Edexcel International GCSE Maths Probability Past Paper Questions and Solutions (4MA1 Higher)

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Edexcel International GCSE Maths A (4MA1) · Higher · Statistics and Probability

Probability: Edexcel 4MA1 Past Paper Questions and Worked Solutions

This page collects every probability question from the Pearson Edexcel International GCSE Mathematics A (4MA1) Higher tier Paper 1H and Paper 1HR, January 2020 to June 2025: 30 questions worth 130 marks. Questions run from easier to harder. Try each one first, then click “Show solution” for a step-by-step method with the reason for every step. All answers are checked against the official Pearson mark schemes.

14 marksHigherJune 2022 · Paper 1HR (4MA1/1HR) · Q1

Here is a biased spinner with four colours. The table gives the probabilities that it lands on red, yellow or green when it is spun once.

Colourredyellowpurplegreen
Probability0.250.20.2

(a) Work out the probability that the spinner lands on red or on yellow. (1)

Yang is going to spin the spinner 300 times.

(b) Work out an estimate for the number of times the spinner will land on purple. (3)

Show solution

Part (a)

  1. Add the probabilities: 0.25 + 0.2

Answer: 0.45

Part (b)

  1. The probabilities add up to 1, so P(purple) = 1 − (0.25 + 0.2 + 0.2) = 1 − 0.65 = 0.35
  2. Estimate = 0.35 × 300

Answer: 105

24 marksHigherJune 2020 · Paper 1H (4MA1/1H) · Q2

Toy cars are made in a factory for 15 hours each day. 5 toy cars are made every 12 seconds. For the toy cars made each day, the probability of a toy car being faulty is 0.002. Work out an estimate of the number of faulty toy cars that are made each day.

Show solution
  1. 15 hours = 15 × 60 × 60 = 54 000 seconds
  2. Number of 12-second periods = 54 000 ÷ 12 = 4500, so cars made = 4500 × 5 = 22 500
  3. Expected faulty cars = 0.002 × 22 500

Answer: 45

32 marksHigherJanuary 2022 · Paper 1HR (4MA1/1HR) · Q2

Each time John plays a game, the probability that he wins the game is 0.65. John is going to play the game 300 times. Work out an estimate for the number of games that John wins.

Show solution
  1. Expected number = probability × number of games = 0.65 × 300

Answer: 195

44 marksHigherJune 2022 · Paper 1H (4MA1/1H) · Q2

Here is a biased 4-sided spinner numbered 1 to 4. The probability that it lands on 1 is 0.26 and on 3 is 0.18. The probability that the spinner will land on 2 is equal to the probability that it will land on 4. Ravina is going to spin the spinner a number of times. Ravina works out that an estimate for the number of times the spinner will land on 3 is 45. Work out an estimate for the number of times the spinner will land on 4.

Show solution
  1. Expected number = probability × number of spins, so 0.18 × spins = 45 and the number of spins = 45 ÷ 0.18 = 250
  2. P(2) + P(4) = 1 − 0.26 − 0.18 = 0.56. They are equal, so P(4) = 0.56 ÷ 2 = 0.28
  3. Estimate for 4 = 0.28 × 250

Answer: 70

54 marksHigherJune 2024 · Paper 1H (4MA1/1H) · Q2

450 students were asked how they travelled to school on Monday. Each student walked or travelled by bus or by car or by bicycle, and used just one method of travel. One of these students is chosen at random. The table shows the probability of each method of travel.

Method of travelwalkbuscarbicycle
Probability0.20x2x0.26

Work out how many of the 450 students travelled by car.

Show solution
  1. The four methods cover every student, so the probabilities must add up to 1.
  2. 0.20 + x + 2x + 0.26 = 1
  3. 3x + 0.46 = 1, so 3x = 0.54 and x = 0.18
  4. P(car) = 2x = 2 × 0.18 = 0.36
  5. Expected number = probability × total = 0.36 × 450

Answer: 162 students

64 marksHigherJune 2024 · Paper 1HR (4MA1/1HR) · Q2

Here is a biased spinner with sections numbered 1 to 5. The table gives the probability that, when the spinner is spun once, it will land on each number.

Number12345
Probability2x0.270.04x0.12

Alexis is going to spin the spinner 400 times. Work out an estimate for the number of times the spinner will land on an odd number.

Show solution
  1. The probabilities of all the outcomes add up to 1: 2x + 0.27 + 0.04 + x + 0.12 = 1
  2. 3x + 0.43 = 1, so 3x = 0.57 and x = 0.19
  3. P(1) = 2x = 0.38
  4. The odd numbers are 1, 3 and 5: P(odd) = 0.38 + 0.04 + 0.12 = 0.54
  5. Estimate = probability × number of spins = 0.54 × 400

Answer: 216

74 marksHigherNovember 2021 · Paper 1H (4MA1/1H) · Q3

There are 54 fish in a tank. Some of the fish are white and the rest of the fish are red. Jeevan takes at random a fish from the tank. The probability that he takes a white fish is 4⁄9.

(a) Work out the number of white fish originally in the tank. (2)

Jeevan puts the fish he took out back into the tank. He puts some more white fish into the tank. Jeevan takes at random a fish from the tank. The probability that he takes a white fish is now 1⁄2.

(b) Work out the number of white fish Jeevan put into the tank. (2)

Show solution

Part (a)

  1. 4⁄9 × 54 = 24

Answer: 24

Part (b)

  1. Let w be the number of white fish added. Then there are 24 + w white fish out of 54 + w.
  2. 24 + w⁄54 + w = 1⁄2, so 48 + 2w = 54 + w
  3. w = 6
  4. Check: 30 white out of 60 = 1⁄2 ✓ (half white means the white fish must equal the 30 red fish)

Answer: 6

85 marksHigherJune 2025 · Paper 1HR (4MA1/1HR) · Q4

Here is a biased 5-sided spinner. When the spinner is spun, it can land on blue, green, purple, red or yellow. The table gives the probability of the spinner landing on each colour.

Colourbluegreenpurpleredyellow
Probability0.120.200.384xx

Sophie spins the spinner once.

(a) Work out the probability that the spinner lands on blue or on green or on purple. (1)

Max spins the spinner 350 times.

(b) Work out an estimate for the number of times the spinner lands on red. (4)

Show solution

Part (a)

  1. “Or” for outcomes that cannot happen together: add the probabilities. 0.12 + 0.20 + 0.38

Answer: 0.70

Part (b)

  1. All the probabilities add up to 1, so red and yellow together = 1 − 0.70 = 0.30
  2. 4x + x = 5x = 0.30, so x = 0.06
  3. P(red) = 4x = 0.24
  4. Estimate = 0.24 × 350

Answer: 84

92 marksHigherJune 2023 · Paper 1HR (4MA1/1HR) · Q5

Each time Evie plays a game against her computer, she will win or lose. For each game, the probability that Evie will win is 0.74. Evie is going to play 300 games against her computer. Work out an estimate for the number of games that Evie will lose.

Show solution
  1. P(lose) = 1 − 0.74 = 0.26
  2. Expected number of losses = 0.26 × 300

Answer: 78

105 marksHigherJune 2021 · Paper 1H (4MA1/1H) · Q6

Grace has a biased 5-sided spinner. The table gives the probabilities that the spinner will land on red, blue or green.

ColourRedBlueGreenOrangePink
Probability0.200.120.08

The probability that the spinner will land on orange is 3 times the probability that the spinner will land on pink.

(a) Work out the probability that the spinner will land on orange. (3)

Grace spins the arrow on the spinner 150 times.

(b) Work out an estimate for the number of times the spinner lands on blue. (2)

Show solution

Part (a)

  1. Orange and pink together: 1 − (0.20 + 0.12 + 0.08) = 1 − 0.40 = 0.60
  2. Orange : pink = 3 : 1, so share 0.60 into 4 parts: 1 part = 0.15
  3. Orange = 3 × 0.15

Answer: 0.45

Part (b)

  1. 0.12 × 150

Answer: 18

114 marksHigherJanuary 2023 · Paper 1H (4MA1/1H) · Q7

Some members of a library were asked to name the type of book that they each liked to read the best. One of the members is chosen at random. The table shows the probability of the type of book that this member answered.

Type of bookcomedyromancemysterythriller
Probability0.240.403xx

48 members answered comedy books. Work out how many of the members answered mystery books.

Show solution
  1. The probabilities add to 1: 3x + x = 1 − 0.24 − 0.40 = 0.36, so x = 0.09 and P(mystery) = 0.27
  2. 48 members is 0.24 of all members, so the total = 48 ÷ 0.24 = 200
  3. Mystery = 0.27 × 200

Answer: 54

124 marksHigherJune 2022 · Paper 1H (4MA1/1H) · Q12

Rudolf goes to the gym. The probability that he will use the treadmill is 0.8. When he uses the treadmill, the probability that he will use the cross trainer is 0.3. When he does not use the treadmill, the probability that he will use the cross trainer is 0.6.

(a) Complete the probability tree diagram for this information. (2)

(b) Work out the probability that Rudolf uses both the treadmill and the cross trainer. (2)

Show solution

Part (a)

  1. Each pair of branches adds to 1: treadmill 0.8 and 0.2; after the treadmill 0.3 and 0.7; without the treadmill 0.6 and 0.4.
treadmillno treadmillcross trainerno cross trainercross trainerno cross trainer0.80.20.30.70.60.4

Answer: 0.8, 0.2; 0.3, 0.7; 0.6, 0.4

Part (b)

  1. Multiply along the treadmill → cross trainer branches: 0.8 × 0.3

Answer: 0.24

137 marksHigherJanuary 2022 · Paper 1HR (4MA1/1HR) · Q13

Hector has a bag that contains 12 counters: 7 green and 5 red. Hector takes at random a counter from the bag, looks at it and puts it back. He then takes at random a second counter, looks at it and puts it back.

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

(b) Work out the probability that both counters are red. (2)

Meghan has a jar containing 15 counters. There are only blue, green and red counters in the jar. Hector is going to take two counters from his bag as before (replacing each one). Meghan is then going to take at random one of the counters from her jar. The probability that the 3 counters each have a different colour is 7⁄24.

(c) Work out how many blue counters there are in the jar. (3)

Show solution

Part (a)

  1. The counter is replaced, so the probabilities stay the same: green 7⁄12, red 5⁄12 on every pair of branches.

Answer: Red first: 5⁄12; after green, the second counter is green 7⁄12 or red 5⁄12

Part (b)

  1. 5⁄12 × 5⁄12

Answer: 25⁄144

Part (c)

  1. Hector’s bag has only green and red. For three different colours, Hector must get one green and one red (in either order) and Meghan must get blue.
  2. P(Hector gets one of each) = 7⁄12 × 5⁄12 + 5⁄12 × 7⁄12 = 70⁄144
  3. Let b be the number of blue counters: 70⁄144 × b⁄15 = 7⁄24
  4. b⁄15 = 7⁄24 × 144⁄70 = 3⁄5

Answer: b = 9

143 marksHigherJune 2023 · Paper 1HR (4MA1/1HR) · Q13

Novak is going to play two games of tennis. The probability that he wins the first game is 0.7. If he wins the first game, the probability that he wins the second game is 0.9. If he loses the first game, the probability that he wins the second game is 0.4.

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

(b) Work out the probability that Novak wins both games of tennis. (2)

Show solution

Part (a)

  1. Each pair of branches adds to 1. After a win: P(lose second) = 1 − 0.9 = 0.1. After a loss: P(lose second) = 1 − 0.4 = 0.6.
winsloseswinsloseswinsloses0.70.30.90.10.40.6First gameSecond game

Answer: 0.1 and 0.6

Part (b)

  1. Win both = win the first and win the second, so multiply along the branches.
  2. 0.7 × 0.9

Answer: 0.63

155 marksHigherJune 2024 · Paper 1H (4MA1/1H) · Q13

Harman has two bags of beads. In bag A, there are 3 white beads and 7 black beads. In bag B, there are 5 white beads and 4 black beads. Harman takes at random a bead from bag A and a bead from bag B.

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

(b) Work out the probability that Harman takes two beads of the same colour. (3)

whiteblackwhiteblackwhiteblackBag ABag B
Show solution

Part (a)

  1. Bag A has 3 + 7 = 10 beads, so P(white) = 3⁄10 and P(black) = 7⁄10.
  2. Bag B has 5 + 4 = 9 beads, so P(white) = 5⁄9 and P(black) = 4⁄9. Bag B’s probabilities are the same on both pairs of branches, because the bead from bag A does not change bag B.
whiteblackwhiteblackwhiteblack3/107/105/94/95/94/9

Answer: Bag A: 3⁄10, 7⁄10. Bag B: 5⁄9, 4⁄9 on both pairs

Part (b)

  1. “Same colour” means white then white, or black then black.
  2. Multiply along the branches: P(white, white) = 3⁄10 × 5⁄9 = 15⁄90
  3. P(black, black) = 7⁄10 × 4⁄9 = 28⁄90
  4. The two outcomes cannot both happen, so add them: 15⁄90 + 28⁄90 = 43⁄90

Answer: 43⁄90

165 marksHigherJanuary 2021 · Paper 1H (4MA1/1H) · Q14

Ding is going to play one game of snooker against each of two of his friends, Marco and Judd. The probability tree diagram gives the probabilities: against Marco, Ding wins with probability 0.6 and loses with probability 0.4. If Ding wins against Marco, he wins against Judd with probability 0.9 (loses 0.1). If Ding loses against Marco, he wins against Judd with probability 0.25 (loses 0.75).

(a) Work out the probability that Ding will win both games. (2)

(b) Work out the probability that Ding will win exactly one of the games. (3)

Show solution

Part (a)

  1. Multiply along win, win: 0.6 × 0.9

Answer: 0.54

Part (b)

  1. Exactly one win: win then lose, or lose then win.
  2. P(W, L) = 0.6 × 0.1 = 0.06 and P(L, W) = 0.4 × 0.25 = 0.1
  3. Add: 0.06 + 0.1

Answer: 0.16

174 marksHigherJanuary 2022 · Paper 1H (4MA1/1H) · Q14

Aika has 2 packets of seeds, packet A and packet B. There are 12 seeds in packet A and 7 of these are sunflower seeds. There are 15 seeds in packet B and 8 of these are sunflower seeds. Aika is going to take at random a seed from packet A and a seed from packet B.

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

(b) Calculate the probability that Aika will take two sunflower seeds. (2)

Show solution

Part (a)

  1. Packet A: sunflower 7⁄12, not sunflower 5⁄12. Packet B: sunflower 8⁄15, not sunflower 7⁄15 (the same on both pairs of branches).

Answer: A: 7⁄12, 5⁄12. B: 8⁄15, 7⁄15 on both pairs

Part (b)

  1. Multiply along sunflower, sunflower: 7⁄12 × 8⁄15 = 56⁄180

Answer: 14⁄45

184 marksHigherJune 2025 · Paper 1H (4MA1/1H) · Q14

Ricardo is going to play one game of badminton and one game of tennis. He can either win or lose each game. The probability that he will win the game of badminton is 0.7. The probability that he will win the game of tennis is 0.4.

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

(b) Work out the probability that Ricardo loses both games. (2)

Show solution

Part (a)

  1. Win and lose are the only outcomes, so each pair of branches adds to 1: P(lose badminton) = 1 − 0.7 = 0.3 and P(lose tennis) = 1 − 0.4 = 0.6.
winlosewinlosewinlose0.70.30.40.60.40.6badmintontennis

Answer: Badminton: 0.7, 0.3. Tennis: 0.4, 0.6 on both pairs

Part (b)

  1. Lose both means lose badminton and lose tennis, so multiply along the branches.
  2. 0.3 × 0.6

Answer: 0.18

195 marksHigherJune 2020 · Paper 1H (4MA1/1H) · Q15

Harry has two fair 5-sided spinners. Spinner A has 2 red and 3 green sections. Spinner B has 4 red and 1 green section. Harry is going to spin each spinner once.

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

(b) Work out the probability that at least one of the spinners will land on green. (3)

Show solution

Part (a)

  1. Spinner A: P(red) = 2⁄5, P(green) = 3⁄5. Spinner B: P(red) = 4⁄5, P(green) = 1⁄5 (the same on both pairs of branches).

Answer: A: 2⁄5, 3⁄5. B: 4⁄5, 1⁄5 on both pairs

Part (b)

  1. “At least one green” is everything except “both red”.
  2. P(both red) = 2⁄5 × 4⁄5 = 8⁄25
  3. P(at least one green) = 1 − 8⁄25

Answer: 17⁄25

208 marksHigherNovember 2021 · Paper 1H (4MA1/1H) · Q15

Magnus and Garry play 2 games of chess against each other. The probability that Magnus beats Garry in any game is 2⁄9. The probability that any game between Magnus and Garry is drawn is 4⁄9. The result of any game is independent of the result of any other game.

(a) Complete the probability tree diagram (each game: Magnus wins, game drawn, Magnus loses). (2)

For each game of chess, the winner gets 2 points and the loser gets 0 points; when the game is drawn, each player gets 1 point.

(b) Work out the probability that, after 2 games, Magnus and Garry have the same number of points. (3)

Magnus and Garry now play a third game of chess.

(c) Work out the probability that, after 3 games, Magnus and Garry have the same number of points. (3)

Show solution

Part (a)

  1. P(Magnus loses) = 1 − 2⁄9 − 4⁄9 = 3⁄9
  2. The games are independent, so every set of second-game branches is the same: 2⁄9, 4⁄9, 3⁄9.

Answer: First game: 2⁄9, 4⁄9, 3⁄9; every second-game set: 2⁄9, 4⁄9, 3⁄9

Part (b)

  1. They have equal points if both games are drawn, or Magnus wins one and loses the other (in either order).
  2. P(D, D) = 4⁄9 × 4⁄9 = 16⁄81
  3. P(W, L) + P(L, W) = 2⁄9 × 3⁄9 + 3⁄9 × 2⁄9 = 12⁄81
  4. Total = 16⁄81 + 12⁄81

Answer: 28⁄81

Part (c)

  1. After 3 games the points are equal if all three are draws, or there is one win, one draw and one loss (in any order).
  2. P(D, D, D) = (4⁄9)3 = 64⁄729
  3. One W, one D, one L in a particular order: 2⁄9 × 4⁄9 × 3⁄9 = 24⁄729. There are 3 × 2 × 1 = 6 orders: 6 × 24⁄729 = 144⁄729
  4. Total = 64⁄729 + 144⁄729

Answer: 208⁄729

214 marksHigherJune 2022 · Paper 1H (4MA1/1H) · Q15

Abraham is going to play a computer game. He can win, draw or lose each game. For any game, the probability that he wins is 0.3, draws is 0.5 and loses is 0.2. When he wins a game, he scores +10 points. When he draws a game, he scores 0 points. When he loses a game, he scores −5 points. Abraham plays 3 games and the points are added to get his total score. Work out the probability that when he has played 3 games his total score is 0 points.

Show solution
  1. Find every combination of 3 results that totals 0. Three draws: 0 + 0 + 0 = 0. One win and two losses: 10 − 5 − 5 = 0. No other combination works.
  2. P(D, D, D) = 0.5 × 0.5 × 0.5 = 0.125
  3. P(one win, two losses) in a particular order = 0.3 × 0.2 × 0.2 = 0.012. The win can be the 1st, 2nd or 3rd game, so there are 3 orders: 3 × 0.012 = 0.036
  4. Total = 0.125 + 0.036

Answer: 0.161

227 marksHigherJanuary 2023 · Paper 1H (4MA1/1H) · Q15

Here are 9 cards. Each card has either a number on it or a letter on it: A, 7, H, F, E, D, G, 5, C. Tomas takes at random one of the cards and keeps it. Tomas then takes at random another card and keeps it.

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

(b) Work out the probability that each of the two cards has a number on it. (2)

(c) Work out the probability that there will be one card with a number on it and one card with a letter on it. (3)

Show solution

Part (a)

  1. There are 2 number cards (7 and 5) and 7 letter cards. The first card is kept, so there are 8 cards left for the second pick.
  2. First card: number 2⁄9, letter 7⁄9
  3. After a number: number 1⁄8, letter 7⁄8. After a letter: number 2⁄8, letter 6⁄8.

Answer: 2⁄9, 7⁄9; then 1⁄8, 7⁄8 or 2⁄8, 6⁄8

Part (b)

  1. 2⁄9 × 1⁄8 = 2⁄72

Answer: 1⁄36

Part (c)

  1. Number then letter, or letter then number: 2⁄9 × 7⁄8 + 7⁄9 × 2⁄8 = 14⁄72 + 14⁄72 = 28⁄72

Answer: 7⁄18

234 marksHigherJanuary 2020 · Paper 1HR (4MA1/1HR) · Q16

Steffi is going to play one game of tennis and one game of chess. The probability that she will win the game of tennis is 0.6. The probability that she will win both games is 0.42. Work out the probability that she will not win either game.

Show solution
  1. The games are independent, so P(win both) = P(win tennis) × P(win chess): 0.6 × P(win chess) = 0.42
  2. P(win chess) = 0.42 ÷ 0.6 = 0.7
  3. P(not win tennis) = 0.4 and P(not win chess) = 0.3
  4. P(win neither) = 0.4 × 0.3

Answer: 0.12

244 marksHigherJune 2024 · Paper 1H (4MA1/1H) · Q16

There are 20 sweets in a box. 15 of the sweets are red and 5 of the sweets are yellow. Fred takes at random 3 sweets from the box. Work out the probability that Fred takes at least one sweet of each colour from the box.

Show solution
  1. “At least one of each colour” fails only if all 3 sweets are the same colour. So find P(all red) and P(all yellow) and subtract from 1.
  2. The sweets are not replaced, so the numbers go down by 1 each time.
  3. P(RRR) = 15⁄20 × 14⁄19 × 13⁄18 = 2730⁄6840
  4. P(YYY) = 5⁄20 × 4⁄19 × 3⁄18 = 60⁄6840
  5. P(at least one of each) = 1 − 2730⁄6840 − 60⁄6840 = 4050⁄6840
  6. Simplify by dividing top and bottom by 90.

Answer: 45⁄76

254 marksHigherJanuary 2020 · Paper 1H (4MA1/1H) · Q18

There are 16 sweets in a bowl: 4 blackcurrant, 5 lemon and 7 orange. Anna, Ravi and Sam each take at random one sweet from the bowl. Work out the probability that the 5 lemon sweets are still in the bowl.

Show solution
  1. All 5 lemons still in the bowl means none of the 3 sweets taken is lemon. There are 11 sweets that are not lemon.
  2. The sweets are not replaced: P = 11⁄16 × 10⁄15 × 9⁄14 = 990⁄3360

Answer: 33⁄112

263 marksHigherJune 2025 · Paper 1HR (4MA1/1HR) · Q19

Kannika has 9 counters with numbers 1, 1, 2, 3, 3, 3, 5, 6, 6. She puts them in a bag, takes a counter at random and does not replace it, then takes a second counter at random. Work out the probability that the sum of the numbers on the two counters is less than 5.

Show solution
  1. List the pairs with a sum less than 5: 1 + 1 = 2, 1 + 2 = 3, 1 + 3 = 4. (2 + 2 is impossible because there is only one 2.)
  2. The counter is not replaced, so the second fraction has 8 on the bottom.
  3. P(1, 1) = 2⁄9 × 1⁄8 = 2⁄72
  4. P(1 and 2 in either order) = 2⁄9 × 1⁄8 + 1⁄9 × 2⁄8 = 4⁄72
  5. P(1 and 3 in either order) = 2⁄9 × 3⁄8 + 3⁄9 × 2⁄8 = 12⁄72
  6. Total = 2 + 4 + 12⁄72 = 18⁄72

Answer: 1⁄4

275 marksHigherJune 2024 · Paper 1HR (4MA1/1HR) · Q21

There are 25 counters in a bag such that 6 counters are blue, x counters are orange, where x > 9, and the rest of the counters are pink. Maalam takes at random two of the counters from the bag. The probability that Maalam takes one orange counter and one pink counter is 22⁄75. Calculate the probability that Maalam takes 2 pink counters from the bag. Show clear algebraic working.

Show solution
  1. Number of pink counters = 25 − 6 − x = 19 − x
  2. One orange and one pink can happen as orange then pink, or pink then orange. The counters are not replaced, so the second fraction has 24 on the bottom.
  3. P(O, P) + P(P, O) = x⁄25 × 19 − x⁄24 + 19 − x⁄25 × x⁄24 = 2x(19 − x)⁄600
  4. Set equal to 22⁄75: 2x(19 − x)⁄600 = 22⁄75, so 2x(19 − x) = 176 and x(19 − x) = 88
  5. 19x − x2 = 88, so x2 − 19x + 88 = 0, which factorises as (x − 8)(x − 11) = 0
  6. x = 8 or x = 11. We are told x > 9, so x = 11 and there are 19 − 11 = 8 pink counters.
  7. P(pink, pink) = 8⁄25 × 7⁄24 = 56⁄600

Answer: 7⁄75

283 marksHigherJune 2025 · Paper 1H (4MA1/1H) · Q21

A box contains 20 counters: 9 are red, 7 are yellow and 4 are green. Alex takes at random three counters from the box. Work out the probability that exactly two of the three counters are the same colour.

Show solution
  1. “Exactly two the same” means NOT all three the same and NOT all three different. It is quicker to work those out and subtract from 1.
  2. P(all red) = 9⁄20 × 8⁄19 × 7⁄18 = 504⁄6840, P(all yellow) = 7⁄20 × 6⁄19 × 5⁄18 = 210⁄6840, P(all green) = 4⁄20 × 3⁄19 × 2⁄18 = 24⁄6840
  3. P(all the same) = 504 + 210 + 24⁄6840 = 738⁄6840
  4. P(all different): one order is 9⁄20 × 7⁄19 × 4⁄18 = 252⁄6840. There are 3 × 2 × 1 = 6 orders of the three colours, so P(all different) = 6 × 252⁄6840 = 1512⁄6840
  5. P(exactly two the same) = 1 − 738⁄6840 − 1512⁄6840 = 4590⁄6840

Answer: 51⁄76

294 marksHigherJune 2020 · Paper 1HR (4MA1/1HR) · Q23

In a bag, there are only 3 blue beads, 4 white beads and x orange beads. Jean is going to take at random two beads from the bag. The probability that Jean will take two beads of the same colour is 3⁄8. Find the total number of beads in the bag. Show clear algebraic working.

Show solution
  1. Total beads = x + 7. The beads are not replaced.
  2. P(same colour) = P(BB) + P(WW) + P(OO) = 3 × 2 + 4 × 3 + x(x − 1)⁄(x + 7)(x + 6) = x2 − x + 18⁄(x + 7)(x + 6)
  3. Set equal to 3⁄8: 8(x2 − x + 18) = 3(x2 + 13x + 42)
  4. 8x2 − 8x + 144 = 3x2 + 39x + 126, so 5x2 − 47x + 18 = 0
  5. Factorise: (5x − 2)(x − 9) = 0. x must be a whole number, so x = 9.
  6. Total = 9 + 7

Answer: 16

305 marksHigherJune 2022 · Paper 1HR (4MA1/1HR) · Q24

Elliot has x counters. Each counter has one red face and one green face. Elliot spreads all the counters out on a table and sees that the number of counters showing a red face is 5. Elliot then picks at random one of the counters and turns the counter over. He then picks at random a second counter and turns the counter over. The probability that there are still 5 counters showing a red face is 19⁄32. Work out the value of x. Show clear algebraic working.

Show solution
  1. The second counter is picked from all x counters, so it could be the same counter again.
  2. There are still 5 red faces if one counter goes red → green and a different counter goes green → red, or if the same counter is turned twice.
  3. Red first, then one of the x − 5 counters that started green: 5⁄x × x − 5⁄x
  4. Green first, then one of the 5 counters that started red: x − 5⁄x × 5⁄x
  5. The same counter twice: x⁄x × 1⁄x = x⁄x2
  6. Total: 10(x − 5) + x⁄x2 = 11x − 50⁄x2 = 19⁄32
  7. Cross-multiply: 32(11x − 50) = 19x2, so 19x2 − 352x + 1600 = 0
  8. Quadratic formula: x = 352 ± √3522 − 4 × 19 × 1600⁄38 = 352 ± √2304⁄38 = 352 ± 48⁄38
  9. x = 8 or x = 10.52…. The number of counters must be a whole number.

Answer: x = 8

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