Year 10 and 10A Probability and counting techniques: free practice test with answers

Sample probability and counting techniques questions with answers

Try each one before you open the answer.

1. Review of probability

   A bag contains 5 red, 3 blue and 4 green marbles. One marble is chosen at random. Find:

   (a) $\Pr(\text{blue})$

   (b) $\Pr(\text{not green})$

   Show the answer

   Key idea: $\Pr(\text{event}) = \dfrac{\text{number of favourable outcomes}}{\text{total number of outcomes}}$, and $\Pr(\text{not } A) = 1 - \Pr(A)$.

   (a)

   1. Total marbles: $5 + 3 + 4 = 12$.
   2. $\Pr(\text{blue}) = \frac{3}{12} = \frac{1}{4}$

   (b)

   1. $\Pr(\text{green}) = \frac{4}{12} = \frac{1}{3}$
   2. $\Pr(\text{not green}) = 1 - \frac{1}{3} = \frac{2}{3}$

   Answer: (a) $\frac{1}{4}$ (b) $\frac{2}{3}$

2. Set notation in Venn diagrams and two-way tables

   Let $\xi = \{1, 2, 3, \ldots, 15\}$, $A = \{\text{multiples of } 3\}$ and $B = \{\text{odd numbers}\}$.

   (a) List the elements of $A \cap B$.

   (b) Find $n(A \cup B)$.

   (c) List the elements of $A' \cap B$.

   Show the answer

   Key idea: $\cap$ means "and" (in both sets), $\cup$ means "or" (in at least one set) and $A'$ means "not in $A$".

   1. $A = \{3, 6, 9, 12, 15\}$ and $B = \{1, 3, 5, 7, 9, 11, 13, 15\}$.

   (a)

   1. Elements in both: $A \cap B = \{3, 9, 15\}$.

   (b)

   1. $n(A \cup B) = n(A) + n(B) - n(A \cap B) = 5 + 8 - 3 = 10$.
   2. Check by listing: $\{1, 3, 5, 6, 7, 9, 11, 12, 13, 15\}$ has 10 elements ✓

   (c)

   1. Odd numbers that are not multiples of 3: $A' \cap B = \{1, 5, 7, 11, 13\}$.

   Answer: (a) $\{3, 9, 15\}$ (b) 10 (c) $\{1, 5, 7, 11, 13\}$

3. The addition rule

   For two events $A$ and $B$, $\Pr(A) = 0.45$, $\Pr(B) = 0.3$ and $\Pr(A \cap B) = 0.12$. Find $\Pr(A \cup B)$.

   Show the answer

   Key idea: The addition rule is $\Pr(A \cup B) = \Pr(A) + \Pr(B) - \Pr(A \cap B)$.

   1. $\Pr(A \cup B) = 0.45 + 0.3 - 0.12$
   2. $= 0.63$

   Answer: 0.63

4. Two-step experiments using arrays

   Two fair six-sided dice are rolled and the numbers are added. Find the probability that the sum is:

   (a) 10

   (b) less than 5.

   Show the answer

   Key idea: Use a $6 \times 6$ array of the 36 equally likely outcomes.

   1. Sum 10: $(4, 6)$, $(5, 5)$, $(6, 4)$, so $\frac{3}{36} = \frac{1}{12}$
   2. Sum 2, 3 or 4: $1 + 2 + 3 = 6$ outcomes, so $\frac{6}{36} = \frac{1}{6}$

   Answer: (a) $\frac{1}{12}$ (b) $\frac{1}{6}$

5. Using tree diagrams

   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 the answer

   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: $0.71$

6. Independent events

   Events $A$ and $B$ are independent, with $\Pr(A) = 0.6$ and $\Pr(B) = 0.25$.

   (a) Find $\Pr(A \cap B)$.

   (b) Find $\Pr(A \cup B)$.

   Show the answer

   Key idea: For independent events, $\Pr(A \cap B) = \Pr(A) \times \Pr(B)$.

   1. $\Pr(A \cap B) = 0.6 \times 0.25 = 0.15$
   2. $\Pr(A \cup B) = 0.6 + 0.25 - 0.15 = 0.7$

   Answer: (a) $0.15$ (b) $0.7$

7. Arrangements

   A school captain team consists of 10 candidates. In how many ways can a President, Vice-President and Treasurer be chosen from the 10 candidates, if no person can hold more than one role?

   Show the answer

   Key idea: Because the three roles are different, order matters. Use the multiplication principle: count the choices at each stage and multiply.

   1. Choose the President: 10 options.
   2. Choose the Vice-President from the remaining 9 candidates: 9 options.
   3. Choose the Treasurer from the remaining 8 candidates: 8 options.
   4. Multiply the choices together:\

   $$10 \times 9 \times 8 = 720$$

   This can also be written using factorial notation as:\

   $$\frac{10!}{(10-3)!} = \frac{10!}{7!} = 720$$

   Answer: 720 ways

8. Selections

   In how many ways can a group of 3 people be selected from 8 people?

   Show the answer

   Key idea: Order does not matter in a selection, so use $^nC_r = \dfrac{n!}{r!(n - r)!}$.

   1. $^8C_3 = \frac{8!}{3! \times 5!} = \frac{8 \times 7 \times 6}{3 \times 2 \times 1}$
   2. $= \frac{336}{6} = 56$

   Answer: 56 ways

Free worksheets for this chapter

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

• Year 10 multi-step probability and tree diagrams worksheet
• Year 10 two-way tables worksheet
• Year 10 probability worksheet
• Year 10A counting principles and arrangements worksheet