Tree diagrams and repeated draws

Multiply along, add across

In 8 of 20 recent exams
1% of the marks
Paper 1 or 2
1 to 4 marks

How it is set in the exam

Paper 2 (calculator) more than Paper 1 (no calculator) (6 exam versions have a Paper 2 (calculator) part, 3 a Paper 1 (no calculator) part), 1–4 marks a part, usually a chain inside a Section B question: complete the tree, show a total, then a conditional or an expected value. Common (8 of 20 exam versions). On Paper 2 (calculator) the branch values are often probabilities found in an earlier part, from a normal or binomial model.

Spot it in the exam

Exam phrases to spot (7)

“a box is chosen at random and a ball is taken from it”
“two are taken without replacement”
“draw a tree diagram”
“complete the tree diagram”
“at least one”
“the first success is on the $k$ -th trial”
“the contents of the bag change after each draw”

In disguise (4)

a second stage described with "if … ; if not, …": "if it rains, the probability that she is late is …"
"one after the other", "in turn": the stages are the draws
"more than $k$ attempts are needed": the first $k$ all fail, a single path
a source is chosen first (one of two machines or suppliers, each with its own rate), then a count among $n$ items from that source is asked: not one $\mathrm{B}(n, p)$ but a tree, whose second-stage branches are binomial probabilities, one with each source's $p$

Variants (2)

structure: two stages (a choice, then an outcome); draws without replacement (two or three); contents that change after a draw; independent trials (first success on trial $k$ ); probabilities in terms of $n$
asked: the branches, a total probability (often "show that"), all the same, the same colour, at least one, exactly one, a first success, or an expression in $n$

Looks like it, but isn't (6)

identical independent trials, a fixed number $n$ of them with the same $p$ , and a number of successes asked on : Binomial probabilities
the result is known and the first stage is asked ("which bag was it?"): Conditional probability
the paths are to be sorted into a table of $\mathrm{P}(X = x)$ for a score or a count: Build the distribution from the experiment
an unknown in the branches and a path probability given: Probabilities in terms of an unknown
the least number of trials for $\mathrm{P}(\text{at least one})$ to pass a target: Find n
two players take turns until one succeeds: Sum to infinity, with a tree for one round

Worked example

Two stages: weather, then how she travels

Paper 1 (no calculator)

Question

On a school day, the probability that it rains in the morning is $\frac{2}{5}$ . If it rains, the probability that Maya takes the bus is $\frac{5}{6}$ . If it does not rain, the probability that she takes the bus is $\frac{1}{4}$ .
(a) Draw a tree diagram to represent this information, showing the probability on every branch. [3]
(b) Show that the probability that Maya takes the bus on a randomly chosen school day is $\frac{29}{60}$ . [3]
(c) Find the probability that, on a randomly chosen school day, it rains or Maya takes the bus. [2]

Try it yourself first, then compare.

Now try one like it

Write your working on paper or on screen. Paper Zero marks it step by step, like the worked example, and shows where each mark went. Free, with no account.

Try one like it

Steps

Draw the stages in order,
with the first-stage probabilities on the first branches (markschemes require them).
Fill the branches.
Each pair (or set) from one point adds to $1$ . Without replacement, the bottom of each later fraction drops by one per draw; the top drops only for a kind that has already been taken (in Example 2.2 a working bulb after a faulty one is still $\frac{26}{29}$ ). Unsimplified branch fractions ( $\frac{3}{6}$ ) are accepted. If the contents change after a draw, write the new contents before the next branches.
Multiply along each path
in the event, then add the paths. A "show that" needs every product and the unsimplified sum written before the given value.
Shortcuts:
at least one $= 1 - \mathrm{P}(\text{none})$ (adding every case also scores, but takes longer); exactly one of three draws has three orders; the first success on trial $k$ is $(1 - p)^{k-1}p$ for independent trials with the same $p$ .
Check:
all the paths of the tree add to $1$ .

Common mistakes

No first-stage probabilities on the tree
markschemes need the container (or weather) branches too.
Treating a draw without replacement as with replacement,
or as a binomial: markschemes then give only the mark for writing the first draw's probability ( $\frac{4}{30}$ in Example 2.2), and nothing more.
Dropping the top of every fraction without replacement
after a red is taken from the bag of $4$ red and $3$ blue in Tree diagrams, blue is still $\frac{3}{6}$ , not $\frac{2}{6}$ . Only the kind just taken loses one.

More mistakes (6)

Adding along a path, or multiplying across the paths.
Along is "and then" (multiply); across is "or" (add).
Forgetting the orders
"exactly one faulty" in three draws is three paths, not one.
The first success with the wrong number of failures
the first success on trial $k$ has $k - 1$ failures before it, $(1 - p)^{k-1}p$ .
Using $(1 - p)^{k-1}p$ without replacement
the formula needs the same $p$ on every trial (Your turn 4).
A "show that" with only the final value
write both products and the unsimplified sum.
Rounding a branch value
before multiplying on .

Easy to confuse with

Next typeProbabilities in terms of an unknown

More in Probability: Venn diagrams, trees and conditional probability

Are the events independent?

Now try one like it

Keep going

Easy to confuse with

More in Probability: Venn diagrams, trees and conditional probability