Conditional probability from exact representations lesson | Mathematical Methods (Year 11, VCE Units 1 & 2, Probability and counting, Two-way tables and conditional probability)

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. The conditioning event must have positive probability

\begin{gathered}\Pr(B)>0\\[4pt]\Pr(A\mid B)=\frac{\Pr(A\cap B)}{\Pr(B)}\end{gathered}

Read after the bar The event after the vertical bar is the known condition and 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 or count, not by the whole space \(\displaystyle \begin{gathered}\Pr(B)>0\\[4pt]\Pr(A\mid B)=\frac{\Pr(A\cap B)}{\Pr(B)}\end{gathered}\)
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

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 the displayed conditional probability

\Pr(A\mid B)=\frac{n(A\cap B)}{n(B)},\qquad n(B)>0

\(\displaystyle \Pr(A\mid B)\text{ has condition }B\)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 B 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 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 a conditional probability?+

The event after the bar is known and must have positive probability. Divide the probability of the overlap by the probability of that condition

\begin{gathered}\Pr(B)>0\\[4pt]\Pr(A\mid B)=\frac{\Pr(A\cap B)}{\Pr(B)}\end{gathered}

Why is the numerator an intersection?+

The required outcome must belong to both the required event and the known conditioning event

A\cap B

Why do the two directions usually give different probabilities?+

They use different reduced sample spaces. Each direction requires its own conditioning event to have positive probability

\begin{aligned}\Pr(A\mid B)&=\frac{\Pr(A\cap B)}{\Pr(B)}\\[4pt]&\Pr(B)>0\\[4pt]\Pr(B\mid A)&=\frac{\Pr(A\cap B)}{\Pr(A)}\\[4pt]&\Pr(A)>0\end{aligned}

Can I use a calculator?+

A calculator is allowed unless a question is marked no calculator

Which year level and chapter is this?+

Year 11, VCE Units 1 & 2, Mathematical Methods, chapter 8: Probability and counting

In this chapter

Finite events, probability assignments and complements Represent a multi-stage finite sample space Compound events, the addition rule and mutual exclusivity Complete and use a two-way probability table Conditional probability from exact representations Distribution of a finite random variable Design and interpret a probability simulation Count restricted selections and calculate exact probabilities Count arrangements with product, sum and restriction rules Test exact independence Exact probability of repeated independent successes Multi-stage probability, multiplication and total probability Probability from a uniform area model Probabilities of sequences of independent events

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Conditional probability from exact representations

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