Two- and three-stage chance experiments
State the draw rule before calculating. Earlier selections determine what remains available
- Track the actual history Remove one physical object after each draw without replacement. Keep every object available with replacement
- Update both counts The numerator counts the required remaining colour. The denominator counts all remaining objects
- Multiply three branches A three-stage path requires three successive outcomes. Use the branch conditional on each previous history \(\displaystyle \Pr(RRB)=\frac35\times\frac24\times\frac23=\frac15\)
- List every favourable order Exactly two reds in three draws includes RRB, RBR and BRR, not just one order
- Add disjoint paths Different complete colour sequences cannot happen in the same trial. Add their probabilities
- Use zero and complements An exhausted colour has probability zero. At least one blue is the complement of all red
- Check independence With uniform replacement, colour probabilities remain unchanged after any history. Without replacement, compare the conditional branches
Worked example
A bag contains three red and two blue counters. Draw three uniformly without replacement. Find the probability of exactly two reds
\[\begin{aligned}&E\\&=\{RRB,RBR,BRR\}\end{aligned}\]
- \(\displaystyle \begin{aligned}&\Pr(RRB)\\&=\frac35\times\frac24\times\frac23\\&=\frac15\end{aligned}\)After two reds, two blue counters remain among three
- \(\displaystyle \begin{aligned}&\Pr(RBR)\\&=\frac35\times\frac24\times\frac23\\&=\frac15\end{aligned}\)After red then blue, two red counters remain among three
- \(\displaystyle \begin{aligned}&\Pr(BRR)\\&=\frac25\times\frac34\times\frac23\\&=\frac15\end{aligned}\)After blue, all three reds remain for draw two
- \(\displaystyle \begin{aligned}&\Pr(E)\\&=\frac15+\frac15+\frac15\\&=\frac35\end{aligned}\)The event includes these three disjoint orders
Common mistakes
Keeping the original denominator
✕\(\displaystyle \begin{aligned}&\Pr(RRR)\\&=\frac35\times\frac25\times\frac15\end{aligned}\)✓\(\displaystyle \begin{aligned}&\Pr(RRR)\\&=\frac35\times\frac24\times\frac13\\&=\frac1{10}\end{aligned}\)
Without replacement, the total drops from five to four to threeCounting only one order
✕\(\displaystyle \begin{aligned}&\Pr(\text{exactly two R})\\&=\frac15\end{aligned}\)✓\(\displaystyle \begin{aligned}&\Pr(\text{exactly two R})\\&=\frac35\end{aligned}\)
Include RRB, RBR and BRRUsing only the last branch
✕\(\displaystyle \begin{aligned}&\Pr(RRB)\\&=\frac23\end{aligned}\)✓\(\displaystyle \begin{aligned}&\Pr(RRB)\\&=\frac15\end{aligned}\)
The last conditional branch must be multiplied by the first two branch probabilitiesDifficulty levels
introCalculate a two-stage path with replacement
standardUpdate two-stage counts without replacement
challengeTrack three-stage removal and combine favourable paths
examCompare two-/three-stage events with and without replacement
Questions about this topic
Does the whole history matter?+
Without replacement, the remaining counts depend on every item already removed, not only the last colour
Why do RRB and RBR differ as outcomes?+
They record different orders, even when their probabilities happen to be equal
Is 1/8 correct for every binary three-stage path?+
Only when the three stages are independent and each branch has probability 1/2.
Can a branch probability be zero?+
Yes. If a colour is exhausted, drawing it next is impossible
When can I use a complement?+
When the opposite event is simpler; for example at least one blue is the complement of all red
Does multiplication assume independence?+
No. A tree multiplies the relevant conditional probability at each stage
Are same-colour counters separate objects?+
Yes. Three red counters are three equally likely physical items when drawing uniformly
What should I check in a tree?+
Sibling probabilities sum to one, counts match the history, each required complete path is included once, and the final probability is between zero and one
Can I use a calculator?+
No. This topic is practised without a calculator
Which year level and chapter is this?+
Year 10 Mathematics, chapter 4: Probability
In this chapter
Describe and order probability valuesSingle-step theoretical probability and complements4A Probability scale and experiments: Experimental probability from observed frequenciesFinite events, probability assignments and complements4A Probability scale and experiments: Design and interpret a probability simulation4B Venn diagrams and sets: Venn regions, set language and probabilityCompound events, the addition rule and mutual exclusivityComplete and use a two-way probability tableConstruct and complete two-way frequency tablesConvert between Venn diagrams and two-way tablesConditional probability from exact representations4D Conditional probability: Conditional probability on Venn diagramsTwo-step sample-space tables with and without replacementRepresent a multi-stage finite sample spaceTree-diagram probability for two-stage experimentsMulti-stage probability, multiplication and total probability4G Independence and simulations: Independence on Venn diagramsTest exact independence4G Independence and simulations: Design and interpret a probability simulation4G Independence and simulations: Experimental probability from observed frequencies