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Year 11 Methods (Unit 1 & 2) Probability

Multi-Stage Experiments

20 practice questions 1 video lesson Theory + worked examples

Master multi-stage experiments for Queensland Year 11 Mathematical Methods (QCAA). When an experiment happens in stages, such as several draws or tosses, a tree diagram maps out every possible path.

You will learn to multiply probabilities along a branch and add across separate paths, handle selections made with and without replacement, and tackle "at least one" questions using the complement — versatile tools tested throughout QCAA probability.

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Theory

In Year 11 Mathematical Methods (QCAA, Unit 1 Topic 5), a multi-stage experiment has several steps modelled with a tree diagram or a sample-space array. You multiply probabilities along a branch and add across separate paths, taking care with without-replacement stages where the second denominator falls by one.

A tree diagram shows each stage of an experiment as a set of branches. The probability of a whole path is the product of the branch probabilities along it — this is the multiplication (AND) rule.

When a compound event can happen along several different paths, the paths are mutually exclusive, so you add their probabilities — the addition (OR) rule for separate paths.

With replacement (or independent stages) the probabilities do not change between stages. Without replacement (dependent stages) the second-stage probabilities change because one item has been removed — the denominator drops by one.

Multiply along, add across. Multiply the branch probabilities to get one path; add the path probabilities to combine several outcomes. All the paths together add to \(1\).
Two draws with replacement from a bag of 3 red and 2 blueA tree with replacement; the four branch products nine, six, six and four twenty-fifths add to one. 3/5 R 3/5 R RR = 9/25 2/5 B RB = 6/25 2/5 B 3/5 R BR = 6/25 2/5 B BB = 4/25
With replacement: \(P(\text{red})=\dfrac{3}{5}\) at every stage. The four path products \(\dfrac{9}{25},\dfrac{6}{25},\dfrac{6}{25},\dfrac{4}{25}\) add to \(1\).
Two draws without replacement from a bag of 3 red and 2 blueA tree without replacement; the second-stage denominator falls from five to four as one ball is kept out. 3/5 R 2/4 R RR = 6/20 2/4 B RB = 6/20 2/5 B 3/4 R BR = 6/20 1/4 B BB = 2/20
Without replacement: after a red is kept out, the second red is \(\dfrac{2}{4}\). The denominator falls from \(5\) to \(4\).

Multiply along a branch (path probability):

\[P(A\text{ and }B)=P(A)\times P(B\mid A)\]
P(A and B)=P(A)×P(B∣A)

Add across separate paths for a compound event:

\[P(\text{event})=\sum(\text{path probabilities})\]
P=∑paths

At least one via the complement:

\[P(\text{at least one})=1-P(\text{none})\]
P(at least one)=1-P(none)
Independent stages: when the stages do not affect each other (with replacement), \(P(A\text{ and }B)=P(A)\times P(B)\).

How to solve a multi-stage problem

  1. Draw the tree: one set of branches per stage, with a probability on each branch.
  2. Adjust for replacement: if there is no replacement, reduce the counts and totals for the later stages.
  3. Multiply along each branch to get the path probabilities.
  4. Add across all paths that satisfy the event; for ‘at least one’, use \(1-P(\text{none})\).
Example 1 — With replacement (multiply along)
A bag holds \(3\) red and \(2\) blue balls. A ball is drawn, its colour noted, and it is replaced; then a second is drawn. Find \(P(\text{both red})\).
Solution

With replacement, \(P(\text{red})=\dfrac{3}{5}\) each time. Multiply along the branch:

\(P(RR)\)\(=\)\(\dfrac{3}{5}\times\dfrac{3}{5}\)
\(=\)\(\dfrac{9}{25}\)

\(P(\text{both red})=\dfrac{9}{25}\).

Two draws with replacement, both redWith replacement the probability of red stays three fifths at each stage. 3/5 R 3/5 R RR = 9/25 2/5 B RB 2/5 B 3/5 R BR 2/5 B BB
Example 2 — Without replacement
From the same bag (\(3\) red, \(2\) blue), two balls are drawn without replacement. Find \(P(\text{both red})\).
Solution

First red: \(\dfrac{3}{5}\). One red is kept out, leaving \(2\) red of \(4\):

\(P(RR)\)\(=\)\(\dfrac{3}{5}\times\dfrac{2}{4}\)
\(=\)\(\dfrac{6}{20}\)
\(=\)\(\dfrac{3}{10}\)

\(P(\text{both red})=\dfrac{3}{10}\).

Two draws without replacement, both redWithout replacement the second red has probability two quarters after one red is kept out. 3/5 R 2/4 R RR = 6/20 2/4 B RB 2/5 B 3/4 R BR 1/4 B BB
Example 3 — Add across paths
Two balls are drawn without replacement from the bag (\(3\) red, \(2\) blue). Find the probability of getting one of each colour.
Solution

Path red then blue:

\(P(RB)\)\(=\)\(\dfrac{3}{5}\times\dfrac{2}{4}\)
\(=\)\(\dfrac{6}{20}\)

Path blue then red:

\(P(BR)\)\(=\)\(\dfrac{2}{5}\times\dfrac{3}{4}\)
\(=\)\(\dfrac{6}{20}\)

Add the two mutually exclusive paths:

\(P(\text{one of each})\)\(=\)\(\dfrac{6}{20}+\dfrac{6}{20}\)
\(=\)\(\dfrac{12}{20}\)
\(=\)\(\dfrac{3}{5}\)

\(P(\text{one of each})=\dfrac{3}{5}\).

One of each colour, two paths addedThe two mutually exclusive paths red-then-blue and blue-then-red each have probability six twentieths. 3/5 R 2/4 R RR 2/4 B RB = 6/20 2/5 B 3/4 R BR = 6/20 1/4 B BB
Example 4 — At least one (complement)
A game is played \(3\) times independently and is won each time with probability \(\dfrac{1}{4}\). Find the probability of at least one win.
Solution

Probability of losing a single game:

\(P(\text{lose})\)\(=\)\(1-\dfrac{1}{4}\)
\(=\)\(\dfrac{3}{4}\)

Probability of losing all three (multiply along):

\(P(\text{no win})\)\(=\)\(\left(\dfrac{3}{4}\right)^3\)
\(=\)\(\dfrac{27}{64}\)

At least one win is the complement:

\(P(\text{at least one win})\)\(=\)\(1-\dfrac{27}{64}\)
\(=\)\(\dfrac{37}{64}\)

\(P(\text{at least one win})=\dfrac{37}{64}\).

Three independent plays with win probability one quarterA three-stage win-lose tree; the all-lose path has probability twenty-seven sixty-fourths. 1/4 W 1/4 W 1/4 W WWW 3/4 L WWL 3/4 L 1/4 W WLW 3/4 L WLL 3/4 L 1/4 W 1/4 W LWW 3/4 L LWL 3/4 L 1/4 W LLW 3/4 L LLL = 27/64

Common pitfalls

Forgetting to reduce for no replacement. Without replacement the second denominator drops by one (\(5\to 4\)) and the chosen colour’s count also drops.
Adding along a branch. Along one path you multiply; you only add when combining different complete paths.
Missing a path. ‘One of each’ includes both red-then-blue and blue-then-red — two paths, not one.
Doing ‘at least one’ the long way. Use \(1-P(\text{none})\) instead of adding every winning case.

Frequently asked questions

How do you find the probability along a branch of a tree?

Multiply the probabilities on the branches that make up the path, for example \(P(RR)=\dfrac{3}{5}\times\dfrac{2}{4}\).

When do you add and when do you multiply?

Multiply along a single path (AND); add across separate mutually exclusive paths (OR).

What changes for drawing without replacement?

The item drawn is not returned, so the later totals and counts fall by one — the second denominator drops from \(5\) to \(4\).

How do you work out ‘at least one’?

Use the complement: \(P(\text{at least one})=1-P(\text{none})\), which is usually far quicker.

What is the difference between with and without replacement?

With replacement the stages are independent and probabilities stay the same; without replacement they change because an item is removed.

Do all the paths of a tree add to 1?

Yes — the products of all complete paths add to \(1\), which is a useful check.

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