Year 9 Probability and statistics: free practice test with answers

Sample probability and statistics questions with answers

Try each one before you open the answer.

1. Review of probability

   A bag contains $4$ red, $7$ blue and $9$ green marbles. One marble is chosen at random. Find:

   (a) $P(\text{blue})$ (b) $P(\text{not green})$ (c) $P(\text{red or green})$

   Show the answer

   Key idea: For equally likely outcomes, $P(\text{event}) = \frac{\text{number of favourable outcomes}}{\text{total number of outcomes}}$.

   1. Total: $4 + 7 + 9 = 20$ marbles
   2. (a) $P(\text{blue}) = \frac{7}{20}$
   3. (b) Not green: $4 + 7 = 11$, so $\frac{11}{20}$ (or $1 - \frac{9}{20}$)
   4. (c) Red or green: $4 + 9 = 13$, so $\frac{13}{20}$

   Answer: (a) $\frac{7}{20}$ (b) $\frac{11}{20}$ (c) $\frac{13}{20}$

2. Venn diagrams and two-way tables

   A group of 40 students were surveyed about whether they play sport and whether they play a musical instrument. The results are shown in the two-way table below.

   	Plays sport	Does not play sport	Total
   Plays instrument	9	6	15
   Does not play instrument	17	8	25
   Total	26	14	40

   Use the table to estimate: (a) $P(\text{plays sport and plays instrument})$ (b) $P(\text{plays sport or plays instrument})$, using inclusive or (c) $P(\text{does not play sport})$

   Show the answer

   Key idea: Relative frequency $= \dfrac{\text{frequency of event}}{\text{total number of observations}}$ estimates probability. For inclusive or, use $P(A \text{ or } B) = P(A) + P(B) - P(A \text{ and } B)$.

   1. There are 40 students in total.
   2. (a) 9 students play both sport and an instrument.\

   $$P(\text{sport and instrument}) = \frac{9}{40}$$

   3. (b) 26 play sport, 15 play instrument, 9 play both.\

   $$P(\text{sport or instrument}) = \frac{26}{40} + \frac{15}{40} - \frac{9}{40} = \frac{32}{40} = \frac{4}{5}$$

   4. (c) 14 students do not play sport.\

   $$P(\text{not sport}) = \frac{14}{40} = \frac{7}{20}$$

   Answer: (a) $\dfrac{9}{40}$, (b) $\dfrac{4}{5}$, (c) $\dfrac{7}{20}$

3. Using arrays for two-step experiments

   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 the answer

   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: $\dfrac{1}{3}$

4. Using tree diagrams

   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 the answer

   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: $\dfrac{1}{2}$

5. Data and sampling

   A news website reports: "We surveyed 800 Australian adults about how often they cook dinner at home."

   (a) What is the sample size?

   (b) What is the population the website wants to describe?

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   Answer: (a) $800$ (b) Australian adults

6. Mean, median and mode

   A gym tracks the number of minutes two groups of members spend exercising in a week. The data for $8$ members in each group is listed below.

   Group A: 20, 35, 40, 45, 50, 60, 65, 75

   Group B: 15, 25, 30, 45, 55, 70, 80, 90

   Find the median for each group, then find the difference between the two medians.

   Show the answer

   Key idea: With $8$ values, the median is the mean of the 4th and 5th values.

   1. Group A values are already in order: 20, 35, 40, 45, 50, 60, 65, 75.\

   $$\text{Median}_A = \frac{45 + 50}{2} = \frac{95}{2} = 47.5$$

   2. Group B values are already in order: 15, 25, 30, 45, 55, 70, 80, 90.\

   $$\text{Median}_B = \frac{45 + 55}{2} = \frac{100}{2} = 50$$

   3. Difference between medians: $50 - 47.5 = 2.5$.

   Answer: The difference between the two medians is $2.5$ minutes.

7. Grouping data into classes

   The times, $t$ minutes, taken by $30$ students to run $1$ km are shown in the grouped frequency table.

   Time (minutes)	Frequency
   $4 \le t < 5$	3
   $5 \le t < 6$	9
   $6 \le t < 7$	12
   $7 \le t < 8$	4
   $8 \le t < 9$	2

   How many students took less than $6$ minutes?

   Show the answer

   Key idea: Add the frequencies of the classes below 6 minutes.

   1. The classes $4 \le t < 5$ and $5 \le t < 6$ are below $6$ minutes
   2. $3 + 9 = 12$

   Answer: $12$

8. Measures of spread: range and interquartile range

   Two basketball players recorded their points scored in $8$ games each.

   Player A: 5, 9, 12, 15, 18, 22, 25, 28

   Player B: 3, 7, 10, 14, 19, 21, 26, 30

   Find the interquartile range (IQR) for each player, then find the difference between their IQRs.

   Show the answer

   Key idea: With $8$ values, split into a lower half (first 4) and upper half (last 4). $Q_1$ is the median of the lower half and $Q_3$ is the median of the upper half.

   1. Player A lower half: 5, 9, 12, 15 and upper half: 18, 22, 25, 28.\

   $$\begin{gathered}Q_1 = \frac{9+12}{2} = 10.5, Q_3 = \frac{22+25}{2} = 23.5 \\ \text{IQR}_A = 23.5 - 10.5 = 13\end{gathered}$$

   2. Player B lower half: 3, 7, 10, 14 and upper half: 19, 21, 26, 30.\

   $$\begin{gathered}Q_1 = \frac{7+10}{2} = 8.5, Q_3 = \frac{21+26}{2} = 23.5 \\ \text{IQR}_B = 23.5 - 8.5 = 15\end{gathered}$$

   3. Difference between IQRs: $15 - 13 = 2$.

   Answer: The difference between the IQRs is $2$.

Free worksheets for this chapter

Worked examples and a free PDF with answers for each topic.

• Year 9 probability worksheet
• Year 9 venn diagrams and two-way tables worksheet
• Year 9 statistics worksheet
• Year 9 comparing data sets worksheet