MathematicsGrade 8–10

Multi-stage experiments and path rules

Draw balls from an urn, with or without replacement. The tree diagram and the path rules show how to calculate probabilities.

Loading simulation …

What is it about?

A multi-stage experiment repeats a simple one, e.g. drawing twice from an urn with 33 red and 22 blue balls. The tree diagram shows every possible course; each path is one outcome.

The branches carry probabilities. With replacement they stay the same; without replacement the urn shrinks (after red: 24\frac24 red, 24\frac24 blue).

  • 1st path rule (product): without replacement P(rr)=35⋅24=310P(rr) = \frac35 \cdot \frac24 = \frac{3}{10}, with replacement 35⋅35=925\frac35 \cdot \frac35 = \frac{9}{25}.
  • 2nd path rule (sum): P(same colour)=310+110=25P(\text{same colour}) = \frac{3}{10} + \frac{1}{10} = \frac25 (without replacement).
  • Complement: P(at least one red)=1−P(bb)=910P(\text{at least one red}) = 1 - P(bb) = \frac{9}{10}.
  • Special cases without replacement: a used-up colour gets a branch with probability 00 (dashed), and the urn runs empty.

In the simulation the relative frequency approaches the probability; the funnel shows the 95 %95\,\% range.

Try it

  1. Draw once and watch the path light up. Draw
  2. Compare both trees side by side. Comparison
  3. Play the calculation for “same colour”. Path rules
  4. Draw 1000 times. Simulation

Tasks

Task 1: Path probabilities

33 red, 22 blue, two draws without replacement: compute all four path probabilities. Load task

Show solution

P(rr)=P(rb)=P(br)=310P(rr) = P(rb) = P(br) = \frac{3}{10}, P(bb)=110P(bb) = \frac{1}{10}; together 11. Show in simulation

Task 2: At least one red

Three draws with replacement: find P(at least one red)P(\text{at least one red}). Load task

Show solution

1−(25)3=117125=93.6 %1 - \left(\frac25\right)^3 = \frac{117}{125} = 93.6\,\%. Show in simulation

Ideas for the classroom

  • Drawing a few times from a real urn first and then letting the simulation do the many repetitions can work well.
  • Common stumbling blocks: keeping the same probabilities without replacement; adding along a path instead of multiplying; treating rbrb and brbr as one outcome.