Home / CIE iGCSE Maths C8.3 Probability of combined events Exam Style Practice Questions- Paper 3

CIE iGCSE Maths C8.3 Probability of combined events Exam Style Practice Questions- Paper 3

CIE iGCSE Maths C8.3 Probability of combined events Exam Style Practice Questions- Paper 3

Question

(a) Li spins a fair 6-sided spinner numbered $1$ to $6$.

On the probability scale, draw an arrow $(\downarrow )$ to show the probability that the spinner lands on the number 2.

(ii) Find the probability that the spinner lands on a prime number.

(iii) Find the probability that the spinner lands on the number 7.

(b) A bag contains 3 red balls and 12 green balls. Li picks a ball at random.

Find the probability that it is a green ball. Give your answer as a fraction in its simplest form.

(c) Li spins two fair 4-sided spinners, each numbered 1 to 4. The two numbers are multiplied to give the score.

Find the probability that the score is:

(i) an even number

(ii) an integer

(iii) at least 10.

(d) A bag contains red discs and blue discs. The probability that a disc picked at random is red is $\frac{1}{5}$. Li picks a disc at random, notes its colour and then replaces it in the bag. She then picks another disc at random.

(i) Complete the tree diagram.

(ii) Work out the probability that both of the discs she picks are blue.

▶️ Answer/Explanation
Solution

(a)(i) Ans: Arrow at $\frac{1}{6}$

The spinner has 6 sides, so the probability of landing on 2 is $\frac{1}{6}$.

(a)(ii) Ans: $\frac{1}{2}$

Prime numbers between 1 and 6: 2, 3, 5. There are 3 primes out of 6, so the probability is $\frac{3}{6} = \frac{1}{2}$.

(a)(iii) Ans: $0$

The spinner is numbered 1 to 6, so landing on 7 is impossible.

(b) Ans: $\frac{4}{5}$

Total balls: $3 + 12 = 15$. Green balls: 12. Probability: $\frac{12}{15} = \frac{4}{5}$.

(c)(i) Ans: $\frac{3}{4}$

Even scores: 2, 4, 6, 8, 12, 16. Favored outcomes: 12. Total outcomes: 16. Probability: $\frac{12}{16} = \frac{3}{4}$.

(c)(ii) Ans: $1$

All possible scores are integers, so the probability is $1$.

(c)(iii) Ans: $\frac{3}{16}$

Scores ≥ 10: 12, 16. Favored outcomes: 3. Probability: $\frac{3}{16}$.

(d)(i)

Probability of red: $\frac{1}{5}$. Probability of blue: $1 – \frac{1}{5} = \frac{4}{5}$.

(d)(ii) Ans: $\frac{16}{25}$

Probability both discs are blue: $\frac{4}{5} \times \frac{4}{5} = \frac{16}{25}$.

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