Probability Worked Solutions — A-Level Maths

Fully worked, step-by-step solutions to A-Level Probability questions. See exactly how to solve problems on sample space, equally likely, complement, mutually exclusive.

sample spaceequally likelycomplementmutually exclusiveaddition ruleconcept
A-Level70 questionsStep-by-step solutions
Question 1
2 markseasy
A fair six-sided die is rolled once. Find the probability of rolling an even number.

Worked solution

  1. Count the outcomes in the sample space

    n(S)=6n(S)=6

    A fair die has six equally likely outcomes.

  2. Count the favourable outcomes

    n(E)=3n(E)=3

    The even numbers are 2, 4 and 6.

  3. Divide favourable by total and simplify

    P(E)=36=12P(E)=\frac{3}{6}=\frac{1}{2}

    Probability is favourable over total.

Answer
12\frac{1}{2}
Question 2
2 markseasy
A fair six-sided die is rolled once. Find the probability of rolling a number greater than 4.

Worked solution

  1. Identify the outcomes greater than 4

    E={5,6}E=\{5,6\}

    Only 5 and 6 exceed 4.

  2. Count sample space and event

    n(S)=6,  n(E)=2n(S)=6,\; n(E)=2

    Six outcomes in total, two favourable.

  3. Form and simplify the probability

    P(E)=26=13P(E)=\frac{2}{6}=\frac{1}{3}

    Simplify the fraction.

Answer
13\frac{1}{3}
Question 3
2 markseasy
A fair spinner is numbered 1 to 8. Find the probability of landing on a multiple of 3.

Worked solution

  1. List the multiples of 3 from 1 to 8

    E={3,6}E=\{3,6\}

    The multiples of 3 in range are 3 and 6.

  2. Count the outcomes

    n(E)=2,  n(S)=8n(E)=2,\; n(S)=8

    Two favourable out of eight sectors.

  3. Form and simplify the probability

    P(E)=28=14P(E)=\frac{2}{8}=\frac{1}{4}

    Simplify.

Answer
14\frac{1}{4}
Question 4
2 markseasy
A bag contains 5 red and 3 blue counters. One counter is taken at random. Find the probability it is red.

Worked solution

  1. Find the total number of counters

    n(S)=5+3=8n(S)=5+3=8

    There are eight counters altogether.

  2. State the number of red counters

    n(R)=5n(R)=5

    Five of them are red.

  3. Write the probability

    P(R)=58P(R)=\frac{5}{8}

    Red counters over total counters.

Answer
58\frac{5}{8}
Question 5
2 markseasy
A bag contains 4 red and 6 green counters. One counter is taken at random. Find the probability it is green.

Worked solution

  1. Find the total number of counters

    n(S)=4+6=10n(S)=4+6=10

    Ten counters in total.

  2. State the number of green counters

    n(G)=6n(G)=6

    Six are green.

  3. Form and simplify the probability

    P(G)=610=35P(G)=\frac{6}{10}=\frac{3}{5}

    Simplify the fraction.

Answer
35\frac{3}{5}

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