Probability
15 free practice questions with explanations
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PassNova has 15 free KS3 Maths practice questions on Probability, each with a clear explanation. Practise them in the browser with instant feedback — 100% free, no sign-up, on any device. Updated for 2026.
Probability: example questions & answers
15 worked examples with answers and explanations below. Practise them in the browser with instant feedback on every answer.
The probability that it rains tomorrow is given as 0.9. Which word best describes this likelihood?
- Avery unlikely
- Bimpossible
- Cvery likely✓
- Devens
Answer: A probability of 0.9 is close to 1, the certain end of the probability scale, so it describes an event that is very likely to happen. Very unlikely would match a probability close to 0, and impossible matches a probability of exactly 0, both far from 0.9. Evens describes a probability of 0.5, where an event is equally likely to happen or not.
A spinner can land on red, blue or green. The probability of landing on red is 0.3 and the probability of landing on blue is 0.45. What is the probability of landing on green?
- A0.75
- B0.25✓
- C0.055
- D0.15
Answer: The probabilities of all possible outcomes must add up to 1, and 0.3 plus 0.45 gives 0.75, so green must be 1 minus 0.75, which is 0.25. Getting 0.055 comes from misplacing the decimal point during the subtraction, and getting 0.15 comes from subtracting the given probabilities incorrectly. Getting 0.75 mistakes the combined probability of red and blue for the probability of green itself.
A fair eight-sided spinner is numbered 1 to 8. Since every number is equally likely, what is the probability of landing on the number 5?
- A1/10
- B1/8✓
- C1/5
- D5/8
Answer: With eight equally likely outcomes, each individual number has a probability of 1 out of 8. A probability of 1 in 10 would only apply to a ten-sided spinner rather than an eight-sided one, and 1 in 5 would apply if there were only five possible outcomes. A probability of 5 in 8 confuses the number on the spinner with a count of favourable outcomes.
A biased six-sided die was rolled 50 times and landed on a 6 a total of 20 times. Based on this experiment, what is the relative frequency of rolling a 6?
- A0.6
- B0.35
- C0.45
- D0.4✓
Answer: Relative frequency is calculated by dividing the number of times the outcome happened by the total number of trials, so 20 divided by 50 gives 0.4. Getting 0.35 or 0.45 comes from a small slip when dividing 20 by 50. Getting 0.6 confuses the number of sixes with the number of rolls that were not sixes.
A fair six-sided die is rolled once. What is the probability of rolling a prime number?
- A1/2✓
- B3/12
- C9/12
- D1/6
Answer: The prime numbers on a die are 2, 3 and 5, which is 3 out of the 6 equally likely outcomes, and 3 out of 6 simplifies to 1/2. Getting 3/12 or 9/12 comes from using twelve possible outcomes by mistake instead of the six sides of the die. Getting 1/6 counts just one prime number instead of three.
A fair coin is flipped twice. What is the probability that both flips land on heads?
- A1/4✓
- B1/3
- C1/2
- D3/4
Answer: Listing all four equally likely outcomes for two flips, HH, HT, TH and TT, shows that only one of them has both heads, giving a probability of 1/4. A probability of 1/2 is the chance of getting heads on a single flip, not on both flips together. A probability of 1/3 does not match any of the four outcomes, and 3/4 is the probability of getting at least one head rather than two.
Two fair coins are tossed together. By listing the full sample space, how many different equally likely outcomes are there?
- A3
- B2
- C8
- D4✓
Answer: Listing every combination of the two coins gives HH, HT, TH and TT, which is 4 equally likely outcomes in total. Getting 2 only counts the outcomes where both coins match. Getting 3 wrongly treats HT and TH as the same outcome, and 8 would only be correct if a third coin were included.
Two fair six-sided dice are rolled and their scores are added together. Using the full sample space of 36 outcomes, what is the probability that the total is 7?
- A1/6✓
- B9/36
- C3/36
- D4/36
Answer: There are 36 equally likely outcomes when two dice are rolled, and exactly 6 of them, such as 1 and 6 or 3 and 4, add up to 7, giving a probability of 6/36, which simplifies to 1/6. Getting 4/36 or 3/36 comes from missing one or more of the six combinations that total 7. Getting 9/36 overcounts the pairs, including some that do not actually add up to 7.
A fair coin is tossed and a fair four-sided spinner numbered 1 to 4 is spun at the same time. How many outcomes are there in the combined sample space?
- A12
- B8✓
- C4
- D16
Answer: The combined sample space is found by multiplying the number of outcomes for each event, so 2 coin outcomes multiplied by 4 spinner outcomes gives 8 combined outcomes. Getting 12 or 16 comes from multiplying by an extra factor by mistake, such as treating the spinner as having six or eight sections instead of four. Getting 4 only counts the spinner's outcomes and ignores the coin completely.
In a class of 30 pupils, a Venn diagram shows that 18 play football, 12 play basketball, and 5 pupils play both sports. How many pupils play football only?
- A13✓
- B5
- C7
- D18
Answer: To find how many pupils play football only, the 5 who play both sports must be removed from the total who play football, so 18 minus 5 gives 13. Choosing 18 forgets to subtract the pupils who also play basketball. Choosing 5 confuses the pupils who play only football with those who play both, and 7 comes from subtracting the wrong pair of numbers.
A two-way grid lists the possible outcomes when a spinner showing red, yellow and blue is spun alongside a coin showing heads or tails. How many cells does a fully completed grid of outcomes contain?
- A18 cells
- B12 cells
- C6 cells✓
- D15 cells
Answer: A two-way grid has one row or column for every spinner colour and one for every coin outcome, so 3 colours combined with 2 coin outcomes gives 6 cells in total. Getting 12 cells or 18 cells comes from mistaking the number of spinner colours or coin outcomes, such as treating the spinner as having four or six sections instead of three. Getting 15 cells comes from a similar miscount of the categories being combined.
A youth club wants to choose a fair way to pick who goes first in a board game. Which method gives every player an equally likely chance of going first?
- Aletting whoever asks first go first every time
- Bletting the club leader choose their favourite player
- Crolling a fair die and letting the highest roller go first✓
- Dletting the tallest player go first every time
Answer: A method is fair when every outcome is equally likely and is not influenced by any player's characteristics, and rolling a fair die gives every player the same chance regardless of height, speed or popularity. Choosing by height always favours the same tall player, so it is not random or fair. Choosing by who asks first or by a leader's preference both depend on factors other than chance, so some players have no real chance at all.
A pupil flips a coin 200 times and records 118 heads. How does this relative frequency compare with the theoretical probability of getting heads on a fair coin?
- Ait is exactly equal to the theoretical probability of 0.5
- Bit is far higher, at around 0.9
- Cit is a little lower than the theoretical probability of 0.5
- Dit is a little higher than the theoretical probability of 0.5✓
Answer: The relative frequency is 118 divided by 200, which is 0.59, and this is a little above the theoretical probability of 0.5 for a fair coin, which is normal for a real experiment. It is not exactly equal, since 118 out of 200 is not exactly half. It is not lower than 0.5, since more than half the flips landed on heads, and 0.59 is nowhere near as high as 0.9.
A fair spinner is divided into 10 equal sections numbered 1 to 10. What is the theoretical probability of spinning a number greater than 7?
- A4/10
- B3/10✓
- C7/10
- D2/10
Answer: The numbers greater than 7 on the spinner are 8, 9 and 10, which is 3 out of the 10 equally likely sections, giving a probability of 3/10. A probability of 7/10 mistakes the numbers up to 7 for the numbers above it. A probability of 2/10 only counts 2 of the 3 correct numbers, and 4/10 counts one number too many.
A standard pack of 52 playing cards is well shuffled and one card is drawn. Since drawing a heart and drawing a spade are mutually exclusive events, what is the probability of drawing a heart or a spade?
- A1/4
- B1/2✓
- C3/4
- D1/13
Answer: For mutually exclusive events, the probability of one or the other happening is found by adding their separate probabilities, and a heart and a spade each have a probability of 1/4, so together they give 1/2. A probability of 1/4 is the chance of drawing just one of the two suits, not either of them. A probability of 1/13 matches drawing one specific card value, and 3/4 would only be correct if three suits were being combined instead of two.