MISSION · Data and probability
Tree Diagrams and Combined Events
Multiply along the branches, add between them
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LEVEL 2 · LEARN
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The idea in one sentence
Draw every way an experiment can unfold, multiply the probabilities along each path, and add the paths that count as success.
Building the tree
Two coin flips. The first splits into heads and tails; each of those splits again.
Four complete paths — HH, HT, TH, TT — each with probability \(\frac{1}{2} \times \frac{1}{2} = \frac{1}{4}\).
Two rules govern the whole diagram:
- Multiply along a branch. That’s the probability of this and then that.
- Add between branches. That’s the probability of this path or that one.
“And means times, or means plus” is the short version, and it holds as long as the paths you add are genuinely separate outcomes, which is exactly what a tree guarantees.
A check worth doing
The probabilities on any single set of branches must add to 1, because something has to happen. And the probabilities of all the complete paths must add to 1 too.
If yours don’t, there’s an arithmetic error or a missing branch, and you’ve found it before it reached the answer.
With and without replacement
A bag holds 3 red and 2 blue counters.
With replacement — you put the first counter back. The second set of branches is identical to the first, still \(\frac{3}{5}\) and \(\frac{2}{5}\). The draws are independent.
Without replacement — you don’t. Now the second branches depend on what happened first. If a red was taken, 2 reds and 2 blues remain, so red is \(\frac{2}{4}\). If a blue was taken, 3 reds and 1 blue remain, so red is \(\frac{3}{4}\).
The tree handles this effortlessly, which is its real advantage — the second layer simply carries different numbers depending on which branch you’re on. Trying to do the same problem in your head is where people go wrong.
\[P(\text{two reds, no replacement}) = \frac{3}{5} \times \frac{2}{4} = \frac{6}{20} = \frac{3}{10}\]“At least one”
\(P(\text{at least one red})\) means adding several paths, and it’s almost always quicker through the complement:
\[P(\text{at least one}) = 1 - P(\text{none})\]For two draws with replacement, \(P(\text{no red}) = \frac{2}{5} \times \frac{2}{5} = \frac{4}{25}\), so at least one red is \(\frac{21}{25}\). One calculation instead of three.
Common mix-ups
- Adding along a branch instead of multiplying.
- Reusing the first probabilities on the second layer when there’s no replacement.
- Missing a path. “One red and one blue” happens two ways — red then blue, and blue then red — so both get added.
- Skipping the total check. The complete paths must sum to 1.
Check yourself
A bag has 4 green and 1 yellow counter. Two are drawn without replacement. What’s the probability of one of each?
Green then yellow is \(\frac{4}{5} \times \frac{1}{4} = \frac{1}{5}\). Yellow then green is \(\frac{1}{5} \times \frac{4}{4} = \frac{1}{5}\). Both count, so the answer is \(\frac{2}{5}\).
LEVEL 3 · REFLECT
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