GCSE Maths Probability Trees With Dependent Events
A probability tree organises a sequence of possible events. The main trap in GCSE questions is forgetting that probabilities can change after the first event, particularly when an object is chosen and not replaced. Understanding the meaning of each branch matters more than memorising an instruction to multiply. This guide uses a simple bag example to show how dependent events work.
What makes two events dependent?
Events are dependent when the outcome of the first event changes the probability of the next. Suppose a bag contains 3 red counters and 2 blue counters. Before any selection, the probability of red is 3/5 and blue is 2/5. If a red counter is taken out and not returned, only 2 red counters remain among 4 counters. The next probability of red becomes 2/4. If a blue counter was removed first, the next probability of red would instead be 3/4. The second set of branches must reflect what happened on the first branch.
Draw the first two levels of the probability tree
The first branches are R with probability 3/5 and B with probability 2/5. From the first R branch, the next branches are R with probability 2/4 and B with probability 2/4. From the first B branch, they are R with probability 3/4 and B with probability 1/4. Every pair of branches leaving the same point must add to 1. These checks help catch the classic mistake of keeping 3/5 and 2/5 on the second level even though an item was removed.
Multiply along a route to find its probability
The probability of two red counters without replacement is P(RR) = 3/5 × 2/4 = 6/20 = 3/10. Multiplication works because the second branch represents the probability of the second outcome given the first outcome. Likewise, the route blue then blue has probability P(BB) = 2/5 × 1/4 = 2/20 = 1/10. Keep the denominator changes visible. If your answer for two reds is greater than the probability of choosing red first, something has gone wrong.
Add separate routes for exactly one red counter
Exactly one red counter means either red then blue or blue then red. The first route has probability P(RB) = 3/5 × 2/4 = 3/10. The second route has P(BR) = 2/5 × 3/4 = 3/10. Add those disjoint possibilities to get 3/10 + 3/10 = 3/5. Check all four possible routes: RR is 3/10, RB is 3/10, BR is 3/10, and BB is 1/10. Their total equals 1 as it should.
See how replacement changes the problem
If the first counter is returned before the second selection, the bag once again contains 3 red and 2 blue counters. The events are now independent under this model, so the probability of red on the second draw remains 3/5 whatever happened first. Then P(RR) = 3/5 × 3/5 = 9/25. Compare that result with 3/10 without replacement. Words such as replaced, kept out or returned completely change the branch probabilities; always identify the selection process before calculating.
Check the tree before answering the final question
Read the event wording carefully. “At least one red” includes RR, RB and BR, and can also be found using 1 − P(BB) = 1 − 1/10 = 9/10. “Exactly one red” excludes RR and gave 3/5 instead. Label all branches, confirm probabilities at each split total 1, multiply down each route and add the routes that satisfy the question. Work neatly enough to see which branch was used. A tutor can help students verbalise why the probabilities change, not only the mechanics of a completed tree.
Frequently asked questions
Why do I multiply along a probability-tree branch?
The route probability combines the first event with the conditional probability of the second event given the first.
How do I know when to add probabilities?
Add mutually exclusive routes that each satisfy the event named in the question.
More GCSE learning resources
Explore these existing guides when a prerequisite skill or another aspect of your GCSE preparation needs attention. Use the links selectively to find the explanation that matches your current problem.
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