Probability, counting and combinations

Probability and statistics

Questions marked Higher are aimed at the Higher tier.

These are original practice questions, not a past paper. This site is independent and not affiliated with AQA or Pearson.

A fair dice is rolled. Find the probability of an even number, and the probability of a number greater than $4$.

2 marks
A spinner has three equal sections, A, B and C. It is spun twice. Find the probability of A both times.

2 marks
Two coins are flipped. List the outcomes and find the probability of one head and one tail.

3 marks
A menu has $3$ starters and $4$ main courses. How many different two-course meals are there?

3 marks
A code uses two characters, each A, B or C, then one digit $1$, $2$ or $3$. Repeats are allowed. How many codes are there?

4 marks
Cards numbered $1$ to $5$ are used. One card is picked. Find the probability that it is prime, the probability that it is not prime, and the probability that both cards are prime if the first card is replaced before the second pick.

4 marks
In how many ways can $4$ different books be arranged on a shelf? In how many ways can $2$ toppings be chosen from $6$, if the order does not matter? A sandwich uses $1$ of $3$ breads and $2$ of $5$ fillings. How many sandwiches are possible? Higher

5 marks
Two fair dice are rolled. How many outcomes are there? Find the probability that the total is $7$, the probability that the total is $9$ or more, the probability of a double, and the probability that both dice show $4$ given that the total is $8$. Higher

6 marks

Mark scheme

Answer. $\dfrac{1}{2}$ and $\dfrac{1}{3}$
- Three of the six outcomes are even.
- Two outcomes are greater than $4$, so $\dfrac{1}{3}$.
Answer. $\dfrac{1}{9}$
- Each spin has probability $\dfrac{1}{3}$.
- $\dfrac{1}{3} \times \dfrac{1}{3} = \dfrac{1}{9}$
Answer. HH, HT, TH, TT. Probability $\dfrac{1}{2}$.
- There are four equally likely outcomes.
- HT and TH are the mixed outcomes, so the probability is $\dfrac{1}{2}$.
Answer. $12$
- $3 \times 4$
- $12$
Answer. $27$
- There are $3$ choices for each of the three positions.
- $3 \times 3 \times 3 = 27$
Answer. $\dfrac{3}{5}$, $\dfrac{2}{5}$ and $\dfrac{9}{25}$
- The primes are $2, 3, 5$.
- The non-primes are $1$ and $4$.
- With replacement, $\dfrac{3}{5} \times \dfrac{3}{5} = \dfrac{9}{25}$.
Answer. $24$ arrangements, $15$ pairs of toppings, and $30$ sandwiches
- $4! = 24$
- $\dfrac{6 \times 5}{2} = 15$
- $3 \times 10 = 30$ sandwiches, since there are $10$ ways to choose two fillings.
Answer. $36$ outcomes. $\dfrac{1}{6}$, $\dfrac{5}{18}$, $\dfrac{1}{6}$ and $\dfrac{1}{5}$.
- There are $6 \times 6 = 36$ outcomes, and $6$ of them total $7$.
- Totals of $9$ or more: $4 + 3 + 2 + 1 = 10$ outcomes, so $\dfrac{10}{36} = \dfrac{5}{18}$.
- There are $6$ doubles.
- Five outcomes total $8$, and only $(4, 4)$ is the required double, so $\dfrac{1}{5}$.