IGCSE · Practice questions · Tree diagrams

Branch by branch.

Six original Cambridge-style questions on tree diagrams: with replacement, independent events, at least one, and Extended problems without replacement.

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

Multiply along, add between paths.

Draw: probabilities beside the branches, outcomes at the ends.
Check: each set of branches adds to 1.
And: multiply along a path.
Or: add the paths that give the event (both orders!).
At least one: 1 minus P(none).
Without replacement (Extended): reduce the totals on the second branches.

Recall

[2 marks]

A fair coin is thrown twice. Find the probability of (a) two heads (b) one head and one tail.

(a) \(\tfrac{1}{2} \times \tfrac{1}{2} = \tfrac{1}{4}\) ✓
(b) HT or TH: \(\tfrac{1}{4} + \tfrac{1}{4} = \tfrac{1}{2}\) ✓

Calculation

[3 marks]

The probability that Ali wins a game of chess is 0.65. He plays two games; the results are independent.

Find the probability that he wins (a) both games (b) neither game (c) at least one game.

(a) \(0.65 \times 0.65 = 0.4225\) ✓
(b) \(0.35 \times 0.35 = 0.1225\) ✓
(c) \(1 - 0.1225 = 0.8775\) ✓

Calculation

[2 marks]

A bag holds 2 white and 3 black balls. A ball is taken, replaced, and a second ball is taken.

Find the probability that the two balls are different colours.

WB: \(\tfrac{2}{5} \times \tfrac{3}{5} = \tfrac{6}{25}\); BW: \(\tfrac{3}{5} \times \tfrac{2}{5} = \tfrac{6}{25}\). ✓
Total \(\tfrac{12}{25}\) ✓

Calculation

[3 marks]

(Extended) A box contains 6 milk and 4 dark chocolates. Two are taken at random without replacement.

Find the probability that (a) both are dark (b) at least one is milk.

(a) \(\tfrac{4}{10} \times \tfrac{3}{9} = \tfrac{12}{90} = \tfrac{2}{15}\) ✓
(b) The only other outcome is both dark: \(1 - \tfrac{2}{15} = \tfrac{13}{15}\) ✓

Analysis

[3 marks]

On her way to work, Zara passes two sets of traffic lights. The probability that the first is red is 0.4 and that the second is red is 0.7, independently.

(a) Find the probability that she is stopped by at least one red light. (b) In 20 journeys, how many times would she expect to meet no red light?

P(no red) \(= 0.6 \times 0.3 = 0.18\). ✓
(a) \(1 - 0.18 = 0.82\) ✓
(b) \(0.18 \times 20 = 3.6\) journeys ✓

Analysis

[5 marks]

(Extended) A bag contains \(x\) red counters and 4 blue counters. Two counters are taken at random without replacement. The probability that both are blue is \(\tfrac{2}{15}\).

Show that \(x^2 + 7x - 78 = 0\), and find the number of red counters.

\(\dfrac{4}{x + 4} \times \dfrac{3}{x + 3} = \dfrac{2}{15}\) ✓
\(12 \times 15 = 2(x + 4)(x + 3)\), so \(90 = x^2 + 7x + 12\). ✓
\(x^2 + 7x - 78 = 0\) ✓
\((x + 13)(x - 6) = 0\), so \(x = 6\) (a number of counters cannot be negative). ✓
There are 6 red counters. ✓

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.