IGCSE Maths · Probability · Core & Extended

Probability of Combined Events

This closes out the Probability unit — covering what happens when two or more events combine. Sample space diagrams, Venn diagrams, the AND/OR rules, and tree diagrams with replacement are shared by both tiers. Tree diagrams without replacement and conditional probability are Extended-only, tagged throughout.

1. Sample Space Diagrams

A sample space diagram lists every possible outcome of a combined event — often as a grid.

Worked example
+123456
1234567
2345678
3456789
45678910
567891011
6789101112
Two fair dice are rolled and their scores added. Find P(the total is 7).
Counting the 7s in the table: there are 6 out of 36 equally likely outcomes.
P(total = 7) = 6/36 = 1/6
Try it — Sample Space Diagrams (using the table above)
  1. Find P(the total is 12).
  2. Find P(the total is less than 5).
  3. Find P(the total is an even number).
Show answers
  1. Only one outcome (6+6): 1/36
  2. Totals of 2,3,4: (1,1),(1,2),(2,1),(1,3),(2,2),(3,1) = 6 outcomes: 6/36 = 1/6
  3. Half the totals are even: 18/36 = 1/2

2. The AND / OR Rules

Mutually exclusive events: P(A or B) = P(A) + P(B)
Independent events: P(A and B) = P(A) × P(B)

Mutually exclusive means A and B can’t both happen at once (like rolling a 2 and a 5 on one die). Independent means one event doesn’t affect the other’s probability (like a coin flip and a separate die roll).

Worked examples Find P(rolling a 2 or a 5) on a fair die.
These outcomes are mutually exclusive (can’t roll both): P(2 or 5) = 1/6 + 1/6 = 1/3

Find P(flipping heads on a coin AND rolling a 6 on a die).
These are independent: P(heads and 6) = 1/2 × 1/6 = 1/12
Try it — AND / OR Rules
  1. Find P(rolling a 1, 2, or 3) on a fair die, using the OR rule.
  2. A coin is flipped and a die is rolled. Find P(tails and an even number).
Show answers
  1. 1/6+1/6+1/6 = 3/6 = 1/2
  2. 1/2 × 1/2 = 1/4

3. Venn Diagrams for Probability

Worked example F B 10 8 6 6 In a class of 30 students, 18 play football (F), 14 play basketball (B), and 8 play both. Find P(a student plays neither sport).
F only: 18−8 = 10.   B only: 14−8 = 6.   Both: 8.   Total accounted for: 10+6+8 = 24
Neither: 30 − 24 = 6 students
P(neither) = 6/30 = 1/5
Worked example — the OR formula from a Venn diagram Using the same data, find P(a student plays football or basketball).
Method 1 (counting): (10+6+8)/30 = 24/30 = 4/5
Method 2 (formula): P(F) + P(B) − P(F and B) = 18/30 + 14/30 − 8/30 = 24/30 = 4/5 — same answer, confirming the general OR rule for events that aren’t mutually exclusive.
Try it — Venn Diagrams (using the same class of 30)
  1. Find P(a student plays football only).
  2. Find P(a student plays exactly one of the two sports).
Show answers
  1. 10/30 = 1/3
  2. (10+6)/30 = 16/30 = 8/15

4. Tree Diagrams — With Replacement

A bag has 3 red and 2 blue counters (5 total). Two counters are drawn with replacement (the first is put back before the second is drawn) — so the probabilities stay the same for both draws.

Worked example 3/5 2/5 Red Blue R (3/5) B (2/5) R (3/5) B (2/5) Find P(both counters are red), and P(exactly one is red).
P(both red) = 3/5 × 3/5 = 9/25
P(exactly one red) = P(R then B) + P(B then R) = (3/5×2/5) + (2/5×3/5) = 6/25+6/25 = 12/25
Try it — Tree Diagrams With Replacement
  1. Using the same bag (3 red, 2 blue, with replacement), find P(both counters are blue).
  2. Find P(at least one counter is red) (hint: use the complement — 1 minus P(no reds)).
Show answers
  1. 2/5×2/5 = 4/25
  2. P(no reds) = P(both blue) = 4/25. P(at least one red) = 1 − 4/25 = 21/25

5. Tree Diagrams — Without Replacement Extended only

Without replacement, the first counter is not put back — so the total (and the count of whichever colour was drawn) changes for the second draw. These are dependent events.

Worked example Using the same bag (3 red, 2 blue), two counters are drawn without replacement. Find P(both red), and P(exactly one red).
P(both red) = 3/5 × 2/4 = 6/20 = 3/10 (after removing 1 red, 2 red remain out of 4 total)
P(red then blue) = 3/5 × 2/4 = 3/10    P(blue then red) = 2/5 × 3/4 = 3/10
P(exactly one red) = 3/10 + 3/10 = 3/5
Try it — Tree Diagrams Without Replacement Extended
  1. Using the same bag (3 red, 2 blue, without replacement), find P(both counters are blue).
  2. A bag has 4 green and 6 yellow balls. Two are drawn without replacement. Find P(both green).
Show answers
  1. 2/5×1/4 = 2/20 = 1/10
  2. 4/10×3/9 = 12/90 = 2/15

6. Conditional Probability Extended only

Conditional probability asks: given that one thing is already known to be true, what’s the probability of something else? Written P(A | B) — “the probability of A, given B.”

P(A | B) = P(A and B) ÷ P(B)
Worked example — using the Venn diagram from Section 3 Using the class of 30 students (18 play football, 14 play basketball, 8 play both), find P(a student plays basketball, given that they play football).
P(B | F) = P(both) / P(F) = (8/30) ÷ (18/30) = 8/18 = 4/9
This makes sense: out of the 18 footballers, 8 of them also play basketball — 8/18 simplifies to 4/9.
Try it — Conditional Probability Extended
  1. Using the same Venn diagram, find P(a student plays football, given that they play basketball).
  2. A bag has 5 red and 3 blue balls. Two are drawn without replacement. Given that the first ball drawn is red, find the probability the second is also red.
Show answers
  1. P(F|B) = (8/30)÷(14/30) = 8/14 = 4/7
  2. After removing 1 red, 4 red remain out of 7 total: 4/7
Common mistakes to watch for
  • Using the simple addition (OR) rule for events that aren’t mutually exclusive — subtract the overlap: P(A or B) = P(A)+P(B)−P(A and B).
  • Forgetting to adjust probabilities for the second draw in a “without replacement” tree diagram — both the count of the relevant colour and the total change.
  • Adding probabilities along a single branch of a tree diagram instead of multiplying — each branch’s probabilities multiply together; only add when combining separate branches (different paths leading to the same outcome).
  • Mixing up P(A|B) with P(B|A) — these are generally different values unless A and B happen to be symmetric in the problem.
  • Forgetting to double-check that all outcomes in a sample space diagram or tree diagram sum to 1 — a good sanity check that nothing’s been miscounted.

7. Quick-Fire Challenge Round

True or False?
  1. For mutually exclusive events, P(A or B) = P(A) + P(B).
  2. For independent events, P(A and B) = P(A) + P(B).
  3. In a tree diagram, you multiply probabilities along a single branch/path.
  4. Without replacement, the probabilities for the second draw stay exactly the same as the first.
Show answers
  1. True.
  2. False — for independent events, P(A and B) = P(A) × P(B), not addition.
  3. True.
  4. False — without replacement, the counts (and total) change for the second draw, since the first item isn’t put back. Ext.
Spot the Error
  1. “P(rolling a 6) or (flipping heads) = 1/6 + 1/2, since I just add probabilities for ‘or’.”
  2. “Bag has 4 red, 6 blue (10 total). Without replacement, P(both red) = 4/10 × 4/10 = 16/100.”
  3. “P(A and B) for independent events A, B with P(A)=0.3, P(B)=0.5: P(A and B) = 0.3+0.5 = 0.8.”
Show answers
  1. This isn’t necessarily wrong on its own since these two events are independent (from different experiments) — but calling it an “or” rule shortcut is misleading; if you want P(6 on die AND heads), that’s the AND rule (multiply): 1/6×1/2=1/12, not an “or” addition. As written (“or”), for two events from different experiments happening together, this needs more care about what’s actually being asked.
  2. Without replacement, the second draw uses updated numbers: 4/10 × 3/9 = 12/90 = 2/15, not 4/10×4/10.
  3. Independent events use multiplication, not addition: P(A and B) = 0.3×0.5 = 0.15, not 0.8.

8. Mixed Practice — Bringing It All Together

Core & Extended
  1. Two fair dice are rolled. Find P(the total is 9), using a sample space diagram.
  2. Find P(rolling a 3 or a 4) on a fair die.
  3. A coin is flipped and a die is rolled. Find P(heads and an odd number).
  4. A bag has 4 red, 6 blue (10 total). Two are drawn with replacement. Find P(both red).
Show answers
  1. Totals of 9: (3,6),(4,5),(5,4),(6,3) = 4/36 = 1/9
  2. 1/6+1/6 = 1/3
  3. 1/2×1/2 = 1/4
  4. 4/10×4/10 = 16/100 = 4/25
Extended-level mixed set Extended only
  1. Using the same bag (4 red, 6 blue), find P(both red) without replacement.
  2. In a group of 40 people, 25 like tea, 20 like coffee, 10 like both. Find P(a person likes coffee, given they like tea).
Show answers
  1. 4/10×3/9 = 12/90 = 2/15
  2. P(C|T) = (10/40)÷(25/40) = 10/25 = 2/5

Practice at Home

A large set again, to close out the Probability unit — take it steadily.

Part A — Sample space diagrams
  1. Two fair dice are rolled. Find P(the total is 8).
  2. Find P(the total is a multiple of 3).
Part B — AND / OR rules
  1. Find P(rolling a 1, 3, or 5) on a fair die.
  2. A coin is flipped twice. Find P(two heads).
  3. A spinner (4 equal sections: red, blue, green, yellow) is spun, and a coin is flipped. Find P(spinner lands red and coin shows tails).
Part C — Venn diagrams
  1. In a group of 25 people, 15 like pizza, 12 like pasta, 6 like both. Find P(likes neither).
  2. Using the same data, find P(likes pizza or pasta).
  3. Find P(likes exactly one of the two foods).
Part D — Tree diagrams (with and without replacement)
  1. A bag has 5 green, 3 orange (8 total). Two are drawn with replacement. Find P(both green).
  2. Using the same bag, find P(both green) without replacement. Ext.
  3. Find P(exactly one green) without replacement. Ext.
Part E — Conditional probability Extended
  1. Using the pizza/pasta data from Part C, find P(likes pasta, given they like pizza).
  2. A bag has 6 red, 4 blue (10 total). Two are drawn without replacement. Given the first is blue, find the probability the second is red.
Part F — Applied / challenge
  1. Two independent machines each have a 5% chance of producing a faulty item on any given run. Find the probability that both produce a faulty item, and separately, the probability that neither does.
  2. A box has 3 working and 2 broken phones. Two phones are selected without replacement. Find the probability that at least one is broken.
  3. In a survey of 50 people, 32 own a car, 18 own a bike, and 8 own neither. Find how many own both, then find P(owns a bike, given they own a car).

Key Vocabulary Recap

Mutually exclusive — events that cannot both happen at once
Independent — events where one doesn’t affect the other’s probability
Sample space — the full list of possible outcomes
Without replacement — an item isn’t put back, changing later probabilities Ext.
Conditional probability — P(A|B), the probability of A given B has occurred Ext.

Syllabus Reference

This note covers Cambridge IGCSE Mathematics (0580) section C8.3 (Core subject content) and section E8.3 (Extended subject content) — Probability of combined events. Sample space diagrams, Venn diagrams, the AND/OR rules, and tree diagrams with replacement are shared by both tiers. Tree diagrams without replacement and conditional probability are Extended-only, marked above. This closes out the Probability unit.

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