Conditional probability is just normal probability inside a smaller world. When the question says “given B“, you throw away every outcome not in B — the B circle becomes your entire universe. Then P(A|B) is simply “what fraction of B is also in A“.
The conditional formula has three useful forms. Learn to switch between them:
Quick independence check: if P(
A|
B) turns out to equal P(
A), then knowing
B didn’t change anything — the events are independent. If P(
A|
B) ≠ P(
A), they’re dependent.
Without replacement — conditional in action
The most common place conditional probability shows up is selecting items without replacement. Each pick changes what’s left, so the second pick’s probabilities are conditional on the first.
Example logic: a bag has 10 balls, 6 red. P(2nd is red | 1st was red) — after removing one red, there are 9 balls left, 5 red, so the answer is 5/9. The probability genuinely changed because the population shrank.
🧠Recipe — any conditional probability question
- Identify the condition: spot the word “given” (or “if”, or an implied prior event). Whatever follows it is your reduced sample space.
- Decide your method: counting (frequencies / two-way table) or the formula (probabilities given).
- Counting route: P(A|B) = n(A ∩ B) / n(B) — restrict to B, count the favourable ones.
- Formula route: P(A|B) = P(A ∩ B) / P(B). Find P(A ∩ B) first if it isn’t given (e.g. via the union formula).
- Check it makes sense: 0 ≤ P(A|B) ≤ 1. For without-replacement, confirm the denominator dropped.
Worked examples
WE 1Conditional probability by counting
In a year group of 40 students, 24 play tennis, 18 swim, and 10 do both. A student is chosen at random. (a) Find the probability they swim, given that they play tennis. (b) Find the probability they play tennis, given that they swim.
(a) P(swim | tennis): reduce to the 24 tennis players
of the 24 tennis players, 10 also swim
P(S|T) = 10/24 = 5/12
(b) P(tennis | swim): reduce to the 18 swimmers
of the 18 swimmers, 10 also play tennis
P(T|S) = 10/18 = 5/9
(a) P(S|T) = 5/12 · (b) P(T|S) = 5/9
notice P(S|T) ≠ P(T|S) — the order matters. The numerator (10, the overlap) is the same both times, but the denominator changes to whichever event is the CONDITION.
WE 2Apply the conditional formula directly
Events A and B satisfy P(A ∩ B) = 0.24 and P(B) = 0.6. Find P(A|B).
use P(A|B) = P(A ∩ B) / P(B)
P(A|B) = 0.24 / 0.6
= 0.4
P(A|B) = 0.4
straight substitution. The numerator is always the INTERSECTION; the denominator is always the CONDITION (the event after the bar).
A bag contains 9 marbles: 4 green and 5 yellow. Two marbles are drawn at random without replacement. (a) Find the probability the second marble is green, given the first was green. (b) Find the probability both marbles are green.
(a) given 1st was green, count what’s left
started with 9 (4 green); removed 1 green
now 8 marbles left, 3 of them green
P(2nd green | 1st green) = 3/8
(b) P(both green) = P(1st green) × P(2nd green | 1st green)
P(1st green) = 4/9
P(both green) = (4/9) × (3/8)
= 12/72 = 1/6
(a) 3/8 · (b) 1/6
part (b) uses the rearranged formula P(A ∩ B) = P(A) × P(B|A). For without-replacement, ALWAYS reduce the counts: 9 → 8 marbles, 4 → 3 green.
WE 4Conditional probability from a two-way table
50 students took an exam. The two-way table shows the results by gender:
Boys: 18 passed, 6 failed (24 total)
Girls: 21 passed, 5 failed (26 total)
Totals: 39 passed, 11 failed (50)
(a) Find P(passed | the student is a girl). (b) Find P(the student is a boy | they failed).
(a) condition = “girl” ⇒ denominator = 26 girls
of 26 girls, 21 passed
P(passed | girl) = 21/26
(b) condition = “failed” ⇒ denominator = 11 who failed
of 11 who failed, 6 are boys
P(boy | failed) = 6/11
(a) 21/26 · (b) 6/11
two-way tables make conditionals easy: the condition picks the ROW or COLUMN total as the denominator; the cell where the row and column meet is the numerator.
WE 5Find an intersection from a conditional
On a winter day, the probability that the sky is cloudy is 0.7. Given that the sky is cloudy, the probability of rain is 0.4. Find the probability that the day is both cloudy AND rainy.
use rearranged formula P(A ∩ B) = P(B) P(A|B)
here: P(cloudy ∩ rain) = P(cloudy) × P(rain | cloudy)
substitute
= 0.7 × 0.4
= 0.28
P(cloudy and rain) = 0.28
the rearranged formula multiplies “probability of the condition” by “conditional probability”. This is exactly how tree-diagram branches work — you multiply along the branches.
WE 6Multi-step — find the intersection first
Events A and B satisfy P(A) = 0.5, P(B) = 0.3, and P(A ∪ B) = 0.65. (a) Find P(A|B). (b) Hence state whether A and B are independent.
(a) Step 1 — find P(A ∩ B) using the union formula
P(A ∪ B) = P(A) + P(B) − P(A ∩ B)
0.65 = 0.5 + 0.3 − P(A ∩ B)
P(A ∩ B) = 0.8 − 0.65 = 0.15
Step 2 — conditional formula
P(A|B) = P(A ∩ B) / P(B) = 0.15 / 0.3
= 0.5
(b) compare with P(A)
P(A|B) = 0.5 and P(A) = 0.5
they are EQUAL ⇒ B doesn’t affect A
(a) P(A|B) = 0.5 · (b) A and B ARE independent
elegant link: P(A|B) = P(A) is exactly the definition of independence. The intersection 0.15 also equals P(A)P(B) = 0.5×0.3 — the same conclusion two ways.
💡 Top tips
- The condition is the denominator: in P(A|B), the event after the bar (B) is always what you divide by.
- Order matters: P(A|B) and P(B|A) are usually different — same numerator (the overlap), different denominators.
- Two-way tables: the condition selects a row/column total as the denominator; the matching cell is the numerator.
- Without replacement: shrink the totals on the second pick — both the population and the count of the chosen type drop by 1.
- P(A|B) = P(A) is the independence test — a neat way to confirm or rule out independence.
âš Common mistakes
- Dividing by n(U) instead of n(B): a conditional probability uses the REDUCED sample space — the denominator is B, not the whole population.
- Swapping P(A|B) and P(B|A): read carefully which event is “given”. They are generally not equal.
- Using P(A)P(B) for the intersection without checking independence: only valid if the events are independent — otherwise use P(B)P(A|B).
- Forgetting to reduce counts without replacement: keeping the denominator at the original total is the single most common slip.
- Treating the comma in “A, B” as “given”: “given” needs the bar |. A comma usually means “and” (intersection).
Next up: Venn Diagrams. You’ve used Venn diagrams informally already — the next note treats them properly: filling in regions from given information, working centre-outwards, and reading off complements, unions, intersections and conditionals straight from the picture. Drawing a Venn diagram is often the fastest route through a probability question.
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