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TMUA Counting and Probability

Counting and probability questions reward a clear system: list outcomes in order, or draw a tree. Work with small cases that you can count by hand.

Paper 1 and Paper 2 topic. Specification codes: M2.5, M7.5, M7.6, M7.7. Last checked 12 November 2026.

Counting

Fill positions one at a time, handling the most restricted position first. Without repetition, each choice reduces the options for the next.

Probability rules

  • Mutually exclusive: P(A or B)=P(A)+P(B)P(A\text{ or }B)=P(A)+P(B).
  • Independent: P(A and B)=P(A)P(B)P(A\text{ and }B)=P(A)P(B).
  • Conditional: P(A∣B)=P(A and B)P(B)P(A\mid B)=\dfrac{P(A\text{ and }B)}{P(B)}.

Without replacement

After each draw, the total and the number of that type both change. Draw a tree and multiply along branches.

Strategy

Decide first whether order matters. Arrangements use products, as in 5×4×35\times4\times3, while selections where order is irrelevant use (nr)\binom nr.

When a question says 'at least one', it is often quicker to find the probability of the opposite event and subtract it from 1. Then check that your answer lies between 0 and 1.

Worked examples

Example 1

How many even three-digit numbers can be made from the digits 1 to 5 with no repeats?

Show the worked solution

The last digit is 2 or 4 (2 ways). Then 4 digits remain for the first place and 3 for the second: 2×4×3=242\times4\times3=24.

Answer: 24

Example 2

A bag has 4 red and 6 blue counters. Two are taken without replacement. Find the probability they are the same colour.

Show the worked solution

P(RR)=410⋅39=1290P(RR)=\tfrac4{10}\cdot\tfrac39=\tfrac{12}{90} and P(BB)=610⋅59=3090P(BB)=\tfrac6{10}\cdot\tfrac59=\tfrac{30}{90}. The sum is 4290=715\tfrac{42}{90}=\tfrac7{15}.

Answer: 715\tfrac7{15}

Example 3

Two fair dice are rolled and at least one shows a 6. Find the probability the total is 7.

Show the worked solution

There are 11 outcomes with at least one 6. Only (6,1)(6,1) and (1,6)(1,6) give a total of 7, so the probability is 211\tfrac2{11}.

Answer: 211\tfrac2{11}

Common mistakes

  • Using the full sample space when the question gives a condition.
  • Keeping the same denominator for the second draw when there is no replacement.
  • Adding probabilities for 'and' or multiplying them for 'or'.
  • Double-counting outcomes that fit two categories.

Test yourself on this topic with our free practice questions.

Quick answers

How do you count possibilities without repeats?

Multiply the number of choices for each position, reducing the choices as items are used up.

What is conditional probability?

The probability of an event given that another has happened. Restrict attention to the outcomes where the condition holds.

How do you solve probability without replacement?

Update the total and the counts after each draw, then multiply along the branches of a tree.

Practise it under timed conditions

A free timed sample paper with a worked solution for every question. No sign-up. For two full papers and a weak-spot drill, see Full Access.

Free sample paper

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