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Cambridge IGCSE Mathematics — 0580 Extended

Topic 8.4: Probability — Tree Diagrams

A tree splits each stage of an experiment into branches. On 0580, probabilities sit by the side of the branches and outcomes sit at the ends. Combined events may be with or without replacement. Two rules run the whole topic: multiply along a branch, add complete branches that match the event.

How to draw a tree

First event on the left, second event from each of those ends. The probabilities on branches that leave the same point must add to 1.

Two-stage tree with probabilities on the branches and outcomes with products at the ends
Read a complete route from left to right, then write its product at the end.

Method

  1. Draw the first split. Label each branch with its probability.
  2. From every first-event end, draw the second split. Without replacement, change those fractions.
  3. Write the outcome at each far end. Multiply along the route to get that end’s probability.
  4. Add the ends that match the event you want.

With replacement

The first counter is put back, so the second fractions copy the first.

A bag has 3 red and 2 blue counters. Two are taken at random, replacing the first. Find \(\mathrm{P}(\text{RR})\).

With-replacement tree for 3 red and 2 blue, two picks, with end products 9/25, 6/25, 6/25 and 4/25
The four end-products add to 1. Second-stage fractions are still \(3/5\) and \(2/5\).

On that tree, show the RR route.

The RR branch highlighted: 3/5 then 3/5, product 9/25
\(\mathrm{P}(\text{RR}) = 3/5 \times 3/5 = 9/25\). Adding \(3/5 + 3/5 = 6/5\) is not a probability.

Without replacement

The first counter stays out. After red, 4 remain: 2 red and 2 blue, so the next red is \(2/4\), not \(3/5\).

The same bag, but the first counter is not replaced. Find \(\mathrm{P}(\text{RR})\).

Without-replacement tree: after red the second fractions are 2/4 and 2/4, so P(RR)=6/20=3/10
\(\mathrm{P}(\text{RR}) = 3/5 \times 2/4 = 6/20 = 3/10\). The second denominator is 4.

Add the matching ends

“Exactly one red” is the RB end or the BR end. Those two complete outcomes cannot both happen, so add their products.

Without replacement, find \(\mathrm{P}(\text{exactly one red})\).

Without-replacement tree with RB and BR ends highlighted, adding to 12/20 = 3/5
\(6/20 + 6/20 = 12/20 = 3/5\).

On any day, \(\mathrm{P}(\text{rain}) = 1/3\). If it rains, \(\mathrm{P}(\text{fishing}) = 3/5\); if it is dry, \(\mathrm{P}(\text{fishing}) = 1/4\). Find \(\mathrm{P}(\text{fishing})\).

Rain then fishing tree with both fishing ends highlighted: 1/5 + 1/6 = 11/30
\(\mathrm{P}(\text{fishing}) = 1/5 + 1/6 = 6/30 + 5/30 = 11/30\).

Paper 2 (non-calculator)

Write products unsimplified first if that helps the add: \(1/5 + 1/6\) needs a common denominator 30. Do not decimalise \(11/30\).

Try this

With the 3-red, 2-blue bag and replacement, find \(\mathrm{P}(\text{same colour})\).

Show answer
Answer

  1. Same colour is RR or BB

    \[ \mathrm{P}(\text{RR}) + \mathrm{P}(\text{BB}) = \dfrac{9}{25} + \dfrac{4}{25} = \dfrac{13}{25} \]

Exam Traps

  • Multiply along a branch; add complete branches. Adding the two fractions on one route (for example \(3/5 + 3/5\)) is the standard tree error.

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