CIE iGCSE Maths C8.3 Probability of combined events - Exam Style Practice Questions- Paper 1-New Syllabus
▶️ Answer/Explanation
(a)
Probabilities on each set of branches must add up to \(1\).
For the first set: “Does not stop” \(= 1 – 0.4 = 0.6\).
For the second set (on both branches): “Stops” \(= 0.7\), and “Does not stop” \(= 1 – 0.7 = 0.3\).
✅ Answer:
First branch (Does not stop): \(0.6\)
Second branches (Top and Bottom pairs): \(0.7\) and \(0.3\) respectively
(b)
To find the probability of multiple independent events happening sequentially, multiply their probabilities along the path.
The path for “Does not stop” and then “Does not stop” gives: \(0.6 \times 0.3 = 0.18\).
✅ Answer: \(0.18\)
▶️ Answer/Explanation
(a) The total ratio parts equal $1 + 4 = 5$. This means the probability of selecting a red counter is $\frac{1}{5}$, and a green counter is $\frac{4}{5}$. Since the counter is put back each time, these probabilities remain exactly identical for the second pick. Write $\frac{1}{5}$ on all ‘Red’ branches and $\frac{4}{5}$ on all ‘Green’ branches.
(b) To find the probability of picking two green counters in a row, we follow the Green-Green path and multiply their individual branch probabilities together: $\frac{4}{5} \times \frac{4}{5} = \frac{16}{25}$ (or $0.64$).
✅ Answer:
(a) $\frac{1}{5}$ on Red branches, $\frac{4}{5}$ on Green branches
(b) $\frac{16}{25}$
(b) To find the probability of picking two green counters in a row, we follow the Green-Green path and multiply their individual branch probabilities together: $\frac{4}{5} \times \frac{4}{5} = \frac{16}{25}$ (or $0.64$).
✅ Answer:
(a) $\frac{1}{5}$ on Red branches, $\frac{4}{5}$ on Green branches
(b) $\frac{16}{25}$