Year 10 Multi-Step Probability and Tree Diagrams Worksheets & Practice

Worked examples

Year 10 multi-step probability and tree diagrams: worked examples

Three examples, from a standard question to a worded problem. Have a go at each one before you open the solution.

1. Example 1 · Two-step tree diagram with equal branches

   A fair coin is flipped twice. Draw a tree diagram and find the probability of getting exactly one Head.

   Show worked solution

   Key idea: Multiply probabilities along each branch to find the probability of each outcome, then add the probabilities of all outcomes that match the event.

   1. List the branches for the first flip: Head (H) with probability $\dfrac{1}{2}$, Tail (T) with probability $\dfrac{1}{2}$.
   2. From each branch, list a second flip: H or T, each with probability $\dfrac{1}{2}$. The four outcomes are HH, HT, TH, TT.
   3. Multiply along branches to find each outcome probability:
      $\begin{aligned} P(\text{HH}) &= \tfrac{1}{2} \times \tfrac{1}{2} = \tfrac{1}{4} \\ P(\text{HT}) &= \tfrac{1}{2} \times \tfrac{1}{2} = \tfrac{1}{4} \\ P(\text{TH}) &= \tfrac{1}{2} \times \tfrac{1}{2} = \tfrac{1}{4} \\ P(\text{TT}) &= \tfrac{1}{2} \times \tfrac{1}{2} = \tfrac{1}{4} \end{aligned}$
   4. Check: $\dfrac{1}{4} + \dfrac{1}{4} + \dfrac{1}{4} + \dfrac{1}{4} = 1$ ✓
   5. Exactly one Head occurs for HT and TH:
      $P(\text{exactly one Head}) = \tfrac{1}{4} + \tfrac{1}{4} = \tfrac{1}{2}$ Answer: $\dfrac{1}{2}$

2. Example 2 · Two-step tree diagram with unequal branches

   A bag contains 3 red marbles and 2 blue marbles. One marble is drawn at random, its colour is noted, and it is put back in the bag. A second marble is then drawn at random. Find the probability that the two marbles drawn are different colours.

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   Key idea: Because the marble is replaced, the probabilities on the second set of branches are the same as on the first. Multiply along branches, then add the matching outcomes.

   1. First draw: $P(\text{Red}) = \dfrac{3}{5}$, $P(\text{Blue}) = \dfrac{2}{5}$.
   2. Second draw (same bag): $P(\text{Red}) = \dfrac{3}{5}$, $P(\text{Blue}) = \dfrac{2}{5}$.
   3. Multiply along branches for all four outcomes:
      $\begin{aligned} P(\text{RR}) &= \tfrac{3}{5} \times \tfrac{3}{5} = \tfrac{9}{25} \\ P(\text{RB}) &= \tfrac{3}{5} \times \tfrac{2}{5} = \tfrac{6}{25} \\ P(\text{BR}) &= \tfrac{2}{5} \times \tfrac{3}{5} = \tfrac{6}{25} \\ P(\text{BB}) &= \tfrac{2}{5} \times \tfrac{2}{5} = \tfrac{4}{25} \end{aligned}$
   4. Check: $\dfrac{9}{25} + \dfrac{6}{25} + \dfrac{6}{25} + \dfrac{4}{25} = \dfrac{25}{25} = 1$ ✓
   5. Different colours are RB and BR:
      $P(\text{different colours}) = \tfrac{6}{25} + \tfrac{6}{25} = \tfrac{12}{25}$ Answer: $\dfrac{12}{25}$

3. Example 3 · Three-step tree diagram: real-life problem

   At a school fundraiser, a student spins a spinner three times. The spinner has four equal sections numbered 1 to 4. A spin is called a win if the spinner lands on 4. The student wins a prize if they get a win on at least two of the three spins. Find the probability that the student wins a prize.

   Show worked solution

   Key idea: For each spin, $P(\text{win}) = \dfrac{1}{4}$ and $P(\text{not win}) = \dfrac{3}{4}$. Use a three-step tree diagram, multiply along branches, and add the probabilities for outcomes with two or three wins.

   1. Label each spin as W (win, $p = \dfrac{1}{4}$) or N (not win, $p = \dfrac{3}{4}$). There are $2^3 = 8$ outcomes.
   2. Outcomes with exactly two wins: WWN, WNW, NWW. Each has probability $\dfrac{1}{4} \times \dfrac{1}{4} \times \dfrac{3}{4} = \dfrac{3}{64}$.
   3. Outcome with exactly three wins: WWW. Its probability is $\dfrac{1}{4} \times \dfrac{1}{4} \times \dfrac{1}{4} = \dfrac{1}{64}$.
   4. Add all matching outcomes:
      $\begin{aligned} P(\text{at least two wins}) &= 3 \times \tfrac{3}{64} + \tfrac{1}{64} \\ &= \tfrac{9}{64} + \tfrac{1}{64} \\ &= \tfrac{10}{64} \\ &= \tfrac{5}{32} \end{aligned}$ Answer: $\dfrac{5}{32}$

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