Conditional probability from exact representations lesson | Mathematical Methods (Year 12, VCE Units 3 & 4, Discrete random variables, Discrete distribution events and transformations)

A membership	\(B\)	\(B^{\prime}\)	Total
\(A\)	\(\frac{3}{26}\)	\(\frac{5}{13}\)	\(\frac{1}{2}\)
\(A^{\prime}\)	\(\frac{2}{13}\)	\(\frac{9}{26}\)	\(\frac{1}{2}\)
Total	\(\frac{7}{26}\)	\(\frac{19}{26}\)	\(1\)

A membership	\(B\)	\(B^{\prime}\)	Total
\(A\)	\(\frac{5}{21}\)	\(\frac{2}{7}\)	\(\frac{11}{21}\)
\(A^{\prime}\)	\(\frac{5}{14}\)	\(\frac{5}{42}\)	\(\frac{10}{21}\)
Total	\(\frac{25}{42}\)	\(\frac{17}{42}\)	\(1\)

A membership	\(B\)	\(B^{\prime}\)	Total
\(A\)	\(\frac{5}{17}\)	\(\frac{5}{17}\)	\(\frac{10}{17}\)
\(A^{\prime}\)	\(\frac{11}{51}\)	\(\frac{10}{51}\)	\(\frac{7}{17}\)
Total	\(\frac{26}{51}\)	\(\frac{25}{51}\)	\(1\)

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 probability must be positive. Counts can replace probabilities when the outcomes are equally likely

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

Read after the bar In Pr(A|B), event B is the known condition and the reduced sample space \(\displaystyle \Pr(A\mid B)\text{ uses }B\)
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 Divide by the probability or count of the conditioning event, not the whole space \(\displaystyle \Pr(A\mid B)=\frac{\Pr(A\cap B)}{\Pr(B)}\)
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 probability of A given B.

\Pr(A\mid B)=\frac{n(A\cap B)}{n(B)}

\(\displaystyle \Pr(A\mid B)\text{ has condition }B\)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 Pr(A|B)?+

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

Why is the numerator an intersection?+

The required outcome must satisfy A and the known condition B at the same time, so it belongs to A ∩ B.

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

They use different reduced sample spaces. The first divides by the size of B and the second divides by the size of A.

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 10: Discrete random variables

In this chapter

Validate, complete and normalise a discrete probability distribution Finite events, probability assignments and complements Distribution of a finite random variable Represent a multi-stage finite sample space Expected value and linear expectation of a discrete random variable Variance, standard deviation and linear variance of a discrete random variable Interval, conditional and μ ± 2σ probabilities from a discrete distribution Distribution of a linear transformation of a discrete random variable Conditional probability from exact representations

Search

Conditional probability from exact representations

Contents

Conditional probability

Keep only the known outcomes

Use the margins to recover missing counts

A joint-probability table is not a list of equal masses

Reverse the quotient by multiplication

Condition on values of a random variable

Renormalise within the positive condition

Construct and explain a complete table, ungraded

Check the regions before the conditional ratios

Check the known event before using a formula

Checks to finish

Conditional probability

Worked example

Common mistakes

Difficulty levels

Questions about this topic

In this chapter