How to use a probability tree for two draws
A probability tree records a sequence of outcomes. Its second set of branches describes what can happen after a particular first outcome, so those probabilities may change from one branch to another. Start with the drawing rule and the contents of the bag before writing any fractions.
Read the rule before drawing the branches
Use this invented practice problem: a bag contains three red marbles and two blue marbles. Two marbles are drawn, one at a time, without replacement. Assume every marble remaining in the bag has the same chance of being selected. What is the probability of drawing exactly one red marble?
Without replacement means the first marble stays out of the bag. There will be four marbles available for the second draw, and their colours depend on what was drawn first. If the question instead says the marble is replaced, return it before calculating the next probabilities. Do not assume a replacement rule that the problem has not supplied.
Build the tree from what remains
Draw two branches from the starting point: red with probability 3/5 and blue with probability 2/5. These cover all possible first-draw colours, so their probabilities sum to 1. Label them as first-draw outcomes to keep the sequence clear.
From the first-red branch, there are two red and two blue marbles left. Its second-draw branches are therefore red with probability 2/4 and blue with probability 2/4. From the first-blue branch, three red and one blue remain. Those branches have probabilities 3/4 and 1/4.
Write a small remaining-contents note beside each branching point: “after red: 2 red, 2 blue” and “after blue: 3 red, 1 blue”. At each point, check that the outgoing probabilities sum to 1. Using 3/5 and 2/5 again for the second draw would describe a different drawing rule.
Multiply along a path, then select the paths
A complete path describes both draws in order. Multiply its first probability by the second probability on that same path. Red then red has probability (3/5) × (2/4) = 3/10. Red then blue has probability (3/5) × (2/4) = 3/10. Blue then red has probability (2/5) × (3/4) = 3/10. Blue then blue has probability (2/5) × (1/4) = 1/10.
Exactly one red marble means red then blue OR blue then red. These are separate outcomes that cannot both occur in the same two-draw trial, so add their probabilities: 3/10 + 3/10 = 6/10 = 3/5. Counting only red then blue would miss the other order.
Check the question’s wording before adding. “Both red” selects just the red-red path, giving 3/10. “At least one red” includes red-red, red-blue and blue-red, giving 9/10. You can also check the latter as 1 minus the probability of no red: 1 − 1/10 = 9/10.
Check the model as well as the arithmetic
All four complete-path probabilities should sum to 1: 3/10 + 3/10 + 3/10 + 1/10 = 1. Every probability should lie between 0 and 1. These checks can catch a missing path or a calculation error, though they do not prove you used the correct replacement rule.
For comparison, if each marble is replaced and mixed back into the bag before the next equally likely draw, the second probabilities remain 3/5 for red and 2/5 for blue. Exactly one red then has probability (3/5 × 2/5) + (2/5 × 3/5) = 12/25. This differs from the without-replacement answer because the contents available for the second draw differ.
For your next tree, note the starting contents and replacement rule, write what remains after each first outcome, label the second branches and list the complete paths. Underline the paths that meet the question. Multiply within each selected path, then add only distinct paths that cannot occur together. If you are unsure whether outcomes are equally likely or how the selection works, resolve that before treating counts as probabilities.
