Exhaustive and mutually exclusive events Worked Solutions — GCSE Maths

Fully worked, step-by-step solutions to GCSE Exhaustive and mutually exclusive events questions. See exactly how to solve problems on probabilities add to 1, exhaustive outcomes, complement rule, P(not A) = 1 - P(A).

probabilities add to 1exhaustive outcomescomplement ruleP(not A) = 1 - P(A)decimal formequally likely outcomes
GCSE Foundation70 questionsStep-by-step solutions
Question 1
1 markeasy
A counter is taken at random from a bag and then put back. The only possible outcomes are red, blue and green. Exactly one of these outcomes must happen. P(red)=12P(\text{red}) = \frac{1}{2}. P(blue)=15P(\text{blue}) = \frac{1}{5}. Work out P(green)P(\text{green}). Give your answer as a fraction in its simplest form.

Worked solution

  1. Use the fact that the probabilities add to 11

    P(red)+P(blue)+P(green)=1P(\text{red}) + P(\text{blue}) + P(\text{green}) = 1

    One of the outcomes must happen and no two can happen together, so the probabilities add to exactly 11. The question gives P(red)=12P(\text{red}) = \frac{1}{2} and P(blue)=15P(\text{blue}) = \frac{1}{5}.

  2. Subtract the given probabilities from 11

    P(green)=1−710=1010−710=310P(\text{green}) = 1 - \frac{7}{10} = \frac{10}{10} - \frac{7}{10} = \frac{3}{10}

    Adding the probabilities given comes to 710\frac{7}{10}, and the rest of the certainty has to belong to green.

  3. State the answer

    310\frac{3}{10}

    So the probability is 310\frac{3}{10}.

Answer
310\frac{3}{10}
Question 2
2 markseasy
A football team plays a match. The only possible outcomes are win, draw and lose. Exactly one of these outcomes must happen. P(win)=25P(\text{win}) = \frac{2}{5}. P(draw)=14P(\text{draw}) = \frac{1}{4}. Work out P(lose)P(\text{lose}). Give your answer as a fraction in its simplest form.

Worked solution

  1. Use the fact that the probabilities add to 11

    P(win)+P(draw)+P(lose)=1P(\text{win}) + P(\text{draw}) + P(\text{lose}) = 1

    One of the outcomes must happen and no two can happen together, so the probabilities add to exactly 11. The question gives P(win)=25P(\text{win}) = \frac{2}{5} and P(draw)=14P(\text{draw}) = \frac{1}{4}.

  2. Subtract the given probabilities from 11

    P(lose)=1−1320=2020−1320=720P(\text{lose}) = 1 - \frac{13}{20} = \frac{20}{20} - \frac{13}{20} = \frac{7}{20}

    Adding the probabilities given comes to 1320\frac{13}{20}, and the rest of the certainty has to belong to lose.

  3. State the answer

    720\frac{7}{20}

    So the probability is 720\frac{7}{20}.

Answer
720\frac{7}{20}
Question 3
1 markeasy
A train arrives at a station. The only possible outcomes are early, on time and late. Exactly one of these outcomes must happen. P(early)=110P(\text{early}) = \frac{1}{10}. P(on time)=34P(\text{on time}) = \frac{3}{4}. Work out P(late)P(\text{late}). Give your answer as a fraction in its simplest form.

Worked solution

  1. Use the fact that the probabilities add to 11

    P(early)+P(on time)+P(late)=1P(\text{early}) + P(\text{on time}) + P(\text{late}) = 1

    One of the outcomes must happen and no two can happen together, so the probabilities add to exactly 11. The question gives P(early)=110P(\text{early}) = \frac{1}{10} and P(on time)=34P(\text{on time}) = \frac{3}{4}.

  2. Subtract the given probabilities from 11

    P(late)=1−1720=2020−1720=320P(\text{late}) = 1 - \frac{17}{20} = \frac{20}{20} - \frac{17}{20} = \frac{3}{20}

    Adding the probabilities given comes to 1720\frac{17}{20}, and the rest of the certainty has to belong to late.

  3. State the answer

    320\frac{3}{20}

    So the probability is 320\frac{3}{20}.

Answer
320\frac{3}{20}
Question 4
2 markseasy
Maya spins a biased spinner. The only possible outcomes are red, blue and green. Exactly one of these outcomes must happen. P(red)=13P(\text{red}) = \frac{1}{3}. P(blue)=16P(\text{blue}) = \frac{1}{6}. Work out P(green)P(\text{green}). Give your answer as a fraction in its simplest form.

Worked solution

  1. Use the fact that the probabilities add to 11

    P(red)+P(blue)+P(green)=1P(\text{red}) + P(\text{blue}) + P(\text{green}) = 1

    One of the outcomes must happen and no two can happen together, so the probabilities add to exactly 11. The question gives P(red)=13P(\text{red}) = \frac{1}{3} and P(blue)=16P(\text{blue}) = \frac{1}{6}.

  2. Subtract the given probabilities from 11

    P(green)=1−12=22−12=12P(\text{green}) = 1 - \frac{1}{2} = \frac{2}{2} - \frac{1}{2} = \frac{1}{2}

    Adding the probabilities given comes to 12\frac{1}{2}, and the rest of the certainty has to belong to green.

  3. State the answer

    12\frac{1}{2}

    So the probability is 12\frac{1}{2}.

Answer
12\frac{1}{2}
Question 5
1 markeasy
Ben flips a biased coin. The only possible outcomes are heads and tails. Exactly one of these outcomes must happen. P(heads)=37P(\text{heads}) = \frac{3}{7}. Work out P(tails)P(\text{tails}). Give your answer as a fraction in its simplest form.

Worked solution

  1. Use the fact that the probabilities add to 11

    P(heads)+P(tails)=1P(\text{heads}) + P(\text{tails}) = 1

    One of the outcomes must happen and no two can happen together, so the probabilities add to exactly 11. The question gives P(heads)=37P(\text{heads}) = \frac{3}{7}.

  2. Subtract the given probabilities from 11

    P(tails)=1−37=77−37=47P(\text{tails}) = 1 - \frac{3}{7} = \frac{7}{7} - \frac{3}{7} = \frac{4}{7}

    Adding the probabilities given comes to 37\frac{3}{7}, and the rest of the certainty has to belong to tails.

  3. State the answer

    47\frac{4}{7}

    So the probability is 47\frac{4}{7}.

Answer
47\frac{4}{7}

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