Tree diagrams (independent & dependent/conditional) – GCSE Maths exam questions with mark scheme

1Grade 1–3FoundationNon-calculator3 marks

A fair coin is flipped twice.

(a) Complete the probability tree diagram. [1]

(b) Work out the probability that the coin lands on heads both times. [2]

(a) $\frac{1}{2}$ on every remaining branchB1
(b) $\frac{1}{2}\times\frac{1}{2}$M1
(b) $\frac{1}{4}$ oeA1
Answer: (a) $\frac{1}{2}$ on every branch (b) $\frac{1}{4}$
Examiner tip: Multiply along the branches — $\frac{1}{2}+\frac{1}{2}=1$ is a very common wrong answer.

2Grade 4–5Non-calculator4 marks

The probability that Kai's bus is late is 0.2. The probability that his train is late is 0.35. The two events are independent.

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

(b) Work out the probability that both the bus and the train are late. [2]

(a) 0.8 for bus not lateB1
(a) 0.35 and 0.65 on both pairs of train branchesB1
(b) $0.2 \times 0.35$M1
(b) 0.07 oeA1
Answer: (a) bus: 0.2, 0.8; train: 0.35, 0.65 on each pair (b) 0.07
Examiner tip: For independent events the second set of branches is the same on both pairs.

3Grade 4–5Non-calculator4 marks

A bag contains 7 red beads and 3 blue beads. Priya takes a bead at random, records its colour and puts it back. She then takes a second bead at random.

(a) Work out the probability that both beads are red. [2]

(b) Work out the probability that the beads are different colours. [2]

(a) $\frac{7}{10}\times\frac{7}{10}$M1
(a) $\frac{49}{100}$ oeA1
(b) $\frac{7}{10}\times\frac{3}{10}+\frac{3}{10}\times\frac{7}{10}$ or $1-\left(\frac{49}{100}+\frac{9}{100}\right)$M1
(b) $\frac{42}{100}$ oeA1
Answer: (a) $\frac{49}{100}$ (b) $\frac{42}{100}=\frac{21}{50}$
Examiner tip: 'Different colours' can happen two ways (red then blue, blue then red) — add both routes.

4Grade 6–7HigherNon-calculator5 marks

There are 5 green sweets and 4 yellow sweets in a bag. Sam takes a sweet at random and eats it. He then takes a second sweet at random and eats it.

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

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

(a) $\frac{5}{9}$, $\frac{4}{9}$ on the first branchesB1
(a) $\frac{4}{8}, \frac{4}{8}$ after green and $\frac{5}{8}, \frac{3}{8}$ after yellowB1
(b) $\frac{5}{9}\times\frac{4}{8}$ or $\frac{4}{9}\times\frac{3}{8}$M1
(b) $\frac{5}{9}\times\frac{4}{8}+\frac{4}{9}\times\frac{3}{8}$M1
(b) $\frac{32}{72}$ oeA1
Answer: (a) 1st: green $\frac{5}{9}$, yellow $\frac{4}{9}$; after green: $\frac{4}{8}$, $\frac{4}{8}$; after yellow: $\frac{5}{8}$, $\frac{3}{8}$ (b) $\frac{32}{72}=\frac{4}{9}$
Examiner tip: The sweet is eaten, so the second-pick denominators are 8, not 9.

5Grade 6–7HigherCalculator3 marks

A string of lights has three bulbs. The probability that any one bulb fails in the first year is 0.04. Bulbs fail independently of each other.

Work out the probability that at least one of the three bulbs fails in the first year. [3]

$0.96$ seen as P(bulb does not fail)M1
$1-0.96^3$ or $1-0.884736$ (or a complete method adding the probabilities of exactly 1, 2 and 3 failures, e.g. $3(0.04)(0.96)^2+3(0.04)^2(0.96)+0.04^3$)M1
0.115(264…) — accept 0.115 or better; 0.12 scores A1 only if 0.115… is seen first (isw)A1
Answer: 0.115264 (0.115 to 3 s.f.)
Examiner tip: 'At least one' is quickest as 1 − P(none fail); adding $0.04+0.04+0.04=0.12$ is a wrong method and scores nothing, even though it looks close.

6Grade 6–7HigherNon-calculator5 marks

Jo plays two games of chess.

The probability that she wins the first game is 0.6. If she wins a game, the probability that she wins the next game is 0.7. If she does not win a game, the probability that she wins the next game is 0.45.

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

(b) Work out the probability that Jo wins exactly one of the two games. [3]

(a) 0.4 for not win in game 1B1
(a) 0.7, 0.3 after win and 0.45, 0.55 after not winB1
(b) $0.6\times0.3$ $(=0.18)$ or $0.4\times0.45$ $(=0.18)$M1
(b) $0.6\times0.3+0.4\times0.45$M1
(b) 0.36 oeA1
Answer: (a) 0.4; 0.7, 0.3; 0.45, 0.55 (b) 0.36
Examiner tip: The second-game probabilities depend on the first result, so the two pairs of branches are different.

7Grade 8–9HigherNon-calculator6 marks

A box contains 6 red pens and 4 blue pens. Three pens are taken from the box at random, without replacement.

(a) Work out the probability that at least one of the pens is blue. [3]

(b) Work out the probability that exactly two of the pens are red. [3]

(a) $\frac{6}{10}\times\frac{5}{9}\times\frac{4}{8}$ $\left(=\frac{120}{720}\right)$M1
(a) $1-\frac{120}{720}$M1
(a) $\frac{600}{720}$ oe, e.g. $\frac{5}{6}$A1
(b) $\frac{6}{10}\times\frac{5}{9}\times\frac{4}{8}$ for one order of two red and one blue (e.g. RRB)M1
(b) $3\times\frac{6\times5\times4}{10\times9\times8}$ or sum of RRB, RBR, BRRM1
(b) $\frac{360}{720}$ oe, e.g. $\frac{1}{2}$A1
Answer: (a) $\frac{5}{6}$ (b) $\frac{1}{2}$
Examiner tip: Exactly two red can happen in 3 orders (RRB, RBR, BRR) — each has the same probability, so multiply by 3.

8Grade 8–9HigherNon-calculator5 marks

A bag contains $n$ beads. 4 of the beads are blue and the rest are white.

Two beads are taken at random from the bag, without replacement. The probability that both beads are blue is $\frac{1}{11}$

(a) Show that $n^2-n-132=0$ [3]

(b) Work out how many white beads are in the bag. [2]

(a) $\frac{4}{n}\times\frac{3}{n-1}$M1
(a) $\frac{12}{n(n-1)}=\frac{1}{11}$ oeM1
(a) $132=n(n-1)$ or $132=n^2-n$ leading to $n^2-n-132=0$ with no errors (shown)A1
(b) $(n-12)(n+11)=0$ or correct substitution into the quadratic formula, or $n=12$ seenM1
(b) 8 (white beads)A1
Answer: (a) shown (b) 8 white beads ($n=12$, reject $n=-11$)
Examiner tip: In a 'show that' the last line must be the given equation exactly with every step shown; in (b) remember to subtract the 4 blue beads from $n$.

9Grade 1–3FoundationNon-calculator5 marks

A café sells a lunch deal. Each lunch deal is one starter and one main.

The starter is either soup or salad.
The main is either fish, chicken or pasta.

(a) Complete the tree diagram to show all the possible lunch deals. [2]

(b) Write down the number of different lunch deals. [1]

Ravi chooses a starter at random. He then chooses a main at random.

(a) at least two of fish, chicken, pasta written on the branches after saladB1
(a) all three of fish, chicken, pasta on the branches after salad (any order)B1
(b) 6 (ft from their completed tree)B1
(c) $\frac{1}{2}\times\frac{1}{3}$ or identifies 1 outcome out of 6 equally likely outcomesM1
(c) $\frac{1}{6}$ oe (accept 0.166… or 0.17; do not accept '1 out of 6' or '1:6')A1
Answer: (a) fish, chicken, pasta after salad (b) 6 (c) $\frac{1}{6}$
Examiner tip: A probability must be written as a fraction, decimal or percentage – '1 in 6' style answers or ratios like 1:6 lose the accuracy mark.

10Grade 4–5Non-calculator5 marks

Customers at a café buy either tea or coffee. Some customers also buy a cake.

The tree diagram shows the probabilities for a customer chosen at random.

(a) Work out the probability that a customer buys tea and a cake. [2]

The café expects 240 customers on Saturday.

(b) Work out an estimate for the number of customers who will buy coffee but no cake. [2]

(c) The tree diagram shows the same probabilities for buying a cake after tea and after coffee.
What assumption does this make? [1]

(a) $0.6\times0.25$M1
(a) 0.15 oe (accept $\frac{3}{20}$, 15%)A1
(b) $240\times0.4\times0.75$ oe, e.g. $0.4\times0.75=0.3$ then $240\times0.3$, or $240\times0.4=96$ then $96\times0.75$M1
(b) 72 (do not accept a probability such as 0.3 or $\frac{72}{240}$ as the final answer)A1
(c) states that whether a customer buys a cake does not depend on (is independent of) whether they buy tea or coffee; accept 'the events are independent' with reference to drink and cakeB1
Answer: (a) 0.15 (b) 72 (c) buying a cake is independent of the drink chosen
Examiner tip: An 'estimate for the number' must be a number of customers – stopping at the probability 0.3 loses the final mark.

11Grade 6–7HigherCalculator5 marks

Lena drives to work. On her route there are three sets of traffic lights, $A$, $B$ and $C$.

The probability that she has to stop at $A$ is 0.3
The probability that she has to stop at $B$ is 0.45
The probability that she has to stop at $C$ is 0.6

Whether she stops at one set of lights is independent of whether she stops at the others.

(a) Work out the probability that Lena does not have to stop at any of the three sets of lights. [2]

(b) Work out the probability that Lena has to stop at exactly two of the three sets of lights. [3]

(a) $0.7\times0.55\times0.4$M1
(a) 0.154 oeA1
(b) one correct product for exactly two stops, e.g. $0.3\times0.45\times0.4$ $(=0.054)$ or $0.3\times0.55\times0.6$ $(=0.099)$ or $0.7\times0.45\times0.6$ $(=0.189)$M1
(b) sum of all three correct products: $0.054+0.099+0.189$M1
(b) 0.342 oeA1
Answer: (a) 0.154 (b) 0.342
Examiner tip: 'Exactly two' has three different routes through the tree – each product needs one 'not stop' probability, and students often find only one route or forget to add.

12Grade 6–7HigherNon-calculator5 marks

Nadia plants two seeds.

Each seed has the same probability of growing. The seeds grow independently of each other.

The probability that both seeds grow is 0.64

(a) Work out the probability that exactly one of the two seeds grows. [3]

Nadia plants three of these seeds.
She says, “The probability that all three seeds grow is more than 0.5”

(b) Is Nadia correct? You must show how you get your answer. [2]

(a) $\sqrt{0.64}=0.8$ for the probability that one seed grows (may be seen on a tree diagram)M1
(a) $0.8\times0.2+0.2\times0.8$ or $2\times0.8\times0.2$ or $1-0.64-0.2^2$M1
(a) 0.32 oeA1
(b) $0.8\times0.8\times0.8$ or $0.8^3$ (ft their 0.8)M1
(b) 0.512 and Yes (correct), e.g. '0.512 > 0.5 so she is correct'; must have correct figure to awardC1
Answer: (a) 0.32 (b) Yes – $0.8^3=0.512>0.5$
Examiner tip: Work backwards first: if both grow with probability 0.64, one seed grows with probability $\sqrt{0.64}=0.8$, not 0.32.

13Grade 8–9HigherNon-calculator5 marks

A box contains 8 sweets.
4 are strawberry, 3 are lemon and 1 is lime.

Tia takes a sweet at random and eats it. She then takes another sweet at random and eats it.

(a) Work out the probability that the two sweets are different flavours. [3]

(b) Given that the two sweets are the same flavour, work out the probability that they are both strawberry. [2]

(a) one correct product of two dependent probabilities, e.g. $\frac{4}{8}\times\frac{3}{7}$ $(=\frac{12}{56})$ or $\frac{3}{8}\times\frac{2}{7}$ $(=\frac{6}{56})$ or $\frac{1}{8}\times\frac{4}{7}$ $(=\frac{4}{56})$M1
(a) $1-\left(\frac{12}{56}+\frac{6}{56}\right)$ or sum of all six 'different' products: $\frac{12+4+12+3+4+3}{56}$M1
(a) $\frac{38}{56}$ oe e.g. $\frac{19}{28}$ (accept 0.678… or 0.68)A1
(b) $\frac{12}{56}\div\frac{18}{56}$ or $\frac{12}{18}$ oe (ft their P(same) from (a) if it is $1-$ their answer)M1
(b) $\frac{2}{3}$ oeA1
Answer: (a) $\frac{19}{28}$ (b) $\frac{2}{3}$
Examiner tip: 'Given that' means the answer is a fraction of the 'same flavour' probability – dividing $\frac{12}{56}$ by $\frac{18}{56}$, not just writing $\frac{12}{56}$.

14Grade 6–7HigherCalculator5 marks

The tree diagram gives information about the students in a year group who took a test.

The probability that a student chosen at random passed the test is 0.73

(a) Work out the value of $x$. [3]

There are 400 students in the year group.

(b) Work out an estimate for the number of students who did not revise and failed the test. [2]

(a) $0.7\times0.85$ $(=0.595)$M1
(a) $0.595+0.3x=0.73$ or $0.3x=0.73-0.595$ $(=0.135)$M1
(a) 0.45 oeA1
(b) $400\times0.3\times(1-0.45)$ or $400\times0.3\times0.55$ oe (ft their $x$, provided $0<x<1$)M1
(b) 66A1
Answer: (a) $x=0.45$ (b) 66
Examiner tip: P(pass) comes from two routes through the tree – set the sum of both products equal to 0.73 rather than using 0.73 on one branch.

15Grade 8–9HigherNon-calculator6 marks

A bag contains 4 red counters and 2 blue counters only.

Zara takes counters at random from the bag, one at a time, without replacement. She stops as soon as she takes a blue counter.

(a) Work out the probability that Zara takes exactly two counters. [2]

(b) Work out the probability that Zara takes exactly three counters. [2]

(a) $\frac{4}{6}\times\frac{2}{5}$M1
(a) $\frac{8}{30}$ oe e.g. $\frac{4}{15}$A1
(b) $\frac{4}{6}\times\frac{3}{5}\times\frac{2}{4}$M1
(b) $\frac{24}{120}$ oe e.g. $\frac{1}{5}$A1
(c) $\frac{4}{6}\times\frac{3}{5}$ (first two red) or $1-\left(\frac{2}{6}+\frac{8}{30}\right)$ (ft their (a)) or $\frac{24}{120}+\frac{4}{6}\times\frac{3}{5}\times\frac{2}{4}$ (exactly 3 + more than 3)M1
(c) $\frac{12}{30}$ oe e.g. $\frac{2}{5}$ (accept 0.4)A1
Answer: (a) $\frac{4}{15}$ (b) $\frac{1}{5}$ (c) $\frac{2}{5}$
Examiner tip: 'At least three' just means the first two counters are both red, so $\frac{4}{6}\times\frac{3}{5}$ is enough – students who use a $1-\ldots$ method often forget P(stops after one) $=\frac{2}{6}$.

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