Combined events — tree diagrams, independent and dependent events
When two events happen sequentially (or in combination), the cleanest way to compute probabilities is the probability tree.
Independent vs dependent events
Independent: the outcome of one event doesn't change the probability of the other.
- Tossing two coins.
- Drawing a ball from a bag with replacement.
Dependent: the outcome of one event changes the probability of the other.
- Drawing two balls without replacement.
- Sequential exam questions where success affects difficulty.
For independent events: P(A and B) = P(A) × P(B).
For dependent events: P(A and B) = P(A) × P(B | A) — multiply by the conditional probability of B given A has happened.
Probability trees
Each branch shows P of going that way. To get P of a path, multiply along the path. To combine paths, add them.
Example: a bag with 4 red and 6 blue balls. Two are drawn without replacement.
Branch probabilities:
- First R: 4/10. First B: 6/10.
- Given first was R: P(R) = 3/9 (one red gone); PB = 6/9.
- Given first was B: P(R) = 4/9; PB = 5/9 (one blue gone).
Probabilities of full paths:
- RR: (4/10)(3/9) = 12/90 = 2/15.
- RB: (4/10)(6/9) = 24/90 = 4/15.
- BR: (6/10)(4/9) = 24/90 = 4/15.
- BB: (6/10)(5/9) = 30/90 = 1/3.
Check: 2/15 + 4/15 + 4/15 + 5/15 = 15/15 = 1. ✓
"At least one" via the complement
Often easier to compute P(none) and subtract from 1.
Worked example: P(at least one red in two without-replacement draws from 4R/6B):
- PBB = 5/15.
- P(at least one R) = 1 − 5/15 = 10/15 = 2/3.
Independent events on a tree
If draws are with replacement, probabilities don't change between branches.
Worked example: P(both red, with replacement, 4R/6B):
- (4/10)² = 16/100 = 4/25.
More than two stages
For three independent events, multiply three probabilities. The tree just has more layers.
Worked example: P(three heads with a fair coin in three tosses) = (1/2)³ = 1/8.
"Exactly one" type problems
Add the probabilities of all paths that result in "exactly one R".
- RB and BR both contribute. (Order matters.)
- For independent events: P(exactly one R) = P(R)P(B') + P(R')PB etc.
Worked example: 4R/6B with replacement, two draws. P(exactly one red).
- (4/10)(6/10) + (6/10)(4/10) = 2 × (24/100) = 48/100 = 12/25.
⚠Common mistakes— Common mistakes (examiner traps)
- Forgetting to subtract the drawn ball in without-replacement problems.
- Multiplying instead of adding when combining different paths leading to the same outcome.
- Treating dependent events as independent. Always read whether replacement happens.
- Wrong denominator on a conditional branch. After removing a ball, the new bag size is n − 1, not n.
- Not simplifying the final fraction.
➜Try this— Quick check
A bag has 3 red and 5 blue balls. Two are drawn without replacement. (a) P(both red). (b) P(at least one red).
(a) (3/8)(2/7) = 6/56 = 3/28. (b) PBB = (5/8)(4/7) = 20/56 = 5/14. P(at least one red) = 1 − 5/14 = 9/14.
AI-generated · claude-opus-4-7 · v3-deep-probability