Probability Tree Exam Questions: How to Solve

Probability tree exam questions ask you to map out every possible outcome of a sequence of events, multiply along the branches to find the probability of each path, and then add the relevant paths for the final answer. You enter the probabilities for each stage, and the calculator below builds the tree and gives you the probability of any combined outcome you select.

Formula

For two independent events A and B, the probability that both occur is:

P(A and B) = P(A) × P(B)

For dependent events, you use conditional probability:

P(A and B) = P(A) × P(B given A)

Where P(A) is the probability of the first event, P(B given A) is the probability of the second event assuming the first has happened, and the result is the joint probability of both. The tree diagram is just a visual way of writing these multiplications.

Worked Example

Using the default values: P(first event) = 0.6, P(second event given first) = 0.7.

1. Branch 1: first event happens (0.6). Then second event happens (0.7). Multiply: 0.6 × 0.7 = 0.42.
2. Branch 2: first event happens (0.6). Second event does not happen (0.3). Multiply: 0.6 × 0.3 = 0.18.
3. Branch 3: first event does not happen (0.4). Second event happens (0.7). Multiply: 0.4 × 0.7 = 0.28.
4. Branch 4: first event does not happen (0.4). Second event does not happen (0.3). Multiply: 0.4 × 0.3 = 0.12.

Check: 0.42 + 0.18 + 0.28 + 0.12 = 1.00. The probability both happen is 0.42, neither happens is 0.12, and exactly one happens is 0.18 + 0.28 = 0.46. The calculator gives the same results.

How Probability Trees Work

A probability tree is a branching diagram where each branch represents a possible outcome of an event. The probability of each outcome is written on its branch. To find the probability of a particular path, multiply the probabilities along that path. To find the probability of a set of paths (like “at least one success”), add the probabilities of each qualifying path.

Trees are especially useful for problems with two or three stages, because they force you to consider all combinations systematically. For more than three stages, trees become unwieldy, and you might prefer formulas or tables.

Choosing the Right Method

Exam questions often specify “draw a tree diagram” or “use a tree diagram.” If not, consider:

• If there are only two or three events, a tree is usually the clearest approach.
• If events are independent (e.g., coin flips), you can multiply directly without a tree, but a tree still helps avoid missing combinations.
• If events are dependent (e.g., drawing cards without replacement), a tree is almost essential because the probabilities change after each step.
• If the question asks for “at least one” or “exactly one,” a tree makes it easy to identify the relevant paths.

Common Mistakes and How to Avoid Them

• Forgetting to multiply along the branch – always multiply, never add along a single path.
• Adding probabilities of paths that are not mutually exclusive – only add paths that represent distinct, non-overlapping outcomes.
• Using the same probability for a dependent event – when drawing without replacement, the second probability must reflect the changed sample space.
• Rounding too early – keep probabilities as fractions or decimals with several places until the final step.
• Misreading “given that” – this indicates conditional probability, so the denominator is the condition, not the whole sample space.

Real Exam-Style Example

A bag contains 3 red and 5 blue marbles. You draw two marbles without replacement. What is the probability both are red?

1. First draw: P(red) = 3/8.
2. Second draw (given first was red): now 2 red and 5 blue left, so P(red) = 2/7.
3. Multiply: (3/8) × (2/7) = 6/56 = 3/28 ≈ 0.1071.

Notice the second probability changed because the first marble was not replaced. This is the classic dependent-event trap.

Limits of the Tree Method

Trees become impractical when there are many events or many outcomes per event. For example, a tree for 10 coin flips would have 2^10 = 1024 paths. In those cases, use the binomial formula or normal approximation. Also, trees require you to know the probabilities at each branch; if you only have summary statistics, a tree cannot help.

Frequently Asked Questions

What is the difference between a tree diagram and a Venn diagram?

A tree diagram shows sequential outcomes and their probabilities along branches, useful for multi-step processes. A Venn diagram shows sets and overlaps of events in a single sample space, useful for understanding unions and intersections but not for sequential probabilities.

How do I handle “at least one” probability questions?

Use the complement: P(at least one) = 1 – P(none). On a tree, find the path where no event occurs, multiply those probabilities, and subtract from 1. This is faster than adding all other paths.

Can a probability tree have more than two branches at each node?

Yes. A node can have any number of branches, as long as the probabilities on all branches from that node sum to 1. For example, rolling a die has six branches, each with probability 1/6.

Do I need to draw the tree in an exam to get full marks?

Usually yes, because examiners award method marks for the diagram and the multiplication. Even if you get the final answer wrong, you can earn partial credit for showing the correct branches and calculations.

What if the events are independent – do I still need a tree?

No, but a tree can still help you visualize. For independent events, you can multiply directly: P(A and B) = P(A) × P(B). The tree just shows the same multiplication in a diagram.

How do I convert a tree diagram into a probability distribution?

List each path’s probability and the outcome it represents. Then group paths that give the same result (e.g., two heads in three flips) and add their probabilities. That gives the probability for each possible number of successes.

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in p1 | Probability of first event | - | 0.6 | 0
in p2 | Probability of second event given first | - | 0.7 | 0
out both | Probability both happen | - | p1 * p2 | 4
out neither | Probability neither happens | - | (1-p1)*(1-p2) | 4
out exactly_one | Probability exactly one happens | - | p1*(1-p2)+(1-p1)*p2 | 4