and the reduced sample space \(\displaystyle \Pr(A\mid B)\text{ uses }B\)
  2. 2Use the intersection The numerator contains outcomes that satisfy both the required event and the condition \(\displaystyle \Pr(A\cap B)\)
  3. 3Use the condition denominator Divide the intersection probability by Pr(B), which must be positive. With equally likely outcomes, the same quotient can be calculated from counts \(\displaystyle \Pr(A\mid B)=\frac{\Pr(A\cap B)}{\Pr(B)},\quad\Pr(B)>0\)
  4. 4Keep directions separate Reversing the events usually changes the reduced sample space and the answer \(\displaystyle \Pr(A\mid B)\ne\Pr(B\mid A)\text{ in general}\)
Year 12, VCE Units 3 & 4, Chapter 14, 4 difficulty levels, Calculator allowedBuild a worksheet

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

\[\Pr(A\mid B)=\frac{\Pr(A\cap B)}{\Pr(B)},\quad\Pr(B)>0\]
  1. Read after the bar In Pr(A|B), event B is the known condition and the reduced sample space \(\displaystyle \Pr(A\mid B)\text{ uses }B\)
  2. Use the intersection The numerator contains outcomes that satisfy both the required event and the condition \(\displaystyle \Pr(A\cap B)\)
  3. Use the condition denominator Divide the intersection probability by Pr(B), which must be positive. With equally likely outcomes, the same quotient can be calculated from counts \(\displaystyle \Pr(A\mid B)=\frac{\Pr(A\cap B)}{\Pr(B)},\quad\Pr(B)>0\)
  4. 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

A finite experiment has 30 equally likely outcomes. Event A contains 12 outcomes, event B contains 15 outcomes, and 6 outcomes are in both events. Find Pr(A|B)

\[\Pr(A\mid B)=\frac{n(A\cap B)}{n(B)}\]
  1. \(\displaystyle \Pr(A\mid B)\text{ has condition }B\)The event after the bar is the reduced sample space
  2. \(\displaystyle n(A\cap B)=6,\quad n(B)=15\)Use the intersection for the numerator and the known event B for the denominator
  3. \(\displaystyle \Pr(A\mid B)=\frac{6}{15}=\frac{2}{5}=0.4\)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 is B, so the denominator must count B rather than A.
Using 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 outcomes
Using a marginal numerator
✕\(\displaystyle \Pr(A\mid B)=\frac{12}{15}=\frac{4}{5}\)✓\(\displaystyle \Pr(A\mid B)=\frac{6}{15}=\frac{2}{5}\)
The numerator must contain outcomes in both A and B, not every outcome in A.

Difficulty 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 mean in Pr(A|B)?+

It means that B is known. The calculation is the probability of A and B divided by the probability of B. The probability of B must be positive

Why is the numerator an intersection?+

The required outcome must be in both A and B, so it belongs to their intersection

Why are Pr(A|B) and Pr(B|A) usually different?+

They divide the same intersection probability by different conditioning probabilities. Use Pr(B) for Pr(A|B) and Pr(A) for Pr(B|A), provided the respective denominator is positive

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 14: Statistical inference

In this chapter

Populations, samples, parameters, statistics and sampling methodsAllocate a stratified random sampleFind a percentage of a quantityCalculate a sample proportion and list its possible valuesExact sampling distribution of a sample proportionFinite events, probability assignments and complementsConditional probability from exact representationsEvaluate factorial and combination expressionsConvert between fractions, decimals and percentagesLarge-sample conditions and the sample-proportion distributionCalculate a standard score or actual valueUse z-score boundaries to estimate normal-model percentages and countsCalculate the standard deviation from the formulaSolve a linear equationSolve a quadratic exactly with the quadratic formulaApproximate confidence intervals for a population proportionRearrange literal equations and recognise equivalent line formsInterpret simulated sampling distributions and interval variationDesign and interpret a probability simulation