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.
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.
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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.