When an item is taken and not put back, the second pick has one fewer item in total and one fewer of the type already taken, so the probabilities change. In a tree diagram, the second set of branches reflects what was removed first. Multiply along the branches as usual, add paths for combined outcomes, and use 1 − P(none) for “at least one”.
Without replacement
- After one pick, the total decreases by 1
- The type that was picked also decreases by 1
- P(A then B) = P(A) × P(B after A)
- P(at least one) = 1 − P(none)
What changes without replacement
If a bag holds 5 red and 3 blue counters, the first pick has 8 counters. If a red is taken and not replaced, 7 counters remain, 4 of them red, so P(second red) is 4/7, not 5/8. The second set of branches on the tree must show these new values.
Calculating with a tree
Multiply along each path. P(red then red) = 5/8 × 4/7 = 20/56 = 5/14. For one of each colour in either order, add the two paths: 5/8 × 3/7 + 3/8 × 5/7 = 30/56 = 15/28. The four outcomes still add up to 1.
“At least one”
Use the complement: P(at least one blue) = 1 − P(no blue) = 1 − P(red, red) = 1 − 5/14 = 9/14. This is much quicker than adding three separate paths.
Three picks
The same method extends to three picks, with the numbers falling at each stage. For a bag of 6 strawberry and 4 lemon sweets, P(three strawberry) = 6/10 × 5/9 × 4/8 = 1/6.
Algebra in without-replacement problems (Higher)
Some questions give the probability and ask for the number of items. If a bag has n counters, 2 of them red, and P(both red) = 1/15, then 2/n × 1/(n − 1) = 1/15, so n(n − 1) = 30. Solving gives n = 6.
Worked examples
Example 1. A bag has 5 red and 3 blue counters. Two are taken without replacement. Find P(red then red).
- First red: 5/8. Second red: 4/7.
- 5/8 × 4/7 = 20/56.
Answer: 5/14.
Example 2. Using the same bag, find P(one of each colour).
- Red then blue: 5/8 × 3/7 = 15/56. Blue then red: 3/8 × 5/7 = 15/56.
- Add: 30/56.
Answer: 15/28.
Example 3. Using the same bag, find P(at least one blue).
- P(no blue) = P(red, red) = 5/14.
- 1 − 5/14.
Answer: 9/14.
Example 4. A bag has 6 strawberry and 4 lemon sweets. Three are taken without replacement. Find P(all three are strawberry).
- 6/10 × 5/9 × 4/8 = 120/720.
Answer: 1/6.
Example 5. A bag has n counters, 2 of which are red. Two are taken without replacement, and P(both red) = 1/15. Find n.
- 2/n × 1/(n − 1) = 1/15, so 2 × 15 = n(n − 1).
- n² − n − 30 = 0, so (n − 6)(n + 5) = 0.
Answer: n = 6.
Common mistakes
- Keeping the same probabilities on the second pick. Reduce both the total and the type removed.
- Only counting one order for “one of each”. Red then blue and blue then red are both needed.
- Adding along a path. Multiply along a path.
- Reducing the wrong count. Only the type that was taken decreases.
- Not simplifying. Simplify fractions such as 20/56 to 5/14.
How without-replacement questions are examined
Without-replacement probability is a Higher tier topic for Edexcel GCSE (1MA1) and appears in the 4MA1 International GCSE too, often with a partly drawn tree. Questions ask you to:
- complete a tree diagram for picks without replacement
- find the probability of two picks of the same type
- find P(one of each) or P(at least one)
- form and solve an equation for a number of items (Higher)
The tree diagrams worksheet includes without-replacement questions with answers, and the probability revision engine gives more.
This extends tree diagrams. Each second-stage branch is a conditional probability.
Frequently asked questions
What does without replacement mean?
The item taken is not put back, so the next pick has fewer items to choose from.
How do the probabilities change without replacement?
The total goes down by one, and the type that was taken goes down by one.
How do you find P(at least one) without replacement?
Work out P(none) using the tree, then subtract it from 1.