Year 10 Probability Worksheets & Practice Tests

Worked examples

Year 10 probability: 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 · Conditional probability from a two-way table

   A group of 80 students were surveyed about whether they play sport and whether they have a part-time job. The results are shown below.

   	Plays sport	Does not play sport	Total
   Has a job	18	12	30
   No job	35	15	50
   Total	53	27	80

   Find the probability that a randomly chosen student plays sport, given that they have a job. Write your answer as a fraction in simplest form.

   Show worked solution

   Key idea: $P(A \mid B) = \dfrac{P(A \cap B)}{P(B)}$, which in a table means: restrict to the row (or column) that matches the given condition, then find the fraction who also satisfy $A$.

   1. The condition is "has a job", so restrict to the "Has a job" row. There are 30 students with a job.
   2. Of those 30, the number who also play sport is 18.
   3. $P(\text{plays sport} \mid \text{has a job}) = \frac{18}{30} = \frac{3}{5}$

   Answer: $\dfrac{3}{5}$

2. Example 2 · Tree diagram with conditional branches

   A bag contains 5 red marbles and 3 blue marbles. A marble is drawn at random and not replaced. A second marble is then drawn. Find the probability that the second marble is red, given that the first marble was blue.

   Show worked solution

   Key idea: When drawing without replacement, the probabilities on the second set of branches depend on what happened on the first draw, making this a conditional probability problem.

   1. The condition is that the first marble drawn was blue. After removing one blue marble, the bag contains 5 red and 2 blue marbles, giving 7 marbles total.
   2. On the second draw, the probability of red is now based on these 7 remaining marbles.
   3. $P(\text{2nd red} \mid \text{1st blue}) = \frac{5}{7}$

   Answer: $\dfrac{5}{7}$

3. Example 3 · Conditional probability in a real-life context

   At a school, $60\%$ of students study French. Of the students who study French, $40\%$ also study Japanese. Of the students who do not study French, $25\%$ study Japanese. A student is chosen at random and is found to be studying Japanese. Find the probability that this student also studies French. Give your answer as a decimal rounded to two decimal places.

   Show worked solution

   Key idea: Use a tree diagram and the conditional probability formula $P(F \mid J) = \dfrac{P(F \cap J)}{P(J)}$.

   1. Define events: $F$ = studies French, $J$ = studies Japanese. Set up the known probabilities: $P(F) = 0.6$, $P(F') = 0.4$, $P(J \mid F) = 0.4$, $P(J \mid F') = 0.25$.
   2. Find the probability of each path that leads to Japanese:
      $\begin{aligned} P(F \cap J) &= 0.6 \times 0.4 = 0.24 \\ P(F' \cap J) &= 0.4 \times 0.25 = 0.10 \end{aligned}$
   3. Find the total probability of studying Japanese:
      $P(J) = 0.24 + 0.10 = 0.34$
   4. Apply the conditional probability formula:
      $P(F \mid J) = \frac{P(F \cap J)}{P(J)} = \frac{0.24}{0.34} = \frac{12}{17} \approx 0.71$

   Answer: $0.71$

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