13.2 Compound Events: 'And', 'Or' & 'At Least'
Key Takeaways
- List all outcomes systematically with a grid, table or tree diagram before counting — most compound-event errors are missing outcomes, not bad arithmetic
- For two events joined by 'and' along branches of a tree, multiply the probabilities: P(A and B) = P(A) x P(B) for independent events
- For mutually exclusive events joined by 'or', add the probabilities: P(A or B) = P(A) + P(B)
- Without replacement, the second draw's denominator drops by one and the composition changes — the draws are no longer independent
- For 'at least one' questions, use the complement: P(at least one) = 1 - P(none), which is almost always shorter than adding cases
From Paper E onwards, ICAS stops asking about single spins and starts combining events: two coins, two draws from a bag, a die and a spinner. Papers G and H explicitly test probabilities involving 'and', 'or' and 'at least'. These questions are where marks are lost — not because the arithmetic is hard, but because students count outcomes unsystematically and miss some. The fix is always the same: list the outcomes first, count second.
Listing Outcomes: Grids, Tables and Trees
For two coins, there are 4 outcomes: HH, HT, TH, TT. For two dice, a 6 x 6 grid gives 36 outcomes. For a bag drawn twice, a tree diagram is clearest because the probabilities can change between draws.
A tree diagram has one set of branches per stage. Write the probability on each branch. Two rules keep every tree correct:
- The branches leaving any point must add to 1.
- The probability of a complete path is found by multiplying along the branches.
Worked example 1: two independent draws (with replacement)
A bag holds 3 red and 2 blue counters. One is drawn, replaced, and another is drawn. Find P(two reds).
- P(red first) = 3/5. Because the counter is replaced, P(red second) = 3/5 again.
- P(red and red) = 3/5 x 3/5 = 9/25
With replacement, the two draws are independent: the first result does not affect the second.
Worked example 2: without replacement
Same bag, but the first counter is kept out. Find P(two reds) now.
- P(red first) = 3/5
- After removing a red, 2 red and 2 blue remain: P(red second) = 2/4 = 1/2
- P(red and red) = 3/5 x 1/2 = 3/10
The classic trap: reusing 3/5 for the second draw. 'Without replacement' always changes both the numerator and the denominator of the next branch. Read that phrase as carefully as any other in the question.
Counting with a grid: two dice
Two fair dice are rolled. What is the probability their total is 9?
A 6 x 6 grid has 36 equally likely outcomes. The pairs that total 9 are (3,6), (4,5), (5,4) and (6,3) — four outcomes — so P(total 9) = 4/36 = 1/9. The grid exposes the standard error: students who list totals (2 through 12) and answer 1/11 have forgotten that totals are not equally likely. There is only one way to make 2 but six ways to make 7, which is why 7 is the most likely total at 6/36 = 1/6. Whenever two dice or two spinners appear, build the grid rather than trusting intuition.
'And' versus 'Or'
The connective word tells you the operation:
- 'And' (both events happen): multiply along a single path. P(red then blue) = 3/5 x 2/4 = 3/10.
- 'Or' (either of several complete paths): multiply along each path first, then add the paths. This works when the paths are mutually exclusive — they cannot both happen on the same trial.
Worked example 3: one of each colour
Same bag, drawn twice without replacement. Find P(one red and one blue, in either order).
Two paths satisfy this:
- Red then blue: 3/5 x 2/4 = 6/20 = 3/10
- Blue then red: 2/5 x 3/4 = 6/20 = 3/10
- Total: 3/10 + 3/10 = 6/10 = 3/5
Note that 'one of each' includes both orders. Answering 3/10 — the result of a single order — is the most common error in this exact question type.
The 'At Least' Strategy
'At least one' means one, two, three... — too many cases to add comfortably. Use the complement instead:
P(at least one) = 1 - P(none)
Worked example 4
A fair coin is tossed 3 times. What is the probability of getting at least one head?
- P(no heads) = P(TTT) = 1/2 x 1/2 x 1/2 = 1/8
- P(at least one head) = 1 - 1/8 = 7/8
Compare the direct method: exactly 1 head (3/8) + exactly 2 heads (3/8) + exactly 3 heads (1/8) = 7/8. Same answer, three times the work. Under a 60-minute ICAS clock, the complement is the professional choice.
Worked example 5: 'at least' without replacement
From the bag of 3 red and 2 blue, two counters are drawn without replacement. Find P(at least one blue).
- P(no blue) = P(two reds) = 3/10 (from example 2)
- P(at least one blue) = 1 - 3/10 = 7/10
Common Traps at Paper G-H Level
- Adding when you should multiply. P(red first and blue second) is one path: multiply. Adding P(red) + P(blue) is never right for 'and'.
- Forgetting both orders for 'one of each' questions.
- Not updating the tree after a without-replacement draw.
- Probabilities above 1. If your answer exceeds 1, you have added overlapping (non-exclusive) events.
- Assuming independence. Weather on two consecutive days, or two draws without replacement, are not independent — the second probability depends on the first outcome.
Quick reference:
| Question wording | Strategy |
|---|---|
| 'A and then B' | multiply along one path |
| 'A or B (exclusive)' | multiply each path, then add |
| 'one of each, either order' | both paths added |
| 'at least one' | 1 - P(none) |
| 'without replacement' | reduce numerator and denominator for later draws |
A bag contains 5 red and 3 green marbles. Two marbles are drawn without replacement. What is the probability that both are red?
The probability that Mia wins a game is 0.4, and games are independent. She plays two games. What is the probability she wins at least one game?