Tree diagrams Worked Solutions — GCSE Maths

Fully worked, step-by-step solutions to GCSE Tree diagrams questions. See exactly how to solve problems on branches from a node sum to 1, three branches, tree diagram, multiply along the branches.

branches from a node sum to 1three branchestree diagrammultiply along the branchesfirst branchindependent events
GCSE Foundation70 questionsStep-by-step solutions
Question 1
1 markeasy
A tree diagram is drawn for two spins of a spinner. The probability that the spinner lands on red is 14\frac{1}{4}. Write down the probability that the spinner does not land on red. Give your answer as a fraction in its simplest form.

Worked solution

  1. Use the rule that the branches from a node add to 11

    P(red)+P(not red)=1P(\text{red}) + P(\text{not red}) = 1

    At the first node the spinner either lands on red or it does not — there is no third possibility. Every set of branches leaving a node must add to exactly 11, so these two probabilities add to 11.

  2. Subtract from 11

    P(not red)=1−14=34P(\text{not red}) = 1 - \frac{1}{4} = \frac{3}{4}

    1−14=341 - \frac{1}{4} = \frac{3}{4}.

  3. State the answer

    34\frac{3}{4}

    So the probability that the spinner does not land on red is 34\frac{3}{4}, and 14+34=1\frac{1}{4} + \frac{3}{4} = 1 as a check.

Answer
34\frac{3}{4}
Question 2
1 markeasy
A tree diagram is drawn for two flips of a biased coin. The probability that the coin lands on heads is 25\frac{2}{5}. Write down the probability that the coin does not land on heads. Give your answer as a fraction in its simplest form.

Worked solution

  1. Use the rule that the branches from a node add to 11

    P(heads)+P(not heads)=1P(\text{heads}) + P(\text{not heads}) = 1

    At the first node the coin either lands on heads or it does not — there is no third possibility. Every set of branches leaving a node must add to exactly 11, so these two probabilities add to 11.

  2. Subtract from 11

    P(not heads)=1−25=35P(\text{not heads}) = 1 - \frac{2}{5} = \frac{3}{5}

    1−25=351 - \frac{2}{5} = \frac{3}{5}.

  3. State the answer

    35\frac{3}{5}

    So the probability that the coin does not land on heads is 35\frac{3}{5}, and 25+35=1\frac{2}{5} + \frac{3}{5} = 1 as a check.

Answer
35\frac{3}{5}
Question 3
2 markseasy
A tree diagram is drawn for two rolls of a biased dice. The probability that the dice lands on six is 16\frac{1}{6}. Write down the probability that the dice does not land on six. Give your answer as a fraction in its simplest form.

Worked solution

  1. Use the rule that the branches from a node add to 11

    P(six)+P(not six)=1P(\text{six}) + P(\text{not six}) = 1

    At the first node the dice either lands on six or it does not — there is no third possibility. Every set of branches leaving a node must add to exactly 11, so these two probabilities add to 11.

  2. Subtract from 11

    P(not six)=1−16=56P(\text{not six}) = 1 - \frac{1}{6} = \frac{5}{6}

    1−16=561 - \frac{1}{6} = \frac{5}{6}.

  3. State the answer

    56\frac{5}{6}

    So the probability that the dice does not land on six is 56\frac{5}{6}, and 16+56=1\frac{1}{6} + \frac{5}{6} = 1 as a check.

Answer
56\frac{5}{6}
Question 4
2 markseasy
A tree diagram is drawn for two spins of a spinner. The probability that the spinner lands on blue is 38\frac{3}{8}. Write down the probability that the spinner does not land on blue. Give your answer as a fraction in its simplest form.

Worked solution

  1. Use the rule that the branches from a node add to 11

    P(blue)+P(not blue)=1P(\text{blue}) + P(\text{not blue}) = 1

    At the first node the spinner either lands on blue or it does not — there is no third possibility. Every set of branches leaving a node must add to exactly 11, so these two probabilities add to 11.

  2. Subtract from 11

    P(not blue)=1−38=58P(\text{not blue}) = 1 - \frac{3}{8} = \frac{5}{8}

    1−38=581 - \frac{3}{8} = \frac{5}{8}.

  3. State the answer

    58\frac{5}{8}

    So the probability that the spinner does not land on blue is 58\frac{5}{8}, and 38+58=1\frac{3}{8} + \frac{5}{8} = 1 as a check.

Answer
58\frac{5}{8}
Question 5
2 markseasy
A tree diagram is drawn for two spins of a spinner. The spinner can land on red, blue or green. The probability that the spinner lands on red is 14\frac{1}{4}. The probability that the spinner lands on blue is 13\frac{1}{3}. Work out the probability that the spinner lands on green. Give your answer as a fraction in its simplest form.

Worked solution

  1. Use the rule that the branches from a node add to 11

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

    The spinner must land on one of red, blue or green, so those three branches leave the same node and must add to exactly 11.

  2. Add the two known branches, then subtract from 11

    14+13=712,1−712=512\frac{1}{4} + \frac{1}{3} = \frac{7}{12}, \quad 1 - \frac{7}{12} = \frac{5}{12}

    The two branches you are given come to 712\frac{7}{12} between them, so the green branch must take up the rest: 1−712=5121 - \frac{7}{12} = \frac{5}{12}.

  3. State the answer

    512\frac{5}{12}

    So the probability of green is 512\frac{5}{12}, and 14+13+512=1\frac{1}{4} + \frac{1}{3} + \frac{5}{12} = 1 as a check.

Answer
512\frac{5}{12}

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