Probability
8 free practice questions with explanations
PassNova has 8 free GCSE Maths Higher 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
8 worked examples with answers and explanations below. Practise them in the browser with instant feedback on every answer.
Two fair dice are rolled. What is the probability of a total of 12?
- A1/12
- B1/18
- C1/6
- D1/36✓
Answer: Only (6,6) gives 12, out of 36 equally likely outcomes, so the probability is 1/36.
P(A) = 0.3. What is P(not A)?
- A0.7✓
- B0.3
- C1.3
- D0
Answer: P(not A) = 1 − P(A) = 1 − 0.3 = 0.7.
A bag has 4 red and 6 green counters. Two are taken without replacement. What is P(both red)?
- A4/15
- B2/15✓
- C1/5
- D4/25
Answer: P(red then red) = 4/10 × 3/9 = 12/90 = 2/15.
The probability of rain is 0.2 each day. What is the probability of no rain on two consecutive days?
- A0.04
- B0.8
- C0.64✓
- D0.4
Answer: P(no rain) = 0.8 each day. Two days: 0.8 × 0.8 = 0.64.
A tin contains 5 toffees and 7 mints. Two sweets are taken at random without replacement. Work out the probability that the two sweets are of different kinds.
- A35/132
- B7/22
- C35/66✓
- D35/72
Answer: There are two orders that give different kinds. Toffee then mint is 5/12 × 7/11 = 35/132, and mint then toffee is 7/12 × 5/11 = 35/132. Adding the two orders gives 70/132, which simplifies to 35/66; using only one order halves the answer.
In a class of 30 students, 18 study French, 14 study German and 7 study both. One of the students who studies French is chosen at random. Work out the probability that this student also studies German.
- A7/18✓
- B7/30
- C3/5
- D1/2
Answer: The student is chosen from the 18 who study French, so that is the denominator. Of those 18, seven also study German, giving 7/18. Dividing by 30 would answer a different question about the whole class.
A bag contains n counters, of which 6 are green. Two counters are taken at random without replacement. The probability that both are green is 1/7. Work out the value of n.
- An = 16
- Bn = 15✓
- Cn = 21
- Dn = 30
Answer: (6/n) × (5/(n − 1)) = 1/7 gives 30/(n² − n) = 1/7, so n² − n − 210 = 0. Factorising gives (n − 15)(n + 14) = 0, and a number of counters must be positive, so n = 15. Checking: 6/15 × 5/14 = 1/7.
For two events E and F, P(E) = 0.45, P(F) = 0.3 and P(E ∩ F) = 0.12. Work out P((E ∪ F)′), the probability that neither event happens.
- A0.63
- B0.25
- C0.13
- D0.37✓
Answer: P(E ∪ F) = 0.45 + 0.3 − 0.12 = 0.63, because the overlap would otherwise be counted twice. The prime symbol means the complement, so P((E ∪ F)′) = 1 − 0.63 = 0.37.