Conditional probability – GCSE Maths exam questions with mark scheme

1Grade 4–5Non-calculator3 marks

The two-way table shows the option chosen by 60 students in Years 10 and 11.

	Art	Music	Drama	Total
Year 10	12	9	7	28
Year 11	10	14	8	32
Total	22	23	15	60

(a) A student is chosen at random. Find the probability that the student chose art. [1]

(b) A Year 11 student is chosen at random. Find the probability that this student chose music. [1]

(a) $\frac{22}{60}$ oeB1
(b) $\frac{14}{32}$ oeB1
(c) $\frac{7}{15}$ oeB1
Answer: (a) $\frac{22}{60}$ (b) $\frac{14}{32}$ (c) $\frac{7}{15}$
Examiner tip: When you are told which group the person comes from, the denominator is that group's total, not 60.

2Grade 6–7HigherNon-calculator3 marks

For two events $A$ and $B$:

$P(A)=0.5$ $P(B)=0.4$ $P(A\cap B)=0.15$

(a) Given that event $B$ happens, work out the probability that event $A$ also happens. [2]

(b) Are events $A$ and $B$ independent? Give a reason for your answer. [1]

(a) $\frac{0.15}{0.4}$M1
(a) 0.375 oe, e.g. $\frac{3}{8}$A1
(b) No, with a correct numerical reason, e.g. $0.5\times0.4=0.2\neq0.15$, or $0.375\neq P(A)=0.5$, or $\frac{0.15}{0.5}=0.3\neq P(B)=0.4$; 'No' with no or wrong reason scores 0C1
Answer: (a) 0.375 (b) No, since $P(A)\times P(B)=0.2\neq0.15$
Examiner tip: 'No' on its own scores 0 — the mark is for the numerical comparison.

3Grade 6–7HigherNon-calculator3 marks

The Venn diagram shows information about 90 members of a gym. $C$ = attends a fitness class, $S$ = uses the swimming pool.

(a) A member who uses the swimming pool is chosen at random. Work out the probability that this member also attends a fitness class. [2]

(b) A member who does not use the swimming pool is chosen at random. Work out the probability that this member attends a fitness class. [1]

(a) $18+30$ $(=48)$ used as denominatorM1
(a) $\frac{18}{48}$ oeA1
(b) $\frac{24}{42}$ oeB1
Answer: (a) $\frac{18}{48}=\frac{3}{8}$ (b) $\frac{24}{42}=\frac{4}{7}$
Examiner tip: Given that the member swims, you only look inside circle $S$ — the denominator is 48, not 90.

4Grade 6–7HigherNon-calculator4 marks

$A$ and $B$ are independent events.

$P(A)=0.3$ $P(A\cap B)=0.12$

(a) Work out $P(B)$. [2]

(b) Work out $P(A\cup B)$. [2]

(a) $0.3\times P(B)=0.12$ or $0.12\div0.3$M1
(a) 0.4A1
(b) $0.3+0.4-0.12$ (ft their $P(B)$), or Venn regions 0.18, 0.12, 0.28 foundM1
(b) 0.58A1
Answer: (a) 0.4 (b) 0.58
Examiner tip: In (b) subtract the overlap — $0.3+0.4=0.7$ counts $A\cap B$ twice.

5Grade 8–9HigherCalculator4 marks

A factory has two machines, $A$ and $B$. Machine $A$ makes 60% of the items and machine $B$ makes the rest.

5% of the items made by machine $A$ are faulty. 8% of the items made by machine $B$ are faulty.

An item is chosen at random and is found to be faulty. Work out the probability that it was made by machine $A$. [4]

$0.6\times0.05$ $(=0.03)$ or $0.4\times0.08$ $(=0.032)$M1
$0.6\times0.05+0.4\times0.08$ $(=0.062)$M1
$\frac{0.03}{0.062}$M1
$\frac{15}{31}$ or 0.484 (0.4838…)A1
Answer: $\frac{15}{31}\approx0.484$
Examiner tip: The answer is not 0.03 — you are told the item IS faulty, so divide by the total probability of a faulty item.

6Grade 8–9HigherNon-calculator4 marks

A bag contains 4 red counters and 6 blue counters. Two counters are taken at random, without replacement.

Given that the two counters are the same colour, work out the probability that they are both red. [4]

$\frac{4}{10}\times\frac{3}{9}$ $\left(=\frac{12}{90}\right)$M1
$\frac{6}{10}\times\frac{5}{9}$ $\left(=\frac{30}{90}\right)$M1
$\frac{12}{90}\div\left(\frac{12}{90}+\frac{30}{90}\right)$M1
$\frac{12}{42}$ oe, e.g. $\frac{2}{7}$A1
Answer: $\frac{2}{7}$
Examiner tip: 'Given that they are the same colour' shrinks the sample space to RR and BB — divide by P(same colour).

7Grade 1–3FoundationNon-calculator4 marks

80 people watched a film at a cinema.

32 of the people were children. The rest were adults.

20 of the children bought popcorn. 18 of the adults bought popcorn.

(a) Use this information to complete the frequency tree. [3]

(b) One of the 80 people is chosen at random.
Write down the probability that this person is an adult who did not buy popcorn. [1]

(a) 32 (child) and 48 (adult) in the first pair of boxesB1
(a) 20 (popcorn) and 12 (no popcorn) on the child branchesB1
(a) 18 (popcorn) and 30 (no popcorn) on the adult branches; ft their adult total $-18$ if 48 is wrongB1
(b) $\frac{30}{80}$ oe, e.g. $\frac{3}{8}$, 0.375; ft their adult/no-popcorn number over 80. Do not accept ratio form 30 : 80 or words such as '30 out of 80'B1
Answer: (a) Child 32 → popcorn 20, no popcorn 12; Adult 48 → popcorn 18, no popcorn 30 (b) $\frac{30}{80}=\frac{3}{8}$
Examiner tip: Each pair of boxes must add up to the box it branches from — check that 32 + 48 = 80, 20 + 12 = 32 and 18 + 30 = 48.

8Grade 4–5Non-calculator4 marks

A box contains 12 chocolates. 7 are milk chocolates and 5 are dark chocolates.

Tom takes a chocolate at random and eats it. He then takes another chocolate at random and eats it.

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

(b) Work out the probability that Tom eats two dark chocolates. [2]

(a) $\frac{6}{11}$ (milk) and $\frac{5}{11}$ (dark) on the second-chocolate branches after milkB1
(a) $\frac{7}{11}$ (milk) and $\frac{4}{11}$ (dark) on the second-chocolate branches after darkB1
(b) $\frac{5}{12}\times\frac{4}{11}$ (ft their tree diagram, provided the second-pick probabilities are less than 1)M1
(b) $\frac{20}{132}$ oe, e.g. $\frac{5}{33}$, or awrt 0.152. $\frac{25}{144}$ (with replacement) scores M0 A0A1
Answer: (a) After milk: milk $\frac{6}{11}$, dark $\frac{5}{11}$; after dark: milk $\frac{7}{11}$, dark $\frac{4}{11}$ (b) $\frac{20}{132}=\frac{5}{33}$
Examiner tip: Once a chocolate is eaten there are only 11 left, so every second-chocolate fraction has denominator 11, and the colour that was eaten goes down by one.

9Grade 4–5Non-calculator4 marks

A pencil case contains 6 blue pens and 4 black pens.

Ali takes a pen at random and does not put it back. He then takes a second pen at random.

Ali says, “The probability that both pens are blue is $\frac{6}{10}\times\frac{6}{10}=\frac{36}{100}$.”

(a) Explain the mistake Ali has made. [1]

(b) Work out the probability that Ali takes two pens of the same colour. [3]

(a) Explanation that the first pen is not replaced, so the second probability should be $\frac{5}{9}$ (accept 'there are only 9 pens left and 5 blue for the second pen' or 'he has treated the picks as independent'). Do not accept just 'he should not multiply'C1
(b) $\frac{6}{10}\times\frac{5}{9}$ $\left(=\frac{30}{90}\right)$ or $\frac{4}{10}\times\frac{3}{9}$ $\left(=\frac{12}{90}\right)$M1
(b) $\frac{6}{10}\times\frac{5}{9}+\frac{4}{10}\times\frac{3}{9}$M1
(b) $\frac{42}{90}$ oe, e.g. $\frac{7}{15}$, 0.466… or 0.467. SC1 for $\frac{52}{100}$ (with-replacement method $\frac{36}{100}+\frac{16}{100}$)A1
Answer: (a) The pen is not replaced, so the second probability is $\frac{5}{9}$, not $\frac{6}{10}$ (b) $\frac{42}{90}=\frac{7}{15}$
Examiner tip: 'Same colour' means two routes — both blue OR both black — so you multiply along each branch and then add the two results.

10Grade 6–7HigherCalculator4 marks

There are 400 students at a college.

55% of the students are female.

$\frac{1}{5}$ of the female students study physics.
$\frac{2}{9}$ of the male students study physics.

A student who studies physics is chosen at random.

Work out the probability that this student is male. [4]

$400\times0.55$ $(=220)$ or $400\times0.45$ $(=180)$ — numbers of female or male students (may be seen in a two-way table or frequency tree)P1
$\frac{1}{5}\times220$ $(=44)$ or $\frac{2}{9}\times180$ $(=40)$ — physics students of one sexP1
$\frac{40}{44+40}$ or $\frac{40}{84}$ — their male physics over their total physics studentsP1
$\frac{10}{21}$ oe, e.g. $\frac{40}{84}$, or awrt 0.476. Do not accept $\frac{40}{400}$ or $\frac{40}{180}$A1
Answer: $\frac{40}{84}=\frac{10}{21}\approx0.476$
Examiner tip: You are told the student studies physics, so the denominator is the 84 physics students, not the 400 students in the college.

11Grade 6–7HigherNon-calculator5 marks

The members of a sports club are asked whether they play tennis and whether they play squash.

$T$ = the member plays tennis $S$ = the member plays squash

The Venn diagram shows the probability that a member chosen at random is in each region. $x$ is a probability.

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

(b) Work out $P(S\,|\,T)$. [2]

(a) $0.2$ oe (from $1-0.35-0.2-0.25$)B1
(b) $\frac{x}{0.35+x}$ with their $x$, e.g. $\frac{0.2}{0.55}$M1
(b) $\frac{4}{11}$ oe, e.g. $\frac{20}{55}$, or 0.36… (0.3636…). Do not accept 0.2 or $\frac{0.2}{0.4}$A1
(c) $\frac{0.35}{0.35+0.25}$ or $\frac{0.35}{0.6}$M1
(c) $\frac{7}{12}$ oe, e.g. $\frac{35}{60}$, or 0.58… (0.5833…)A1
Answer: (a) $x=0.2$ (b) $\frac{4}{11}$ (c) $\frac{7}{12}$
Examiner tip: $S'$ means everything outside circle $S$ — including the 0.25 outside both circles — so the denominator in (c) is 0.35 + 0.25.

12Grade 6–7HigherCalculator6 marks

The probability that Sam’s bus is late on Monday is 0.2.

If the bus is late on one day, the probability that it is late the next day is 0.35.
If the bus is not late on one day, the probability that it is late the next day is 0.15.

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

(b) Work out the probability that Sam’s bus is late on Tuesday. [2]

(a) 0.8 for 'not late' on MondayB1
(a) Tuesday branches all correct: 0.35, 0.65 after late; 0.15, 0.85 after not lateB1
(b) $0.2\times0.35+0.8\times0.15$ (ft their tree)M1
(b) 0.19 oeA1
(c) $\frac{0.2\times0.35}{\text{their }0.19}$, e.g. $\frac{0.07}{0.19}$M1
(c) $\frac{7}{19}$ oe, or awrt 0.368. Do not accept 0.07 or 0.35A1
Answer: (a) Monday: late 0.2, not late 0.8; Tuesday after late: 0.35, 0.65; after not late: 0.15, 0.85 (b) 0.19 (c) $\frac{7}{19}\approx0.368$
Examiner tip: In (c) you already know the bus was late on Tuesday, so divide P(late both days) = 0.07 by P(late on Tuesday) = 0.19, not by 1.

13Grade 8–9HigherNon-calculator5 marks

A bag contains $n$ sweets. 3 of the sweets are orange.

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

The probability that neither sweet is orange is $\frac{7}{15}$.

(a) Show that $4n^2-49n+90=0$ [3]

(b) Work out the number of sweets in the bag at the start. You must show your working. [2]

(a) $\frac{n-3}{n}\times\frac{n-4}{n-1}$ oeM1
(a) $\frac{(n-3)(n-4)}{n(n-1)}=\frac{7}{15}$ leading to $15(n-3)(n-4)=7n(n-1)$ oe, e.g. $15n^2-105n+180=7n^2-7n$M1
(a) Correct algebra with no errors via $8n^2-98n+180=0$ to $4n^2-49n+90=0$ (the '$=0$' must be seen)A1
(b) $(4n-9)(n-10)=0$ or correct use of the quadratic formula, e.g. $\frac{49\pm\sqrt{2401-1440}}{8}$M1
(b) 10 only ($\frac{9}{4}$ rejected, as $n$ must be a whole number). SC1 for $n=10$ by trial with no algebraA1
Answer: (a) $\frac{n-3}{n}\times\frac{n-4}{n-1}=\frac{7}{15}\Rightarrow 15(n^2-7n+12)=7n^2-7n\Rightarrow 8n^2-98n+180=0\Rightarrow 4n^2-49n+90=0$ (b) 10
Examiner tip: 'Neither orange' means both sweets come from the $n-3$ non-orange ones, and after one is eaten there are $n-4$ of them left out of $n-1$.

14Grade 6–7HigherNon-calculator4 marks

60 students were asked which languages they study.

$F$ = studies French $G$ = studies German $S$ = studies Spanish

The Venn diagram shows the number of students in each region.

(a) A student who studies exactly two of these languages is chosen at random.
Work out the probability that this student studies French and Spanish. [2]

(b) A student who studies German is chosen at random.
Work out the probability that this student also studies at least one other of these languages. [2]

(a) $5+9+4$ $(=18)$ used as the denominatorM1
(a) $\frac{9}{18}$ oe, e.g. $\frac{1}{2}$. Do not accept $\frac{12}{60}$ or $\frac{9}{60}$A1
(b) $\frac{5+4+3}{8+5+4+3}$ or $\frac{12}{20}$ seen, or $1-\frac{8}{20}$M1
(b) $\frac{12}{20}$ oe, e.g. $\frac{3}{5}$, 0.6A1
Answer: (a) $\frac{9}{18}=\frac{1}{2}$ (b) $\frac{12}{20}=\frac{3}{5}$
Examiner tip: 'Exactly two' excludes the centre region (all three languages), so the 3 students in the middle are not in the denominator for (a).

15Grade 8–9HigherNon-calculator5 marks

Two fair six-sided dice are rolled.

(a) Given that at least one of the dice shows a 6, work out the probability that the total score is 10 or more. [3]

Hana says,
“Knowing that at least one dice shows a 6 makes no difference to the probability that the total is 10 or more.”

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

(a) 11 outcomes with at least one 6 identified, e.g. a sample space with the row and column for 6 marked, or 11 used as a denominatorM1
(a) 5 favourable outcomes identified: (4, 6), (6, 4), (5, 6), (6, 5), (6, 6)M1
(a) $\frac{5}{11}$ cao. Do not accept $\frac{5}{36}$, or $\frac{5}{12}$ or $\frac{6}{12}$ ((6, 6) counted twice)A1
(b) $P(\text{total} \geq 10)=\frac{6}{36}$ oe, e.g. $\frac{1}{6}$M1
(b) No, with a correct comparison, e.g. $\frac{5}{11}\neq\frac{1}{6}$ or 'knowing there is a 6 makes it more likely' (ft their answer to (a) if not equal to $\frac{1}{6}$)C1
Answer: (a) $\frac{5}{11}$ (b) No: $P(\text{total}\geq10)=\frac{6}{36}=\frac{1}{6}$, which is not equal to $\frac{5}{11}$
Examiner tip: The outcome (6, 6) appears only once in the sample space, so there are 11 outcomes with a 6, not 12.

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