Mathematics · Probability
On probability trees, when do the denominators change on the second branch?
Whenever the first item is not replaced. That single phrase in the question controls every number on the second set of branches, and finding it should be the first thing you do.
Spotting it
Read for the words. "Without replacement", "does not put it back", "takes two sweets", "selects two students" — all mean the first item is gone before the second is chosen.
"With replacement", "puts it back", "rolls the dice twice", "spins again" mean nothing changes between trials.
If a question is silent, physical sense usually decides it. Taking two counters from a bag means you are holding the first one, so it is gone. Rolling a die twice obviously does not use the die up.
What changes, and what people forget
A bag holds 5 red and 3 blue counters, 8 in total. Two are taken without replacement.
First pick: P(red) = 5/8, P(blue) = 3/8.
Second pick, if the first was red: 4 red and 3 blue remain, 7 total. So P(red) = 4/7, P(blue) = 3/7.
Second pick, if the first was blue: 5 red and 2 blue remain, 7 total. P(red) = 5/7, P(blue) = 2/7.
Both the numerator and the denominator change, and the numerator only changes on the branch matching what was taken. Students reliably remember to reduce the total to 7 and then forget to reduce the count of the colour they drew. Check that each pair of branches still sums to 1 — 4/7 + 3/7 = 1 — and that error becomes impossible to miss.
Using the tree
Along the branches, multiply. Between separate paths, add.
P(both red) = 5/8 × 4/7 = 20/56.
P(one of each) needs two paths, red-then-blue and blue-then-red:
5/8 × 3/7 + 3/8 × 5/7 = 15/56 + 15/56 = 30/56Forgetting the second path is the other big mark-loser. "One of each" does not specify an order, so both orders count.
The shortcut worth knowing
For "at least one", go the long way round: P(at least one red) = 1 − P(no reds).
With three or four selections, listing every path containing a red is slow and error-prone, while there is only ever one path with none. One calculation instead of six, and far less to get wrong.
Leave your answers as fractions over the original denominator unless the question says otherwise. Cancelling to lowest terms is fine, but converting to a rounded decimal early loses accuracy that later parts of the question may depend on.
Working from a past paper? Mark your attempt against the published mark scheme before reading an answer like this one. Finding out where you lost the marks is worth more than seeing a correct solution, because the correct solution rarely tells you what you did instead.
Related questions
Ordered by how close they are to this one: same topic first, then the same subject.
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