IMAT Logical Reasoning · Numerical Reasoning

Counting Systematically, Overlaps and Chance

List possibilities in a fixed order, multiply independent choices, subtract the overlap when groups share members, and find a probability as favourable cases over all cases.

All Numerical Reasoning notes

Why this matters

Systematic listing (codes, coins, scoring tables, seating plans) and overlapping groups appear in the Cambridge papers; probability appeared in 2012 as a screening-test question and in 2024 as a prize draw.

Concept 1 of 3: Systematic listing and the multiplication rule

Counting goes wrong when you list at random: you miss some cases and repeat others. Fix an order (smallest first, or one position at a time) and the list becomes complete and checkable. When a choice is made in independent stages, the number of outcomes is the product of the choices at each stage.

Definition

Systematic listing: fix the first item, run through every option for the next, then move the first item on. Useful when there are conditions that a formula cannot handle.
Multiplication rule: $a$ ways for one stage and $b$ for the next give $a \times b$ outcomes.
Arrangements of $n$ different items in a row: $n! = n \times (n-1) \times \dots \times 1$ .
Order matters (codes, rankings): $n \times (n-1) \times \dots$ for as many places as you fill.
Order does not matter (choosing a team): divide by the number of orders, $\binom{n}{r} = \dfrac{n!}{r!\,(n-r)!}$ .

Choosing r items from n

\text{ordered: } \frac{n!}{(n-r)!} \qquad \text{unordered: } \binom{n}{r} = \frac{n!}{r!\,(n-r)!}

$n$ number of items available
$r$ number chosen

Worked example

A café offers 3 starters, 4 main courses and 2 desserts. How many different three-course meals are there? Separately, in how many ways can you choose 2 of your 5 friends to come to the cinema?

Practice this conceptself-check · 4 quick reps

Does order matter?

A code 123 differs from 321, so order matters. A pair of friends AB is the same as BA, so order does not. Using the ordered count for a selection doubles (or more) the true answer.

Concept 2 of 3: Overlapping groups and Venn reasoning

When two groups share members, adding their sizes counts the shared members twice. Subtract the overlap once and you get the number in at least one group. A quick Venn diagram, filling the overlap first and working outwards, keeps the regions straight.

Definition

Either or both: $n(A \cup B) = n(A) + n(B) - n(A \cap B)$ .
Neither = total $-\ n(A \cup B)$ .
Only A = $n(A) - n(A \cap B)$ .
Smallest possible overlap: $n(A) + n(B) - \text{total}$ , if that is positive.
Fill a Venn diagram from the centre outwards: overlap first, then the "only" regions, then "neither".

Two overlapping groups

n(A \cup B) = n(A) + n(B) - n(A \cap B)

$A \cup B$ in A or B or both
$A \cap B$ in both A and B

Worked example

Of 50 students, 32 study French, 21 study Spanish and 9 study both. How many study neither language? How many study French but not Spanish?

Practice this conceptself-check · 3 quick reps

Adding two groups counts the overlap twice

45 with one condition and 30 with another do not make 75 people with a condition if some have both. Subtract the overlap once:

n(A) + n(B) - n(A \cap B)

Concept 3 of 3: Probability in context

Probability is a fraction: the number of ways the event can happen over the number of equally likely outcomes. For events that happen one after the other, multiply, remembering that taking an item away changes what is left. For tests and screening, imagine a round number of people (say 1000) and count how many land in each group: counts are much harder to get wrong than conditional probabilities.

Definition

$P(\text{event}) = \dfrac{\text{favourable outcomes}}{\text{all equally likely outcomes}}$ , between 0 and 1.
$P(\text{not } A) = 1 - P(A)$ .
Independent events: $P(A \text{ and } B) = P(A) \times P(B)$ .
Without replacement, the second probability is out of one fewer item.
Either A or B (no overlap): add. With overlap: add and subtract the overlap, as with groups.
Natural frequencies: turn percentages into counts out of a round total, then read the answer as count over count.

Two events in turn

P(A \text{ then } B) = P(A) \times P(B \text{ given } A)

$P(B \text{ given } A)$ probability of B once A has happened; equals $P(B)$ if they are independent

Worked example

In a town of 5,000 people, 2% have a certain condition. A test is positive for 90% of people who have it and for 5% of people who do not. Of the people who test positive, what fraction have the condition?

Practice this conceptself-check · 3 quick reps

Without replacement, the second fraction changes

After one red counter is removed, there is one red fewer and one counter fewer. Squaring the first probability treats the draws as if the counter had been put back.

A positive test is not the same as having the condition

When a condition is rare, most positive results can come from the large healthy group. Count people in each group out of a round total before answering.

Summary — formulas & gotchas at a glance

A revision cheat-sheet for the formulas and gotchas above. Click any concept name to jump back to its full explanation.

Formulas (3)

Systematic listing and the multiplication rule
Choosing r items from n
$\text{ordered: } \frac{n!}{(n-r)!} \qquad \text{unordered: } \binom{n}{r} = \frac{n!}{r!\,(n-r)!}$
Overlapping groups and Venn reasoning
Two overlapping groups
$n(A \cup B) = n(A) + n(B) - n(A \cap B)$
Probability in context
Two events in turn
$P(A \text{ then } B) = P(A) \times P(B \text{ given } A)$

Watch out for (4)

Does order matter?
→ Systematic listing and the multiplication rule
Adding two groups counts the overlap twice
→ Overlapping groups and Venn reasoning
Without replacement, the second fraction changes
→ Probability in context
A positive test is not the same as having the condition
→ Probability in context

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