Probabilities on branches; multiply along a path: A tree diagram shows each stage as a set of branches with their probabilities.
To find the probability of a particular path, multiply along its branches.
With replacement, the probabilities are the same at each stage.
[Diagram: math-prob-tree] - Available in full study mode
IB-style question — with replacement
A bag has 3 red and 2 white balls.
A ball is drawn, replaced, then another is drawn.
Find the probability both are red.
Step by step
- P(red) is 3/5 each draw (replaced).
- Multiply along the red–red path.
Final answer
P(both red) = 9/25.
Branches at each stage sum to 1: Check each split: the branch probabilities leaving a point should add to 1 (e.g. 3/5 + 2/5 = 1).
Second-stage branches change: Without replacement, the item isn't put back, so the second-stage probabilities use reduced totals (one fewer item, and one fewer of the type drawn).
IB-style question — without replacement
From the same bag (3 red, 2 white), two balls are drawn without replacement.
Find the probability both are red.
Step by step
- First red 3/5; second red now 2/4.
- Multiply.
Final answer
P(both red) = 3/10.
Update BOTH numbers: After drawing a red, reds drop and the total drops: 3/5 then 2/4 — not 3/5 then 3/4.
IB-style question — algebraic tree (no replacement)
A bag contains x red counters and 4 white counters. Two counters are drawn without replacement. The probability that both are red is 1⁄3.
Find x.
Step by step
- Down the 'red then red' branch (the second draw has one fewer red and one fewer total).
- Cross-multiply and expand into a quadratic.
- Factor; reject the negative root (a count can't be negative).
Final answer
x = 6 red counters (check: (6⁄10)(5⁄9) = 1⁄3 ✓).
[Diagram: math-prob-tree] - Available in full study mode
Memorize terms 3x faster
Smart flashcards show you cards right before you forget them. Perfect for definitions and key concepts.
Add the paths that match the event: If several paths satisfy the event, find each path (multiply along it) and add them.
For 'at least one', it's often faster to do 1 − P(none).
IB-style question — one of each colour
From the bag (3 red, 2 white), two are drawn without replacement.
Find the probability of one red and one white (in any order).
Step by step
- Two matching paths: red-then-white and white-then-red.
- Add the paths.
Final answer
P(one of each) = 3/5.
'At least one' → complement: For 'at least one red', do 1 − P(no red) — one product instead of adding several paths.