IGCSE · Practice questions · Introduction to probability

Count, then divide.

Six original Cambridge-style questions on single-event probability: the scale, equally likely outcomes, P(not A), and probabilities from tables.

Original questions All questions on this page are original work, written in the Cambridge IGCSE style. They are not from past papers. They test the same concepts and skills the syllabus rewards.

The method

Wanted outcomes over all outcomes.

Scale: 0 impossible, 1 certain; nothing outside 0 to 1.
Equally likely: P = number wanted \(\div\) total number.
Not happening: \(\text{P(not A)} = 1 - \text{P(A)}\) (Extended: \(\text{P}(A')\)).
All outcomes: the probabilities add up to 1, so a missing one is 1 minus the rest.
Format: fraction, decimal or percentage, never "3 out of 10".

Recall

[3 marks]

A fair six-sided dice is rolled. Find the probability of rolling (a) a 5 (b) an even number (c) a 7.

(a) \(\tfrac{1}{6}\) ✓
(b) 2, 4 or 6: \(\tfrac{3}{6} = \tfrac{1}{2}\) ✓
(c) Impossible: 0 ✓

Recall

[1 marks]

The probability that a bus is late is 0.12. Find the probability that the bus is not late.

\(1 - 0.12 = 0.88\) ✓

Calculation

[2 marks]

A bag contains 5 red, 8 white and 7 black beads. One bead is taken at random. Find the probability that it is (a) white (b) not black.

Total \(= 20\) beads. ✓
(a) \(\tfrac{8}{20} = \tfrac{2}{5}\) ✓
(b) \(\tfrac{5 + 8}{20} = \tfrac{13}{20}\) ✓

Calculation

[3 marks]

A biased four-sided spinner is numbered 1 to 4. P(1) = 0.3, P(2) = 0.25, and 3 and 4 are equally likely.

(a) Find P(3). (b) Find the probability that the spinner lands on an odd number.

(a) \(0.3 + 0.25 + 2x = 1\), so \(2x = 0.45\) and P(3) \(= 0.225\) ✓
(b) Odd means 1 or 3: \(0.3 + 0.225 = 0.525\) ✓

Analysis

[3 marks]

The table shows the pets owned by 60 people (each owns exactly one).

	Cat	Dog
Adult	12	18
Child	14	16

One person is chosen at random. Find the probability that the person is (a) a child (b) an adult who owns a dog (c) a cat owner.

(a) \(14 + 16 = 30\): \(\tfrac{30}{60} = \tfrac{1}{2}\) ✓
(b) \(\tfrac{18}{60} = \tfrac{3}{10}\) ✓
(c) \(12 + 14 = 26\): \(\tfrac{26}{60} = \tfrac{13}{30}\) ✓

Analysis

[4 marks]

A box contains 40 chocolates. Each is milk, dark or white. A chocolate is chosen at random: P(milk) = 0.45 and P(dark) = 0.3.

(a) How many white chocolates are in the box?

(b) 4 more dark chocolates are added to the box. Find the new probability of choosing a dark chocolate.

(a) P(white) \(= 1 - 0.45 - 0.3 = 0.25\), so \(0.25 \times 40 = 10\) white. ✓
(b) Dark at first: \(0.3 \times 40 = 12\). Now 16 dark out of 44. ✓
P(dark) \(= \tfrac{16}{44} = \tfrac{4}{11}\) ✓

Mark this once you have attempted all six and checked your working. It records a Practiced badge on the topic and adds a one-time bonus. Revealing the solutions alone does not count.