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Multi-stage probability, multiplication and total probability

Start a set of 8
Step 1 of 5
Two-stage red or blue probability tree, with branch probabilities conditional on the first outcome and ordered terminal paths RR, RB, BR and BB First branches: R 1/2, B 1/2. After R: R 1/2, B 1/2. After B: R 3/4, B 1/4.FirstSecondR1/2R1/2RRB1/2RBB1/2R3/4BRB1/4BB
First stage
\(\displaystyle \Pr(R_1)=\Pr(B_1)=\frac12\)
After red
\(\displaystyle \Pr(R_2\mid R_1)=\Pr(B_2\mid R_1)=\frac12\)
After blue
\(\displaystyle \Pr(R_2\mid B_1)=\frac34,\quad\Pr(B_2\mid B_1)=\frac14\)

Each of two stages has mutually exclusive and exhaustive outcomes R (red) and B (blue). How is the probability of the ordered path R then B calculated?

Year 11, VCE Units 1 & 2, Chapter 8, 4 difficulty levels, No calculatorBuild a worksheet

Multi-stage probability trees

A probability tree separates stages into branches. Update any changed state, multiply down one path, and add path probabilities when different terminal paths describe the same event

\[\Pr(\text{path})=\prod\Pr(\text{branch}),\quad\Pr(E)=\sum_{p\in E}\Pr(p)\]
  1. Update the state Without replacement, reduce the selected category and the total before the next stage \(\displaystyle \Pr(R_2\mid R_1)=\frac{2}{4}\quad\text{after 3 red and 2 blue items start}\)
  2. Check sibling branches The branch probabilities leaving any node must add to one \(\displaystyle \frac{2}{5}+\frac{3}{5}=1\)
  3. Multiply down a path All branches on one ordered path are required together \(\displaystyle \Pr(RB)=\frac{1}{2}\times\frac{1}{3}=\frac{1}{6}\)
  4. Add alternative paths Disjoint terminal paths that satisfy one event are added \(\displaystyle \Pr(E)=\frac{1}{6}+\frac{1}{4}=\frac{5}{12}\)

Worked example

A two-stage trial has mutually exclusive and exhaustive first outcomes A and B. The second outcome is C or D, where D means not C. The tree labels every branch probability, conditional on its preceding outcome. Find the probability that C occurs on the second stage

Two-stage probability tree: A and B at the first stage, then C or D, where D means not C. Branches are A 2/5, B 3/5, C after A 1/2, D after A 1/2, C after B 1/3 and D after B 2/32/53/51/21/21/32/3StartABACADBCBD

If needed, scroll sideways to follow each complete path

\[\Pr(C)\]
  1. \(\displaystyle \Pr(A\cap C)=\frac{2}{5}\times\frac{1}{2}=\frac{1}{5}\)Multiply the first-stage probability by the conditional probability after A
  2. \(\displaystyle \Pr(B\cap C)=\frac{3}{5}\times\frac{1}{3}=\frac{1}{5}\)Multiply the first-stage probability by the conditional probability after B
  3. \(\displaystyle \Pr(C)=\Pr(A\cap C)+\Pr(B\cap C)=\frac{1}{5}+\frac{1}{5}=\frac{2}{5}\)The two paths to C are disjoint alternatives, so add their probabilities

Common mistakes

Adding along a path
✕\(\displaystyle \Pr(AC)=\frac{2}{5}+\frac{1}{2}=\frac{9}{10}\)✓\(\displaystyle \Pr(AC)=\frac{2}{5}\times\frac{1}{2}=\frac{1}{5}\)
A path requires both stages, so its branch probabilities are multiplied
Multiplying alternative paths
✕\(\displaystyle \Pr(C)=\frac{1}{5}\times\frac{1}{5}=\frac{1}{25}\)✓\(\displaystyle \Pr(C)=\frac{1}{5}+\frac{1}{5}=\frac{2}{5}\)
Different paths to the same event are disjoint alternatives, so their probabilities are added
Reusing counts after removal
✕\(\displaystyle \Pr(R_2\mid R_1)=\frac{3}{5}\)✓\(\displaystyle \Pr(R_2\mid R_1)=\frac{2}{4}=\frac{1}{2}\)
Without replacement, one red item has been removed, so both the red count and total count change

Difficulty levels

introCalculate one path in a two-stage tree with replacement
standardUpdate counts without replacement and add two alternative paths
challengeUse state-dependent branches, three stages or a reverse conditional after a total probability
examRecognise the valid tree plan and calculate a marginal probability from its paths

Questions about this topic

When are branch probabilities multiplied?+

Multiply the branch probabilities along one ordered path because every branch on that path is required

When are path probabilities added?+

Add path probabilities when the paths are disjoint alternatives that all satisfy the same event

What changes without replacement?+

The selected item is not returned, so the relevant category count and total count change before the next branch

Can I use a calculator?+

No. This topic is practised without a 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 complementsRepresent a multi-stage finite sample spaceCompound events, the addition rule and mutual exclusivityComplete and use a two-way probability tableConditional probability from exact representationsDistribution of a finite random variableDesign and interpret a probability simulationCount restricted selections and calculate exact probabilitiesCount arrangements with product, sum and restriction rulesTest exact independenceExact probability of repeated independent successesMulti-stage probability, multiplication and total probabilityProbability from a uniform area modelProbabilities of sequences of independent events