Build the distribution from the experiment

List the outcomes that qualify

In 5 of 20 recent exams
Paper 1, no calculator
1 to 6 marks

How it is set in the exam

Paper 1 (no calculator) mostly (4 of its 5 exam versions), 1–6 marks per part: usually the later parts of a Section B question, after a table or tree part and sometimes before an expected-value part; once it opens the question. Typical (5 of 20 exam versions).

Spot it in the exam

Exam phrases to spot (6)

“the die is rolled twice”
“the score is …”
“show that $\mathrm{P}(S = s) = \ldots$ ”
“complete the probability distribution”
“the total of the two rolls is at least …”
“ $X$ and $Y$ are independent”

In disguise (4)

a table with its entries missing, and the experiment described above it
the number of successes in two, three or four rounds, on : list the orders by hand
"describe, in words, the event …": say what happens in the experiment
an unknown number on a face, fixed by the largest possible total or by one of the probabilities

Variants (2)

the experiment: two fair dice (a sample space diagram); a biased die or spinner used twice; two independent random variables; a few independent rounds with a score; counts read from a tree
asked: table entries, the probability of an event, an unknown face or probability, the event in words

Looks like it, but isn't (3)

the table is given and $k$ is unknown: Probability distribution tables
on , a fixed number $n$ of independent trials with the same probability, and the probability of a count: Binomial probabilities
draws without replacement, and a probability asked for (not a distribution): Tree diagrams and repeated draws

Worked example

A biased spinner used twice: a total of at least 7

Paper 1 (no calculator)

Question

A spinner has four sectors, numbered $1$ , $2$ , $3$ and $6$ . It lands on $1$ with probability $\frac{1}{2}$ , and on each of the other three numbers with probability $\frac{1}{6}$ . Priya spins it twice, and $T$ is the total of her two numbers.
(a) Find $\mathrm{P}(T = 4)$ . [3]
(b) Priya wins a prize if $T$ is at least $7$ . Find the probability that she wins a prize. [6]

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Which method?

The experiment	Tool
two fair dice, or any equally likely outcomes	a sample space diagram: count the cells, divide by the total
a biased die or spinner used twice, or two independent variables	list the ordered pairs that qualify; multiply along each pair; count both orders of an unequal pair
the same trial repeated a few times: a count of successes, or a score that depends only on it, on ; a total over rounds with three or more results, on either paper	the probability of one order $\times$ the number of orders: $\binom{n}{r}$ when the outcome is $r$ of one result and $n - r$ of another; when it mixes three different results, list the orders ( $2 + 1 + 0$ in $3$ rounds has $6$ orders, not $3$ ). On , a count of successes (or a score that depends only on it) is Binomial probabilities. The orders are still counted, by hand or with the GDC's binomial coefficients, when each round has three or more results ( $2$ wins and $1$ draw in $3$ games: $\binom{3}{2} = 3$ orders)
draws without replacement, or stages that change	a tree (Tree diagrams); collect the paths by the value of $X$

Steps

Define the variable:
what is counted or scored, and its possible values. A value that can't happen is left out (or given probability $0$ ).
List the outcomes
for each value (or for the event asked): ordered pairs for two stages, a sample space diagram for equally likely outcomes.
Find each probability.
Equally likely: count ÷ total. Independent stages with unequal probabilities: multiply, and count every order, because $(1, 4)$ and $(4, 1)$ are different outcomes.
Add
the probabilities of the outcomes that qualify, and nothing else.
Check
that the whole table adds up to $1$ , or find the last entry as $1$ minus the others (an accepted route).
An unknown face
comes from an extreme (the largest total uses the two largest faces) or from matching one probability.

Common mistakes

One order only
for an unequal pair: $(1, 6)$ and $(6, 1)$ are two outcomes, so $\mathrm{P}(T = 7) = 2 \times \frac{1}{2} \times \frac{1}{6}$ , not $\frac{1}{2} \times \frac{1}{6}$ . Over three questions, two correct and one wrong have $3$ orders.
Counting a double twice
$(2, 2)$ happens in one order only.
Including an outcome that doesn't qualify
(a total of $6$ in "at least $7$ "): the method mark is lost (), and so is the final answer. Only the marks for each correct qualifying total survive.

More mistakes (6)

Treating the totals as equally likely.
Two dice give $11$ possible totals, but they are not equally likely: count the $36$ cells.
Merging repeated faces.
A die with faces $1, 1, 2, 3$ has four equally likely faces, and the two $1$ s are two different outcomes: $\mathrm{P}(1) = \frac{2}{4}$ , not $\frac{1}{3}$ .
Adding where you should multiply
"the first spin is $1$ and the second is $6$ " is a product; the different outcomes that qualify are then added.
Writing the event in symbols when asked for words
" $X \le 1$ " scores nothing. A real markscheme accepted the event written as a sentence about the total score, but not the same event written in symbols. Describing what happens in the experiment is the safest answer: "Tom answers at most one question correctly".
An unknown face with no justification
when the face is fixed by a probability, a correct value with no working has scored . Show the probability it matches, and rule out the other possibilities. When it is fixed by the largest total, write the equation anyway ( $4 + n = 11$ ).
A table that doesn't add up to $1$
a counting slip. Check before using it for $\mathrm{E}(X)$ .

Easy to confuse with

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