Every probability of combined events question from 10 Cambridge IGCSE Mathematics (0580) past papers (May/June 2025 to May/June 2026), syllabus section 8.3. 11 questions covering Core and Extended tiers, ordered from Core to Extended. Each question is labelled with its question paper code (for example 0580/12/M/J/26) and every final answer matches the official Cambridge mark scheme. Try each question first, then click “Show solution” to check your working. This page is part of our IGCSE Maths 0580 Probability past paper questions.
0580 Probability of Combined Events past paper questions: Core (Papers 1 and 3)
May/June 2026 · Paper 1 · 0580/12/M/J/26 · Q22
A bag contains 7 pink blocks and 5 green blocks. Kay picks a block at random, notes its colour and puts it back. She does this a second time.
(a) Complete the tree diagram. (b) Use the tree diagram to work out the probability that the two blocks are green.
Show solution
- There are 7 + 5 = 12 blocks. P(pink) = 7⁄12 and P(green) = 5⁄12.
- The block is put back, so the second pick has the same probabilities.
- Every pair of branches is pink 7⁄12 and green 5⁄12
- Follow green then green and multiply along the branches.
- P(green, green) = 5⁄12 × 5⁄12 = 25⁄144
May/June 2026 · Paper 3 · 0580/33/M/J/26 · Q20
A fair spinner has six equal sections numbered 1 to 6. Azra spins the spinner twice.
(a) Complete the tree diagram, with branches “lands on 5” and “does not land on 5” for each spin.
(b) Find the probability the spinner lands on 5 both times.
Show solution
- One of the 6 equal sections is a 5, so P(lands on 5) = 1⁄6.
- The two branches from each point add up to 1: P(does not land on 5) = 1 − 1⁄6 = 5⁄6
- Write 1⁄6 and 5⁄6 on the branches for both spins.
- Follow the “lands on 5” branch both times and multiply along the branches.
- P(5 then 5) = 1⁄6 × 1⁄6 = 1⁄36
Oct/Nov 2025 · Paper 1 · 0580/13/O/N/25 · Q26
There are 2 sets of traffic lights on Li’s journey to work. The probability he stops at the first set is 0.4. The probability he stops at the second set is 0.7.
(a) Complete the tree diagram. (b) Work out the probability that Li does not stop at both sets of traffic lights.
Show solution
- On each pair of branches the two probabilities add up to 1.
- First set: stops 0.4, so does not stop = 1 − 0.4 = 0.6
- Second set (on both pairs of branches): stops 0.7, so does not stop = 1 − 0.7 = 0.3
- “Does not stop at both sets” means he does not stop at the first set and does not stop at the second set.
- Multiply along those branches: 0.6 × 0.3 = 0.18
May/June 2025 · Paper 1 · 0580/11/M/J/25 · Q19
A box contains 10 counters. The counters are either red or green. The ratio red counters : green counters = 1 : 4. Shareen picks a counter at random, notes its colour and puts it back in the box. She then picks a second counter at random.
(a) Complete the tree diagram.
(b) Find the probability that both counters are green.
Show solution
- The ratio 1 : 4 has 1 + 4 = 5 parts. 10 counters ÷ 5 = 2 counters per part.
- Red = 1 × 2 = 2 counters and green = 4 × 2 = 8 counters.
- P(red) = 2⁄10 = 1⁄5 and P(green) = 8⁄10 = 4⁄5.
- The counter is put back, so the second pick has the same probabilities as the first.
- Every red branch is 1⁄5 and every green branch is 4⁄5.
- Follow the green branch, then the green branch again, and multiply along the path.
- P(green, green) = 4⁄5 × 4⁄5 = 16⁄25
0580 Probability of Combined Events past paper questions: Extended (Papers 2 and 4)
May/June 2026 · Paper 2 · 0580/22/M/J/26 · Q18
Marwa has 10 numbered tiles: 1, 2, 2, 3, 3, 4, 4, 4, 4, 4.
(a) Marwa picks one tile at random, replaces it, then picks a second tile at random. Find the probability that the two tiles have the same even number.
(b) From the 10 tiles, Marwa now picks three tiles at random without replacement. Find the probability that the sum of the numbers on the three tiles is 11.
Show solution
- “The same even number” means both 2 or both 4.
- P(2) = 2⁄10 because there are two 2s. P(4) = 5⁄10 because there are five 4s.
- The tile is replaced, so the probabilities stay the same for the second pick.
- P(2, 2) = 2⁄10 × 2⁄10 = 4⁄100. P(4, 4) = 5⁄10 × 5⁄10 = 25⁄100.
- Add: 4⁄100 + 25⁄100 = 29⁄100
- Find which three tiles add up to 11: the only way is 3 + 4 + 4.
- Without replacement, one order is 3 then 4 then 4: 2⁄10 × 5⁄9 × 4⁄8 = 40⁄720.
- The 3 could be picked first, second or third, and each order has the same probability.
- Total = 3 × 40⁄720 = 120⁄720 = 1⁄6
May/June 2026 · Paper 2 · 0580/23/M/J/26 · Q27
Juan has a bag containing 2 red counters, 3 green counters and 5 blue counters. Juan picks two counters from the bag at random, without replacement. Find the probability that the two counters are different colours.
Show solution
- It is easier to find P(same colour) and subtract from 1.
- There are 10 counters. Without replacement, there are 9 left for the second pick.
- P(red, red) = 2⁄10 × 1⁄9 = 2⁄90. P(green, green) = 3⁄10 × 2⁄9 = 6⁄90. P(blue, blue) = 5⁄10 × 4⁄9 = 20⁄90.
- P(same) = 2 + 6 + 20⁄90 = 28⁄90.
- P(different) = 1 − 28⁄90 = 62⁄90 = 31⁄45
Feb/Mar 2026 · Paper 4 · 0580/42/F/M/26 · Q18(c)
Of 90 delayed trains, 36 were delayed for more than 30 and up to 45 minutes, and 6 for more than 45 minutes. Two of the trains with delays of more than 30 minutes are chosen at random. Find the probability that just one of these trains is delayed for 45 minutes or less.
Show solution
- There are 36 + 6 = 42 trains delayed by more than 30 minutes. 36 of them are delayed 45 minutes or less.
- Two trains are chosen, so the second choice is from 41 trains (without replacement).
- Just one ≤ 45 minutes can happen in two orders.
- First ≤ 45, second > 45: 36⁄42 × 6⁄41 = 216⁄1722.
- First > 45, second ≤ 45: 6⁄42 × 36⁄41 = 216⁄1722.
- Add: 432⁄1722. Divide the top and bottom by 6: 72⁄287
Oct/Nov 2025 · Paper 2 · 0580/21/O/N/25 · Q20
Bag A contains 5 white balls and 3 black balls. Bag B contains 3 white balls and 1 black ball.
(a) Two balls are picked at random from bag B without replacement. Find the probability that both balls are black.
(b) The balls are replaced into bag B. Kyle picks a ball at random from each bag. (i) Complete the tree diagram. (ii) Find the probability that the two balls are the same colour.
(c) The balls are replaced into their bags. Jo picks a ball at random from bag A and places it into bag B. She then picks a ball at random from bag B. Find the probability that she picks a black ball from bag B.
Show solution
- Bag B has only 1 black ball. Once it is taken, there are no black balls left for the second pick.
- P(both black) = 0
- Bag A has 5 + 3 = 8 balls: white 5⁄8 and black 3⁄8
- Bag B has 3 + 1 = 4 balls. The two bags are separate, so bag B’s probabilities are the same on both pairs of branches: white 3⁄4 and black 1⁄4
- “Same colour” means both white or both black.
- P(white, white) = 5⁄8 × 3⁄4 = 15⁄32. P(black, black) = 3⁄8 × 1⁄4 = 3⁄32.
- Add: 15⁄32 + 3⁄32 = 18⁄32 = 9⁄16
- Bag B now has 5 balls. What is in it depends on the ball Jo moved.
- If she moved a white ball (probability 5⁄8), bag B has 4 white and 1 black, so P(black) = 1⁄5.
- If she moved a black ball (probability 3⁄8), bag B has 3 white and 2 black, so P(black) = 2⁄5.
- P(black) = 5⁄8 × 1⁄5 + 3⁄8 × 2⁄5 = 5⁄40 + 6⁄40 = 11⁄40
Oct/Nov 2025 · Paper 2 · 0580/23/O/N/25 · Q11
Sarah rolls a fair 6-sided dice twice. Find the probability she rolls a number greater than 4 both times.
Show solution
- Greater than 4 means a 5 or a 6: 2 outcomes out of 6.
- P(greater than 4) = 2⁄6 = 1⁄3 for each roll.
- The two rolls are independent, so multiply: 1⁄3 × 1⁄3 = 1⁄9
Oct/Nov 2025 · Paper 2 · 0580/23/O/N/25 · Q25
Mahir picks one number at random from 5, 10 and 15. He then picks one number at random from 4, 5 and 6. He adds the two numbers.
(a) Complete the sample space diagram. (b) Given that the total of the two numbers is odd, find the probability that one of the numbers added is 15.
Show solution
- Add each first number to each second number.
- 5 + 4, 5 + 5, 5 + 6: 9, 10, 11
- 10 + 4, 10 + 5, 10 + 6: 14, 15, 16. 15 + 4, 15 + 5, 15 + 6: 19, 20, 21.
- “Given that the total is odd” means only look at the odd totals: 9, 11, 15, 19, 21. That is 5 outcomes.
- Of these, the totals that used the number 15 are 19 (15 + 4) and 21 (15 + 6): 2 outcomes.
- P = 2⁄5
May/June 2025 · Paper 2 · 0580/21/M/J/25 · Q17
(a) A bag contains 6 red marbles, 3 green marbles and 1 blue marble. Two marbles are picked at random from the bag with replacement. Find the probability that both marbles are green.
(b) Another bag contains 4 red counters and 2 yellow counters. Two counters are picked at random from this bag without replacement.
(i) Complete the tree diagram. The first branches are Red 4⁄6 and Yellow 2⁄6; the second branch Red then Red is 3⁄5.
(ii) Find the probability that one of the two counters is yellow.
Show solution
- There are 6 + 3 + 1 = 10 marbles, so P(green) = 3⁄10.
- With replacement, the probability is the same the second time.
- P(green, green) = 3⁄10 × 3⁄10 = 9⁄100
- Without replacement, there are only 5 counters left for the second pick.
- After a red: 3 red and 2 yellow are left, so Red → Yellow = 2⁄5.
- After a yellow: 4 red and 1 yellow are left, so Yellow → Red = 4⁄5 and Yellow → Yellow = 1⁄5.
- Red → Yellow 2⁄5; Yellow → Red 4⁄5; Yellow → Yellow 1⁄5
- “One yellow” can happen two ways: red then yellow, or yellow then red.
- P(R, Y) = 4⁄6 × 2⁄5 = 8⁄30. P(Y, R) = 2⁄6 × 4⁄5 = 8⁄30.
- Add: 8⁄30 + 8⁄30 = 16⁄30 = 8⁄15
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