Year 9 Probability Worksheets & Practice Tests

Worked examples

Year 9 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 · Tree diagram with replacement

   A bag contains one red (R) marble and one blue (B) marble. A marble is drawn at random, its colour is recorded, and it is placed back in the bag. A second marble is then drawn. List all possible outcomes using a tree diagram and find the probability of drawing two marbles of the same colour.

   Show worked solution

   Key idea: When drawing with replacement, the bag is returned to its original state before the second draw, so the probabilities on each branch stay the same.

   1. Draw the first branch: outcomes R or B, each with probability $\dfrac{1}{2}$.
   2. From each first outcome, draw a second branch: outcomes R or B, each again with probability $\dfrac{1}{2}$ (bag reset).
   3. List all four equally likely outcomes: RR, RB, BR, BB.
   4. Identify outcomes where both marbles match: RR and BB.
   5. Calculate the probability:
      $P(\text{same colour}) = \frac{2}{4} = \frac{1}{2}$ Answer: $\dfrac{1}{2}$

2. Example 2 · Two-step experiment without replacement

   A box contains cards labelled 1, 2 and 3. One card is drawn at random and not replaced. A second card is then drawn. Using a table, list all possible outcomes and find the probability that the sum of the two cards is greater than 4.

   Show worked solution

   Key idea: Without replacement, the first card drawn cannot be drawn again, so it does not appear as a second draw.

   1. Set up a table with the first draw as rows and the second draw as columns, leaving out pairs where both cards are the same (since the first card is not replaced):

   First \ Second	1	2	3
   1	,	(1, 2)	(1, 3)
   2	(2, 1)	,	(2, 3)
   3	(3, 1)	(3, 2)	,

   2. Count the total number of possible outcomes: $6$.
   3. Find the sums: $(1,2)\to3$, $(1,3)\to4$, $(2,1)\to3$, $(2,3)\to5$, $(3,1)\to4$, $(3,2)\to5$.
   4. Outcomes with sum greater than 4: $(2,3)$ and $(3,2)$, giving 2 outcomes.
   5. Calculate the probability:
      $P(\text{sum} > 4) = \frac{2}{6} = \frac{1}{3}$ Answer: $\dfrac{1}{3}$

3. Example 3 · Real-life probability: choosing without replacement

   At a school lunch stall, a box holds 3 ham rolls and 2 cheese rolls. Layla picks one roll at random and eats it. Her friend Marcus then picks one roll at random from the remaining rolls. Find the probability that Layla gets a ham roll and Marcus also gets a ham roll.

   Show worked solution

   Key idea: Because Layla eats her roll, the total number of rolls decreases and the remaining mix changes for Marcus, so this is a without-replacement situation.

   1. Find the probability Layla picks a ham roll:
      $P(\text{Layla gets ham}) = \frac{3}{5}$
   2. If Layla has taken a ham roll, 2 ham rolls and 2 cheese rolls remain (4 rolls total). Find the probability Marcus picks a ham roll:
      $P(\text{Marcus gets ham} \mid \text{Layla got ham}) = \frac{2}{4} = \frac{1}{2}$
   3. Multiply the probabilities along the branch (both events must occur):
      $P(\text{both get ham}) = \frac{3}{5} \times \frac{1}{2} = \frac{3}{10}$ Answer: $\dfrac{3}{10}$

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