Tree Diagrams and Multi-Step Probability
A free Statistics and Data Analysis lesson from the “Probability and Distributions” unit, with a worked example and practice problems including step-by-step solutions.
A tree diagram organizes multi-step chance processes. Multiply probabilities along a path, then add paths that lead to the same outcome. This lesson builds the habit of reading the context first, choosing the right statistical tool, calculating carefully, and then writing what the result means. By the end, students should be able to do the computation and explain why that computation answers the question.
What you'll learn
- Build a tree diagram
- Multiply along branches
- Add across disjoint paths
Worked example
Problem. If P(A) = 0.35, what is P(not A)?
- Worked Example: First identify exactly what the question is asking: If P(A) = 0.35, what is P(not A)?
- Use inverse operations to isolate the unknown, and keep both sides balanced at every step.
- Complements add to 1.
- Subtract P(A) from 1.
Answer: 0.65
Practice problems
1. Practice case A: If P(A) = 0.35, what is P(not A)?
Show solution
- Warm-up: First identify exactly what the question is asking: If P(A) = 0.35, what is P(not A)?
- Use inverse operations to isolate the unknown, and keep both sides balanced at every step.
- Complements add to 1.
- Subtract P(A) from 1.
- Check the result by substituting or estimating: the response should match 0.65 and make sense in the original problem.
Answer: 0.65
2. Practice case B: Events A and B are disjoint. If P(A) = 0.2 and P(B) = 0.5, what is P(A or B)?
Show solution
- Warm-up: First identify exactly what the question is asking: Events A and B are disjoint. If P(A) = 0.2 and P(B) = 0.5, what is P(A or B)?
- Use inverse operations to isolate the unknown, and keep both sides balanced at every step.
- For disjoint events, add the probabilities.
- There is no overlap to subtract.
- Check the result by substituting or estimating: the response should match 0.7 and make sense in the original problem.
Answer: 0.7
3. Practice case C: Among 15 students in a group, 5 play soccer. What is P(plays soccer | in this group)?
Show solution
- Warm-up: First identify exactly what the question is asking: Among 15 students in a group, 5 play soccer. What is P(plays soccer | in this group)?
- Choose the operation or relationship that matches the wording, then carry it out one clear step at a time.
- The condition sets the denominator to this group.
- 5/15 simplifies to 1/3.
- Check the result by substituting or estimating: the response should match 1/3 and make sense in the original problem.
Answer: 1/3
4. Practice case D: If two events are independent, then P(B | A) equals:
Choices: 0 · P(A) - P(B) · P(B) · P(A) + P(B)
Show solution
- Warm-up: First identify exactly what the question is asking: If two events are independent, then P(B | A) equals:
- Compare each answer choice with the calculation or rule, and eliminate choices that do not satisfy the condition.
- Independence means knowing A happened does not change B.
- So P(B | A) = P(B).
- Verify the selected choice by checking that it satisfies the original prompt and that the other choices fail the same test.
Answer: P(B)
5. Practice case E: A fair coin is flipped 3 times. What is the probability of HHH?
Show solution
- Warm-up: First identify exactly what the question is asking: A fair coin is flipped 3 times. What is the probability of HHH?
- For probability, count favorable outcomes and total outcomes carefully before writing the ratio.
- Each heads has probability 1/2.
- (1/2)^3 = 1/8.
- Check the result by substituting or estimating: the response should match 1/8 and make sense in the original problem.
Answer: 1/8
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