Two events

Fill the Venn diagram, use the addition rule

In 6 of 20 recent exams
Paper 1 or 2
4 to 6 marks

How it is set in the exam

Paper 1 (no calculator) mostly, as a short Section A question of 4–6 marks in two or three parts; on Paper 2 (calculator) as the opening parts of a longer question, often with counts. Parts are 1–3 marks each. Common (6 of 20 exam versions, 4 of them on Paper 1 (no calculator)). It is also the first step of most Conditional probability and Are the events independent? parts, which follow it in the same question.

Spot it in the exam

Exam phrases to spot (6)

“find $\mathrm{P}(A \cup B)$ ”
“find $\mathrm{P}(A \cap B)$ ”
“find the probability that neither $A$ nor $B$ occurs”
“find $\mathrm{P}(A \cap B')$ ”
“complete the Venn diagram”
“find the least possible value of $\mathrm{P}(A \cap B)$ ”

In disguise (3)

a context with "at least one of the two" (the union) or "exactly one of" (the two "only" regions)
counts instead of probabilities: "of the $n$ members, … do both", with the probability asked only at the end
percentages instead of probabilities: the same diagram, with $100$ as the total, and the answer in the form the question uses

Variants (2)

given: the singles and the intersection; the singles and the union; the singles and "neither"; the singles and "exactly one"; or counts
asked: the union, the intersection, "neither", an "only" region, exactly one, or the least and greatest possible overlap

Looks like it, but isn't (5)

a probability given that something has happened: Conditional probability
a region is given in terms of an unknown ( $x$ or $k$ ), and independence or a conditional fixes it: Probabilities in terms of an unknown
"are $A$ and $B$ independent?": Are the events independent?
independence is given and only $\mathrm{P}(A \cap B)$ is asked: multiply (Independent events)
"at least one" over repeated draws or stages: Tree diagrams and repeated draws ( $1 - \mathrm{P}(\text{none})$ )

Worked example

Union, neither and exactly one from the singles and the intersection

Paper 1 (no calculator)

Question

Events $A$ and $B$ are such that $\mathrm{P}(A) = 0.42$ , $\mathrm{P}(B) = 0.35$ and $\mathrm{P}(A \cap B) = 0.14$ .
(a) Find $\mathrm{P}(A \cup B)$ . [2]
(b) Find the probability that neither $A$ nor $B$ occurs. [2]
(c) Find the probability that exactly one of $A$ and $B$ occurs. [2]

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Steps

Put the intersection in first:
it is given, or the addition rule gives it, $\mathrm{P}(A \cap B) = \mathrm{P}(A) + \mathrm{P}(B) - \mathrm{P}(A \cup B)$ (with "neither" given, $\mathrm{P}(A \cup B) = 1 - \mathrm{P}(\text{neither})$ first). If the fact given is another region, such as "exactly one", let the intersection be $x$ : the "only" regions are $\mathrm{P}(A) - x$ and $\mathrm{P}(B) - x$ , and the fact gives one linear equation in $x$ .
Fill the "only" regions:
$\mathrm{P}(A) - \mathrm{P}(A \cap B)$ and $\mathrm{P}(B) - \mathrm{P}(A \cap B)$ .
Neither
$= 1 - \mathrm{P}(A \cup B)$ . Check that the four regions add to $1$ (or to the total).
Read off
the region or sum of regions asked for, and state it as the answer (or label it on the diagram as that probability). With counts, divide by the total last.
Bounds for the overlap:
the least overlap makes the union as big as possible, and $\mathrm{P}(A \cup B) \le 1$ , so $\mathrm{P}(A \cap B) \ge \mathrm{P}(A) + \mathrm{P}(B) - 1$ (and $\ge 0$ ). The greatest overlap puts the smaller event inside the larger, so $\mathrm{P}(A \cap B) \le$ the smaller of $\mathrm{P}(A)$ and $\mathrm{P}(B)$ . Write the reason with each bound.

Common mistakes

Writing $\mathrm{P}(A)$ in the " $A$ only" region.
$0.42$ is the whole circle; the "only" region is $0.42 - 0.14$ .
Adding the singles for the union
when the events overlap. $\mathrm{P}(A) + \mathrm{P}(B)$ is only the union for mutually exclusive events.
"Neither" as $1 - \mathrm{P}(A) - \mathrm{P}(B)$
that removes the overlap twice ( $0.23$ instead of $0.37$ in Example 1.1).

More mistakes (5)

"Exactly one" for "at least one"
(or the reverse): $0.49$ against $0.63$ in Example 1.1.
A value left unlabelled on the diagram
the markscheme gives unless it is stated as the answer or labelled as that probability.
A bound with no reason
" $\mathrm{P}(A \cup B) \le 1$ " or " $A$ inside $B$ " has to be written.
The least overlap given as $0$ , or as $\mathrm{P}(A)\,\mathrm{P}(B)$
when $\mathrm{P}(A) + \mathrm{P}(B) > 1$ the events must overlap, by at least $\mathrm{P}(A) + \mathrm{P}(B) - 1$ ( $0.15$ in Example 1.3, not $0$ and not $0.315$ ). The product is the overlap only when the events are independent.
Counts and probabilities mixed
in one diagram: work in counts, divide by the total at the end.

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