Tree Diagrams in Probability

Lessons

Senior 2 · Mathematics

Tree Diagrams in Probability

4 lessons

Lesson 3 of 4

Finding probabilities with tree diagrams

25 min

When the branches are not equally likely

A basket at the market holds 4 ripe and 2 unripe avocados. If you pick two, the paths "ripe, ripe" and "unripe, unripe" are not equally likely, so counting paths no longer works. Instead, write the probability on every branch and let the tree do the arithmetic.

The two rules

Multiply along a path. The probability of a whole path (this happens AND then that happens) is the product of the probabilities on its branches.
Add between paths. If an event can happen along several paths (this path OR that path), add their probabilities.

Two checks: the branches leaving any one point add up to 1, and the probabilities of all the complete paths add up to 1.

Worked example 1: with replacement

From the basket of 4 ripe (R) and 2 unripe (U) avocados, one is picked, looked at and put back, then a second is picked. Each pick has P(R) = 46 and P(U) = 26.

Figure 2. Tree diagram for two picks with replacement. Diagram: Ganzaa.

P(both ripe) = 46 × 46 = 1636 = 49.
P(one of each) = P(RU) + P(UR) = 836 + 836 = 1636 = 49.
P(both unripe) = 26 × 26 = 436 = 19.
Check: 1636 + 836 + 836 + 436 = 3636 = 1.

Worked example 2: without replacement

Now the first avocado is not put back. The second-stage branches change, because the basket now holds only 5 avocados.

If the first was ripe, 3 ripe and 2 unripe are left: P(R) = 35, P(U) = 25.
If the first was unripe, 4 ripe and 1 unripe are left: P(R) = 45, P(U) = 15.

So P(both ripe) = 46 × 35 = 1230 = 25; P(one of each) = 46 × 25 + 26 × 45 = 830 + 830 = 1630 = 815; P(both unripe) = 26 × 15 = 230 = 115. Check: 1230 + 1630 + 230 = 1.

Worked example 3: a real-life tree

In April the probability that it rains in Musanze on a given morning is 0.7. If it rains, the probability that the school bus is late is 0.4; if it does not rain, it is 0.1. Find the probability that the bus is late.
Two paths lead to "late": rain and late, 0.7 × 0.4 = 0.28; no rain and late, 0.3 × 0.1 = 0.03. P(late) = 0.28 + 0.03 = 0.31.

Tip

AND means multiply along; OR means add across. Write the probability on every branch before you calculate anything. In "without replacement" problems, the second-stage denominators go down by one.

✦

Practice: try each one before you read the solution

1

A fair coin is tossed and a fair die is thrown. Find P(head and 6).

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Solution: 12 × 16 = 112.

2

A box holds 3 blue and 2 white pens. One is taken and replaced, then another is taken. Find P(both white) and P(different colours).

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Solution: P(both white) = 25 × 25 = 425. P(different) = 35 × 25 + 25 × 35 = 625 + 625 = 1225.

3

Repeat the first part of question 2 without replacement.

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Solution: 25 × 14 = 220 = 110.

4

A bean seed germinates with probability 0.8. Two seeds are planted. Find P(exactly one germinates) and P(at least one germinates).

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Solution: exactly one: 0.8 × 0.2 + 0.2 × 0.8 = 0.32. At least one: 1 − P(neither) = 1 − 0.2 × 0.2 = 0.96.

Remember

Multiply along a path; add the paths that make up the event.
The branches from any point, and all the complete paths, add up to 1.
Without replacement, the second-stage probabilities change.

Sources and credits

Introductory Statistics (OpenStax), 3.6 Tree and Venn Diagrams by Barbara Illowsky, Susan Dean and contributors, OpenStax, via LibreTexts, licensed under CC BY 4.0. Adapted by Ganzaa.

The Rwandan examples, practice items and exercises are written by Ganzaa.

1When the branches are not equally likely

2The two rules

Worked example 1: with replacement

Worked example 2: without replacement

Worked example 3: a real-life tree

Practice: try each one before you read the solution

When the branches are not equally likely

The two rules