Probability Tree Practice Questions

This guide provides practice questions and worked solutions for probability tree diagrams, a visual method for calculating probabilities of combined events. You’ll learn how to construct a tree, multiply along branches, and add across final outcomes, with examples covering independent and dependent events.

Formula

The probability of a sequence of events along a path is the product of the probabilities on each branch. For two independent events A and B, the probability of both occurring is:

P(A and B) = P(A) × P(B)

Where P(A) is the probability of the first event, and P(B) is the probability of the second event. For mutually exclusive final outcomes, add their probabilities.

Worked Example

Using the calculator defaults: P(A) = 0.6, P(B) = 0.5.

1. P(both) = 0.6 × 0.5 = 0.30
2. P(neither) = 0.4 × 0.5 = 0.20
3. P(exactly one) = 0.6 × 0.5 + 0.4 × 0.5 = 0.30 + 0.20 = 0.50

These match the calculator outputs.

Building a Probability Tree

• Start with a root point and draw branches for each possible outcome of the first event.
• Label each branch with its probability.
• From the end of each first branch, draw branches for the second event, labeled with conditional probabilities.
• Multiply along branches to get the probability of each combined outcome.

Practice Questions

1. A bag contains 5 red and 3 blue marbles. Two marbles are drawn without replacement. Draw a tree and find the probability of getting one of each color.

2. A coin is tossed twice. What is the probability of getting at least one head?

3. In a class, 70% of students pass math and 60% pass English, independently. What is the probability a random student passes both?

4. A spinner has 3 equal sections: A, B, C. Spun twice, what is the probability of getting A then B?

5. A box has 4 defective and 6 working bulbs. Two bulbs are picked without replacement. Find the probability that both are defective.

Solutions

1. P(red then blue) = (5/8)×(3/7)=15/56; P(blue then red) = (3/8)×(5/7)=15/56; total = 30/56 = 15/28.

2. P(at least one head) = 1 – P(TT) = 1 – (1/2×1/2)=3/4.

3. P(pass both) = 0.7 × 0.6 = 0.42.

4. P(A then B) = (1/3)×(1/3)=1/9.

5. P(both defective) = (4/10)×(3/9)=12/90=2/15.

Common Mistakes

• Forgetting to multiply probabilities along branches—always multiply, not add.
• Using the same probability for the second draw when the events are dependent; adjust for the reduced sample size.
• Adding probabilities of outcomes that are not mutually exclusive.
• Misreading “without replacement” as “with replacement”.

When to Use a Tree Diagram

Use a tree diagram when an experiment has two or more stages and you need the probability of a specific combination or sequence of outcomes. It is especially useful for dependent events where conditional probabilities change after each step. For independent events, the tree still works but the calculations simplify to direct multiplication.

Limits of Tree Diagrams

Tree diagrams become unwieldy with many stages or many outcomes per stage. For more than three stages, the number of branches grows exponentially. In such cases, use formulas for binomial or hypergeometric distributions. Also, tree diagrams assume that all branches are exhaustive and mutually exclusive, which they are if constructed correctly.

Frequently Asked Questions

How do you calculate probability using a tree diagram?

Multiply the probabilities along the branches that represent the desired outcome. If multiple paths give the same final result, add their probabilities.

What is the difference between with and without replacement?

With replacement means the item is returned to the set before the next draw, so probabilities stay constant. Without replacement means the item is not returned, so the denominator and possibly the numerator change for subsequent draws.

Can a tree diagram be used for three events?

Yes, draw three layers of branches. Multiply along each path to get the probability of that sequence, then add paths that match the condition.

Why do we multiply probabilities in a tree?

Multiplication comes from the definition of conditional probability: P(A and B) = P(A) × P(B|A). For independent events, P(B|A) = P(B).

How do you find the probability of at least one success?

It is often easier to subtract the probability of zero successes from 1. For example, P(at least one head) = 1 – P(no heads).

What if the tree has many branches—any shortcut?

For repeated independent trials, use the binomial probability formula instead of drawing a large tree. For dependent draws without replacement, use combinations or the hypergeometric formula.

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in p1 | Probability of first event | - | 0.6 | 0
in p2 | Probability of second event | - | 0.5 | 0
out both | Probability of both events | - | p1 * p2 | 2
out neither | Probability of neither event | - | (1-p1)*(1-p2) | 2
out exactly_one | Probability of exactly one | - | p1*(1-p2)+(1-p1)*p2 | 2