Probability, sample spaces, tree diagrams & conditional probability – IGCSE Maths exam questions with mark scheme

1Grade 1–3FoundationCalculator4 marks

A spinner can land on red, blue, green or yellow. The table shows some of the probabilities.

Colour	red	blue	green	yellow
Probability	0.25	0.3	$x$	$2x$

(a) Work out the probability that the spinner lands on green. [2]

(b) Work out the probability that the spinner lands on red or blue. [1]

(a) $0.25+0.3+x+2x=1$ or $3x=1-0.55$ or $3x=0.45$ (oe)M1
(a) $0.15$ (oe, e.g. $\tfrac{3}{20}$)A1
(b) $0.55$ (oe)B1
(c) $0$ (oe, e.g. $\tfrac{0}{4}$; do not accept 'impossible' alone, 'none', 0 : 4 or '0 out of 4')B1
Answer: (a) 0.15 (b) 0.55 (c) 0
Examiner tip: Probabilities of all the outcomes add to 1 – remember the yellow section is $2x$, so you divide 0.45 by 3, not 2.

2Grade 1–3Calculator5 marks

Ella has two fair spinners. Spinner A is labelled 1, 2, 3. Spinner B is labelled 1, 2, 3, 4.

She spins both spinners and multiplies the two numbers to get her score.

(a) List all the possible outcomes, as pairs (A, B), in a systematic way, or show them in a sample space diagram of scores. [2]

(b) Find the probability that her score is an even number. [2]

(a) at least 8 of the 12 outcomes/scores correct with no repeatsB1
(a) all 12 correct: (1,1)…(3,4), scores 1,2,3,4,2,4,6,8,3,6,9,12B1
(b) 12 outcomes identified as denominator, or 8 even scores counted (ft their table)M1
(b) $\dfrac{8}{12}=\dfrac{2}{3}$ (oe)A1
(c) $\dfrac{3}{12}=\dfrac{1}{4}$ (oe; scores 8, 9, 12)B1
Answer: (a) 12 outcomes (b) $\frac{2}{3}$ (c) $\frac{1}{4}$
Examiner tip: Work through Spinner A one value at a time against every value on Spinner B so you neither miss nor repeat an outcome.

3Grade 4–5Calculator4 marks

Tomas thinks his four-sided dice is biased. He rolls it 250 times. The table shows his results.

Score	1	2	3	4
Frequency	45	70	45	90

(a) Work out an estimate for the probability that the dice lands on 4. [1]

(b) Give a reason why Tomas's results suggest the dice may be biased. [1]

Tomas is going to roll the dice another 600 times.

(a) $\dfrac{90}{250}$ (oe, e.g. 0.36)B1
(b) e.g. a fair dice would give about 62.5 of each score, but 4 came up 90 times (allow any valid comparison with the expected 62.5 or probability 0.25)C1
(c) $\dfrac{70+90}{250}\times600$ or $0.64\times600$M1
(c) $384$A1
Answer: (a) 0.36 (b) e.g. score 4 occurs far more than the 62.5 expected (c) 384
Examiner tip: Use the relative frequency from the experiment, not the theoretical 0.5, when asked for an estimate based on these results.

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4Grade 4–5Calculator4 marks

The Venn diagram shows probabilities for the members of a sports club.

$T$ = members who play tennis, $S$ = members who swim.

(a) Find the value of $x$. [1]

A member is chosen at random.

(b) Find the probability that the member swims. [1]

The club has 240 members.

(d) Work out the number of members who neither play tennis nor swim. [1]

(a) $0.3$ (oe)B1
(b) $0.45$ (oe; ft $0.15+x$)B1
(c) $0.65$ (oe; ft $0.35+x$)B1
(d) $48$B1
Answer: (a) 0.3 (b) 0.45 (c) 0.65 (d) 48
Examiner tip: For 'swims' include the overlap region 0.15 as well as 'swim only'.

5Grade 4–5HigherCalculator7 marks

Kofi travels to college by bus and then by train.

The probability that the bus is late is 0.2. The probability that the train is late is 0.15. These events are independent.

(a) Complete the probability tree diagram. [2]

(b) Work out the probability that exactly one of the bus and the train is late. [3]

Kofi travels to college on 200 days.

(a) 0.8 on the bus 'not late' branchB1
(a) 0.15 and 0.85 correctly placed on both pairs of train branchesB1
(b) $0.2\times0.85$ or $0.8\times0.15$ (ft their tree)M1
(b) $0.2\times0.85+0.8\times0.15$M1
(b) $0.29$ (oe)A1
(c) $0.2\times0.15\times200$ or $0.03\times200$M1
(c) $6$A1
Answer: (a) 0.8; 0.15, 0.85, 0.15, 0.85 (b) 0.29 (c) 6
Examiner tip: 'Exactly one' means two separate paths – multiply along each branch and then add the two results.

6Grade 6–7HigherCalculator5 marks

A bag contains 5 red counters, 3 blue counters and 2 green counters.

Priya takes a counter at random and does not replace it. She then takes a second counter at random.

(a) Work out the probability that both counters are blue. [2]

(b) Work out the probability that both counters are the same colour. [3]

(a) $\dfrac{3}{10}\times\dfrac{2}{9}$M1
(a) $\dfrac{6}{90}=\dfrac{1}{15}$ (oe)A1
(b) one of $\dfrac{5}{10}\times\dfrac{4}{9}$, $\dfrac{3}{10}\times\dfrac{2}{9}$, $\dfrac{2}{10}\times\dfrac{1}{9}$M1
(b) $\dfrac{20}{90}+\dfrac{6}{90}+\dfrac{2}{90}$M1
(b) $\dfrac{28}{90}=\dfrac{14}{45}$ (oe, 0.311 or better)A1
Answer: (a) $\frac{1}{15}$ (b) $\frac{14}{45}$
Examiner tip: Without replacement, both the numerator and the denominator drop by one for the second pick.

7Grade 6–7HigherCalculator4 marks

Asha rolls a fair six-sided dice three times.

(a) Work out the probability that she rolls at least one six. [3]

Asha says, 'The probability of at least one six is $\tfrac{1}{6}+\tfrac{1}{6}+\tfrac{1}{6}=\tfrac{1}{2}$.'

(b) Explain why Asha is wrong. [1]

(a) $\left(\dfrac{5}{6}\right)^3$ or $\dfrac{125}{216}$ (probability of no sixes)M1
(a) $1-\left(\dfrac{5}{6}\right)^3$M1
(a) $\dfrac{91}{216}$ (oe, 0.421 or better)A1
(b) e.g. the events are not mutually exclusive (she could get a six on more than one roll), so adding counts outcomes with two or three sixes more than onceC1
Answer: (a) $\frac{91}{216}$ (b) the three events are not mutually exclusive
Examiner tip: For 'at least one', work out 1 − P(none) – it avoids listing every case.

8Grade 8–9HigherCalculator5 marks

A box contains $n$ counters. 4 of the counters are red.

Jamal takes two counters at random from the box without replacement.

The probability that both counters are red is $\dfrac{2}{15}$.

(a) Show that $n^2-n-90=0$. [3]

(b) Find the number of counters in the box. You must show your working. [2]

(a) $\dfrac{4}{n}\times\dfrac{3}{n-1}$M1
(a) $\dfrac{12}{n(n-1)}=\dfrac{2}{15}$M1
(a) correct completion to $n^2-n-90=0$ with $2n(n-1)=180$ or $n^2-n=90$ seenA1
(b) $(n-10)(n+9)=0$ or correct use of the quadratic formulaM1
(b) $n=10$ only (reject $-9$)A1
Answer: (b) 10
Examiner tip: In a 'show that' every algebraic step must be written down – you cannot just quote the quadratic.

9Grade 8–9HigherCalculator5 marks

A bag contains 4 red sweets, 3 blue sweets and 2 yellow sweets.

Leo takes three sweets at random from the bag, one after the other, without replacement.

(a) Show that the probability that he takes a red, then a blue, then a yellow sweet, in that order, is $\dfrac{1}{21}$. [2]

(b) Work out the probability that the three sweets are all different colours. [2]

(a) $\dfrac{4}{9}\times\dfrac{3}{8}\times\dfrac{2}{7}$M1
(a) $\dfrac{24}{504}$ shown to equal $\dfrac{1}{21}$A1
(b) $6\times\dfrac{1}{21}$ or recognises 6 ordersM1
(b) $\dfrac{6}{21}=\dfrac{2}{7}$ (oe)A1
(c) $\dfrac{5}{7}$ (oe; ft $1-$ their (b))B1
Answer: (b) $\frac{2}{7}$ (c) $\frac{5}{7}$
Examiner tip: The three colours can come out in $3\times2\times1=6$ different orders, each with the same probability.

10Grade 4–5Calculator4 marks

A fair six-sided dice is rolled once.

Event $A$: the score is even. Event $B$: the score is greater than 3.

Leon says, 'P($A$ or $B$) $= \tfrac{1}{2}+\tfrac{1}{2}=1$.'

(a) (i) Explain why Leon is wrong. [1]

(ii) Find the correct value of P($A$ or $B$). [1]

Two fair six-sided dice are rolled and the scores are added.

Maya says, 'There are 11 possible totals from 2 to 12, so the probability of a total of 7 is $\tfrac{1}{11}$.'

(b) (i) Explain why Maya is wrong. [1]

(ii) Find the probability of a total of 7. [1]

(a)(i) $A$ and $B$ are not mutually exclusive / 4 and 6 are in both events so they are counted twiceC1
(a)(ii) $\dfrac{4}{6}=\dfrac{2}{3}$ (oe; outcomes 2, 4, 5, 6)B1
(b)(i) the totals are not equally likely (e.g. 7 can be made in 6 ways but 2 in only 1 way)C1
(b)(ii) $\dfrac{6}{36}=\dfrac{1}{6}$ (oe)B1
Answer: (a)(ii) $\frac{2}{3}$ (b)(ii) $\frac{1}{6}$
Examiner tip: You can only add probabilities when the events cannot happen together, and you can only count outcomes when they are equally likely.

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