Elite Mathematics1. May 2023 P1HR · Q5
Show worked solution
- Probability of losing\[ 1-0.74=0.26 \]
- Estimate for 300 games\[ 300\times0.26=78 \]
Final answer
\(78\)
IGCSE · 4MA1 · Topic
Probability Toolkit: past paper questions & worked solutions
Every Probability Toolkit question from the 4MA1 papers in our bank, 2020–2026, sorted from easiest to hardest. Try each one first, then open the worked solution.
42questions
163marks in total
61%of papers (33 of 54) include it
2/31/9easy / medium / hard
Questions, easiest first
Level is estimated from marks: 1–2 easy, 3–4 medium, 5+ hard.
2. Jan 2022 P1HR · Q2
Show worked solution
- Expected number = probability × number of games\[ 0.65\times300=195 \]
Final answer
\(195\)
3. Jun 2026 P2H · Q20
Show worked solution
- Solution | Step 1
Let \(g=P(G)\). With replacement, the draws are independent. \[P(GG)=1-P(\text{at least one red})=1-\frac{39}{64}=\frac{25}{64}.\]
The complementary event to at least one red is two greens.
- Solution | Step 2\[g^2=\frac{25}{64}\quad\Rightarrow\quad g=\frac58,\qquad P(R)=1-\frac58=\frac38.\]
Multiply identical independent green probabilities and take the nonnegative root.
- Solution | Step 3\[P(\text{exactly one red})=P(RG)+P(GR)=\frac38\frac58+\frac58\frac38=\frac{30}{64}=\frac{15}{32}.\]
The two possible orders are mutually exclusive and must both be included.
Final answer
\(\dfrac{15}{32}\)
4. May 2026 4WM1H · Q11
Show worked solution
- Let r red be added
Then \(2r\) blue are added and the total is \(40+3r\).
- Use P(green)\[\frac{25}{40+3r}=\frac{5}{11}\Rightarrow 275=200+15r\Rightarrow r=5\]
- Blue added\[2r=10\]
Final answer
\(10\)
5. Nov 2025 P2H · Q9
Show worked solution
- Red counters in each bag
Bag A: \(120\times0.45=54\). Bag B: \(80\times0.3=24\).
- Probability of red from bag C
Bag C has \(54+24=78\) red counters out of \(120+80=200\).
\[ \frac{78}{200}=0.39 \]
Final answer
\(0.39\)
6. Jun 2025 P2H · Q11
Show worked solution
- Counters that are not orange
\(17+28=45\) counters are not orange, and \(P(\text{not orange})=1-\frac49=\frac59\).
- Total number of counters\[ \frac59\times T=45 \;\Rightarrow\; T=81 \]
- Orange counters\[ 81-45=36 \]
Final answer
\(36\)
7. Jun 2025 P1HR · Q19
Show worked solution
- Pairs with a sum less than 5
The counters are \(1,1,2,3,3,3,5,6,6\). The only possible sums under \(5\) are \(1+1\), \(1+2\) and \(1+3\).
- First counter, then second (no replacement)\[ P(1,1)=\frac29\times\frac18=\frac{2}{72} \]\[ P(1\text{ and }2)=2\times\frac29\times\frac18=\frac{4}{72},\qquad P(1\text{ and }3)=2\times\frac29\times\frac38=\frac{12}{72} \]
- Add the cases\[ \frac{2+4+12}{72}=\frac{18}{72}=\frac14 \]
Final answer
\(\dfrac14\)
8. Jun 2025 P1H · Q21
Show worked solution
- There are\[ \binom{20}{3}=1140 \]
- ways to choose 3 counters
ways to choose 3 counters.
- Exactly two counters are the
Exactly two counters are the same colour:
\[ \binom{9}{2}(11)+\binom{7}{2}(13)+\binom{4}{2}(16) \]\[ =36(11)+21(13)+6(16) \]\[ =396+273+96=765 \] - Calculate probability
So the probability is
\[ \frac{765}{1140}=\frac{51}{76} \]
Final answer
\(\frac{51}{76}\)
9. May 2025 4WM1HR · Q17
Show worked solution
- The counters are\[ 1,\ 1,\ 2,\ 3,\ 3,\ 3,\ 5,\ 6,\ 6 \]
- There are\[ \binom{9}{2}=36 \]
- ways to choose two counters
ways to choose two counters.
- Calculate probability
For a sum less than \(5\), the pairs are:
\[ (1,1),\quad (1,2),\quad (1,3) \] - Counting the repeated counters gives\[ 1+2+6=9 \]
- Calculate probability
So
\[ P(\text{sum}<5)=\frac{9}{36}=\frac14 \]
Final answer
\(\frac14\).
10. May 2025 4WM1H · Q22
Show worked solution
- Total ways to choose counters
Total ways to choose \(3\) counters:
\[ \binom{20}{3}=1140 \] - Exactly two red
Exactly two red:
\[ \binom{9}{2}\binom{11}{1}=396 \] - Exactly two yellow
Exactly two yellow:
\[ \binom{7}{2}\binom{13}{1}=273 \] - Exactly two green
Exactly two green:
\[ \binom{4}{2}\binom{16}{1}=96 \] - Favourable outcomes
Favourable outcomes:
\[ 396+273+96=765 \]\[ P=\frac{765}{1140}=\frac{51}{76} \]
Final answer
\(\frac{51}{76}\)
11. Jun 2024 P2HR · Q14
Show worked solution
- Find the probability of a 6
The dice and the coin are independent:
\[ P(6)\times0.8=0.24 \;\Rightarrow\; P(6)=0.3 \] - 6 and Tails
\(P(\text{Tails})=1-0.8=0.2\), so
\[ 0.3\times0.2=0.06 \]
Final answer
\(0.06\)
12. Jun 2023 P2H · Q2
Show worked solution
- Probability of pink\[ 1-(0.32+0.13+0.28)=1-0.73=0.27 \]
- Estimate for 200 spins\[ 200\times0.27=54 \]
Final answer
\(54\)
13. Jun 2023 P2HR · Q15
Show worked solution
- Find the probability of a head\[ P(H)^3=\frac{27}{64} \;\Rightarrow\; P(H)=\sqrt[3]{\frac{27}{64}}=\frac34 \]
- Three tails
\(P(T)=1-\frac34=\frac14\), so
\[ P(TTT)=\left(\frac14\right)^3=\frac{1}{64} \]
Final answer
\(\dfrac{1}{64}\)
14. Jun 2026 P1HR · Q5
Show worked solution
- Solution | Step 1\[0.45+0.31+5x+x=1.\]
The four colours are mutually exclusive and exhaust all counters, so their probabilities sum to 1.
- Solution | Step 2\[6x=1-0.76=0.24,\qquad x=0.04.\]
Collect the unknown probabilities and solve the linear equation.
- Solution | Step 3\[P(\text{red})=5(0.04)=0.20,\qquad N_{\text{red}}=400(0.20)=80.\]
Multiply the red proportion by the total number of counters.
Final answer
\(80\) red counters
15. Jun 2026 P1H · Q7
Show worked solution
- Solution | Step 1\[0.12+x+2x+0.16=1\implies3x=0.72\implies x=0.24.\]
The four exhaustive outcome probabilities sum to 1.
- Solution | Step 2\[P(\text{even})=P(2)+P(4)=0.24+0.16=0.40.\]
The events 2 and 4 are mutually exclusive.
- Solution | Step 3\[\text{Expected number}=500\times0.40=200.\]
Multiply the number of spins by the probability of an even result.
Final answer
\(200\)
16. May 2026 4WM1H · Q7
Show worked solution
- Probabilities add to 1\[0.12+x+2x+0.16=1\Rightarrow 3x=0.72\Rightarrow x=0.24\]
- Even numbers are 2 and 4\[P(\text{even})=0.24+0.16=0.40\]
- Expected number\[500\times0.40=200\]
Final answer
\(200\)
17. May 2025 4WM1H · Q3
Show worked solution
- Calculate probability
The probabilities add to \(1\).
\[ 0.14+0.23+2x+0.18+7x=1 \]\[ 0.55+9x=1 \]\[ x=0.05 \] - The probability of blue is\[ 7x=7(0.05)=0.35 \]
- Expected number of blue results
Expected number of blue results:
\[ 500\times0.35=175 \]
Final answer
\(175\)
18. Nov 2024 P2H · Q5
Show worked solution
- Probabilities of 1 and 4 together\[ 1-(0.14+0.17+0.21)=1-0.52=0.48 \]
- They are equal\[ P(4)=0.48\div2=0.24 \]
- Estimate for 400 spins\[ 400\times0.24=96 \]
Final answer
\(96\)
19. May 2024 P1H · Q2
Show worked solution
- Find x\[ 0.20+x+2x+0.26=1 \;\Rightarrow\; 3x=0.54 \;\Rightarrow\; x=0.18 \]
- Probability of travelling by car\[ 2x=0.36 \]
- Number of students\[ 450\times0.36=162 \]
Final answer
\(162\)
20. May 2024 P1HR · Q2
Show worked solution
- Find x\[ 2x+0.27+0.04+x+0.12=1 \;\Rightarrow\; 3x=0.57 \;\Rightarrow\; x=0.19 \]
- Probability of an odd number (1, 3 or 5)\[ 2x+0.04+0.12=0.38+0.04+0.12=0.54 \]
- Estimate for 400 spins\[ 400\times0.54=216 \]
Final answer
\(216\)
21. Nov 2023 P1H · Q4
Show worked solution
- (a) Probabilities add to 1\[ 0.58+2x+x=1 \;\Rightarrow\; 3x=0.42 \;\Rightarrow\; x=0.14 \]
- (b) Estimate for blue in 250 spins\[ 250\times0.58=145 \]
Final answer
(a) \(x=0.14\) (b) \(145\)
22. Nov 2023 P1H · Q20
Show worked solution
- Let the probability of taking
Let the probability of taking a 10 cent coin be \(p\).
- The mean is 10 cents
The mean is 10 cents only when both coins are 10 cent coins.
\[ p^2=\frac{49}{121} \]\[ p=\frac{7}{11} \] - Calculate probability
So the probability of taking a 20 cent coin is
\[ 1-\frac{7}{11}=\frac{4}{11} \] - The mean is 15 cents
The mean is 15 cents when one coin is 10 cents and the other is 20 cents.
- There are two possible orders
There are two possible orders:
\[ P(10,20)+P(20,10) \]\[ =\frac{7}{11}\cdot\frac{4}{11}+\frac{4}{11}\cdot\frac{7}{11} \]\[ =\frac{56}{121} \]
Final answer
\(\frac{56}{121}\)
23. Jun 2023 P2HR · Q19
Show worked solution
- Adults who booked breakfast and dinner
\(30-4=26\) adults booked at least one meal.
\[ n(B\cap D)=19+15-26=8 \] - Choose two of them (without replacement)\[ \frac{8}{30}\times\frac{7}{29}=\frac{56}{870}=\frac{28}{435} \]
Final answer
\(\dfrac{28}{435}\)
24. Jun 2022 P1HR · Q1
Show worked solution
- (a) Red or yellow\[ 0.25+0.2=0.45 \]
- (b) Probability of purple\[ 1-(0.25+0.2+0.2)=0.35 \]
- (b) Estimate for 300 spins\[ 300\times0.35=105 \]
Final answer
(a) \(0.45\) (b) \(105\)
25. Jun 2022 P1H · Q2
Show worked solution
- Probabilities of landing on 2 and on 4
The probabilities add to \(1\), and \(P(2)=P(4)\):
\[ 1-(0.26+0.18)=0.56,\qquad P(4)=0.56\div2=0.28 \] - Number of spins
An estimate of \(45\) for the number 3 means
\[ 0.18\times n=45 \;\Rightarrow\; n=250 \] - Estimate for landing on 4\[ 250\times0.28=70 \]
Final answer
\(70\)
26. Jun 2022 P1H · Q15
Show worked solution
- Ways to score a total of 0
Three draws (\(0+0+0\)), or one win and two losses (\(10-5-5=0\)) in any order.
- Three draws\[ 0.5^3=0.125 \]
- One win and two losses (3 orders)\[ 3\times0.3\times0.2\times0.2=0.036 \]
- Add the cases\[ 0.125+0.036=0.161 \]
Final answer
\(0.161\)
27. Jan 2022 P2HR · Q1
Show worked solution
- (a) Probabilities add to 1\[ P(\text{Nilgiri})=1-(0.38+0.24+0.16)=1-0.78=0.22 \]
- (b) Darjeeling or Rize\[ 0.24+0.16=0.40 \]
Final answer
(a) \(0.22\) (b) \(0.4\)
28. Nov 2021 P1H · Q3
Show worked solution
- Calculate probability
(a) Let the number of white fish be \(w\).
\[ \frac{w}{54}=\frac49 \]\[ w=54\times\frac49=24 \] - (b) There are still 54
(b) There are still 54 fish, and now
\[ \frac{\text{white fish}}{54}=\frac12 \]\[ \text{white fish}=27 \] - Jeevan added\[ 27-24=3 \]
Final answer
(a) \(24\) white fish, (b) \(3\) white fish.
29. Jan 2020 P2H · Q6
Show worked solution
- Use the fact that probabilities add to 1
Let \(P(\text{red})=r\), so \(P(\text{green})=2r\).
\[ r+0.24+2r+0.31=1 \;\Rightarrow\; 3r=0.45 \;\Rightarrow\; r=0.15 \] - Expected number of red beads\[ 180\times0.15=27 \]
Final answer
\(27\)
30. Jan 2020 P1H · Q18
Show worked solution
- Count the sweets that are not lemon
There are \(16\) sweets and \(5\) are lemon, so \(11\) are not lemon.
- Three picks without replacement, none lemon
All \(5\) lemon sweets stay in the bowl only if Anna, Ravi and Sam each take a non-lemon sweet:
\[ \frac{11}{16}\times\frac{10}{15}\times\frac{9}{14}=\frac{990}{3360}=\frac{33}{112} \]
Final answer
\(\dfrac{33}{112}\)
31. May-Nov 2020 P1H · Q2
Show worked solution
- Seconds in a working day\[ 15\times60\times60=54\,000\text{ seconds} \]
- Toy cars made each day\[ \frac{54\,000}{12}\times5=22\,500 \]
- Expected number of faulty cars\[ 22\,500\times0.002=45 \]
Final answer
\(45\)
32. Specimen 4WM1H · Q3
Show worked solution
- Find the remaining probability\[ 1-0.26-0.18=0.56 \]
- Split the equal probabilities
The probabilities for \(2\) and \(4\) are equal, so
\[ P(2)=P(4)=\frac{0.56}{2}=0.28 \] - Find the probability of 1 or 2\[ P(1\text{ or }2)=0.26+0.28=0.54 \]
- Calculate the estimate\[ 0.54\times 250=135 \]
Final answer
\(135\).
33. May-Nov 2020 P1HR · Q23
Show worked solution
- The total number of beads
The total number of beads is
\[ x+7 \] - The probability of choosing two
The probability of choosing two beads of the same colour is
\[ \frac{\binom{3}{2}+\binom{4}{2}+\binom{x}{2}}{\binom{x+7}{2}} \] - Evaluate fraction
This is equal to \(\frac38\), so
\[ \frac{3+6+\frac{x(x-1)}{2}}{\frac{(x+7)(x+6)}{2}}=\frac38 \]\[ \frac{x^2-x+18}{(x+7)(x+6)}=\frac38 \]\[ 8(x^2-x+18)=3(x+7)(x+6) \]\[ 8x^2-8x+144=3x^2+39x+126 \]\[ 5x^2-47x+18=0 \]\[ (x-9)(5x-2)=0 \] - Calculate value
Since \(x\) must be a whole number,
\[ x=9 \] - Calculate value
So the total number of beads is
\[ 9+7=16 \]
Final answer
\(16\) beads
34. Nov 2025 4WM1H · Q5
Show worked solution
- Calculate value\[ P(\text{tennis or cricket})=0.2+0.35=0.55 \]
- Calculate value
Also,
\[ 2x+0.2+x+0.35=1 \]\[ 3x=0.45 \]\[ x=0.15 \] - Calculate value
So
\[ P(\text{football})=2x=0.30 \]\[ 700\times 0.30=210 \]
Final answer
\(0.55\), \(210\) students.
35. Jun 2025 P1HR · Q4
Show worked solution
- (a) Blue, green or purple\[ 0.12+0.20+0.38=0.70 \]
- (b) Find x
Red and yellow share the rest: \(4x+x=1-0.70=0.30\), so \(5x=0.30\) and \(x=0.06\).
- (b) Probability of red\[ 4x=4\times0.06=0.24 \]
- (b) Estimate for 350 spins\[ 350\times0.24=84 \]
Final answer
(a) \(0.7\) (b) \(84\)
36. May 2025 4WM1HR · Q3
Show worked solution
- Let the yellow probability be
Let the yellow probability be \(x\). Then the red probability is \(4x\).
\[ 0.12+0.20+0.30+4x+x=1 \]\[ 5x=0.38 \]\[ x=0.076 \] - Calculate probability
So
\[ P(\text{red})=4x=0.304 \]\[ P(\text{blue, green or purple})=0.12+0.20+0.30=0.62 \] - Expected red
Expected red:
\[ 350\times 0.304=106.4 \]
Final answer
\(0.62\), and about \(106\).
37. May 2024 P1HR · Q21
Show worked solution
- There are 25 counters
There are 25 counters.
- Blue counters
Blue counters:
\[ 6 \] - Orange counters
Orange counters:
\[ x \] - Pink counters
Pink counters:
\[ 25-6-x=19-x \] - The probability of taking one
The probability of taking one orange and one pink in either order is
\[ \frac{x}{25}\cdot\frac{19-x}{24} +\frac{19-x}{25}\cdot\frac{x}{24} \]\[ =\frac{2x(19-x)}{25\cdot24} \] - Evaluate fraction
This is equal to \(\frac{22}{75}\), so
\[ \frac{2x(19-x)}{600}=\frac{22}{75} \]\[ \frac{x(19-x)}{300}=\frac{22}{75} \]\[ x(19-x)=88 \]\[ 19x-x^2=88 \]\[ x^2-19x+88=0 \]\[ (x-8)(x-11)=0 \] - Solve inequality
Since \(x>9\),
\[ x=11 \] - Calculate value
So the number of pink counters is
\[ 19-11=8 \] - The probability of taking 2
The probability of taking 2 pink counters is
\[ \frac{8}{25}\cdot\frac{7}{24}=\frac{56}{600} \]\[ =\frac{7}{75} \]
Final answer
\(\frac{7}{75}\)
38. Jun 2022 P1HR · Q24
Show worked solution
- There are counters. Initially 5
There are \(x\) counters. Initially 5 show red, so \(x-5\) show green.
- List probability cases
After two turns, there are still 5 red faces if:
- the same counter is picked
- the same counter is picked twice, or - one red counter and one green counter are picked.
- The probability of picking the
The probability of picking the same counter twice is
\[ \frac{1}{x} \] - The probability of picking one
The probability of picking one red and one green in either order is
\[ \frac{5}{x}\cdot\frac{x-5}{x}+\frac{x-5}{x}\cdot\frac{5}{x} \]\[ =\frac{10(x-5)}{x^2} \] - Evaluate fraction
So
\[ \frac{1}{x}+\frac{10(x-5)}{x^2}=\frac{19}{32} \]\[ \frac{11x-50}{x^2}=\frac{19}{32} \]\[ 32(11x-50)=19x^2 \]\[ 19x^2-352x+1600=0 \]\[ (x-8)(19x-200)=0 \] - Calculate value
Since \(x\) must be a whole number,
\[ x=8 \]
Final answer
\(x=8\)
39. Jan 2022 P2H · Q6
Show worked solution
- Split the ratio
The probabilities for green and red are
\[ 0.20+0.35=0.55 \] - Split the ratio
So
\[ P(\text{pink or white})=1-0.55=0.45 \] - Split the ratio
Pink : white \(=2:1\), so white is \(\frac13\) of \(0.45\).
\[ P(\text{white})=\frac13\times 0.45=0.15 \] - Split the ratio
The red probability is \(0.35\), and there are \(28\) red sweets.
\[ \text{total sweets}=\frac{28}{0.35}=80 \] - Number of white sweets
Number of white sweets:
\[ 0.15\times 80=12 \]
Final answer
\(12\)
40. May 2021 P1H · Q6
Show worked solution
- (a) Set up with the missing probabilities
Let \(P(\text{pink})=x\), so \(P(\text{orange})=3x\).
\[ 0.20+0.12+0.08+3x+x=1 \;\Rightarrow\; 4x=0.60 \;\Rightarrow\; x=0.15 \] - (a) Probability of orange\[ 3x=0.45 \]
- (b) Estimate for blue\[ 150\times0.12=18 \]
Final answer
(a) \(0.45\) (b) \(18\)
41. May 2021 P2H · Q23
Show worked solution
- Let the number of red
Let the number of red pens be \(r\).
- List probability cases
Then the number of black pens is
\[ r+3 \] - Calculate value
and
\[ N=2r+3 \] - Evaluate fraction
So
\[ r=\frac{N-3}{2} \] - Evaluate fraction
and
\[ r+3=\frac{N+3}{2} \] - The probability of black followed
The probability of black followed by red is
\[ \frac{(N+3)/2}{N}\cdot \frac{(N-3)/2}{N-1}=\frac{9}{35} \]\[ \frac{N^2-9}{4N(N-1)}=\frac{9}{35} \]\[ 35(N^2-9)=36N(N-1) \]\[ 35N^2-315=36N^2-36N \]\[ N^2-36N+315=0 \]\[ (N-15)(N-21)=0 \]
Final answer
\(N=15\) or \(N=21\)
42. Jan 2021 P2HR · Q21
Show worked solution
- There are beads altogether
There are \(n\) beads altogether.
- Red beads
Red beads:
\[ 6 \] - Blue beads
Blue beads:
\[ n-6 \] - The probability of taking two
The probability of taking two beads of the same colour is
\[ P(RR)+P(BB)=\frac9{17} \] - Use trigonometry
So
\[ \frac6n\cdot\frac5{n-1} +\frac{n-6}{n}\cdot\frac{n-7}{n-1} =\frac9{17} \]\[ \frac{30+(n-6)(n-7)}{n(n-1)}=\frac9{17} \] - Expand
Expand:
\[ (n-6)(n-7)=n^2-13n+42 \] - Use trigonometry
So
\[ \frac{n^2-13n+72}{n(n-1)}=\frac9{17} \] - Cross multiply
Cross multiply:
\[ 17(n^2-13n+72)=9n(n-1) \]\[ 17n^2-221n+1224=9n^2-9n \]\[ 8n^2-212n+1224=0 \] - Divide by 4
Divide by 4:
\[ 2n^2-53n+306=0 \] - Factor
Factor:
\[ (2n-17)(n-18)=0 \] - Use trigonometry
So
\[ n=\frac{17}{2}\quad\text{or}\quad n=18 \] - The number of beads must
The number of beads must be a whole number, so
\[ n=18 \]
Final answer
\(\boxed{18}\)
More IGCSE Maths 4MA1 topics
- Number Toolkit 5
- Set Notation and Venn Diagrams 34
- Prime Factors, HCF and LCM 26
- Powers, Roots and Standard Form 35
- Fractions 18
- Percentages 44
- Compound Interest and Depreciation 35
- Fractions, Decimals and Percentages 11
- Rounding, Estimation and Bounds 35
- Surds 13
- Using a Calculator 7
- Ratio Toolkit 14
- Direct & Inverse Proportion 18
- Algebra Toolkit 8
- Algebraic Roots & Indices 49
- Expanding Brackets 21
- Factorisation 23
- Completing the Square 18
- Algebraic Fractions 28
- Rearranging Formulas 22
- Algebraic Proof 37
- Solving Linear Equations 14
- Solving Quadratics 6
- Solving Inequalities 44
- Simultaneous Equations 38
- Forming and Solving Equations 6
- Sequences 35
- Functions 26
- Linear Graphs y = mx + c 48
- Graphs of Functions 31
- Real-Life Graphs 7
- Transformations of Graphs 22
- Differentiation 29
- Standard and Compound Units 42
- Angles in Polygons and Parallel Lines 22
- Bearings, Scale Drawing and Constructions 13
- Circle Theorems 33
- Area and Perimeter 31
- Circles, Arcs and Sectors 32
- Volume and Surface Area 30
- Congruence, Similarity and Geometrical Proof 11
- Area and Volume of Similar Shapes 23
- Right-Angled Triangles: Pythagoras and Trigonometry 36
- Sine Rule, Cosine Rule and Area of a Triangle 21
- 3D Pythagoras and Trigonometry 19
- Vectors 28
- Transformations 11
- Statistics 64
- Histograms 27
- Cumulative Frequency 27
- Combined and Conditional Probability 35
Dr Eslam Ahmed
Elite Mathematics · Questions classified by topic, solutions checked by Dr Eslam