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.
Multiply along the branches, add the end-paths. Without replacement, the second-pick probabilities change.
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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).
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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 ✓).
On the without-replacement tree, the second-draw probabilities have one fewer of the chosen colour and one fewer total — multiply along 'red then red' and set equal to 1⁄3.
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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.