Conditional probability is the chance of one event happening given that another has already happened — written P(A|B), “A given B“. The “given” information shrinks your sample space: you only look at the cases where B happened, then ask what fraction of those also have A. This is exactly what powers “without replacement” problems, where each pick changes the odds for the next.
📘 What you need to know
P(A|B) = the probability of Agiven thatB has happened.
Formula: P(A|B) = P(A ∩ B)P(B) — in the formula booklet.
Rearranged: P(A ∩ B) = P(B)P(A|B) = P(A)P(B|A).
Given B reduces the sample space to just the outcomes in B; count what fraction also has A.
Without replacement: the total shrinks after each pick, so later probabilities are conditional.
With counts: P(A|B) = n(A ∩ B)n(B) — out of the B group, how many are also in A.
If independent, P(A|B) = P(A) — the condition makes no difference.
The idea: a shrinking sample space
The word “given” tells you to ignore everything outside B. Your new total is the B group; you count how many of those are also A.
P(A|B) — restrict to B, then find the A part
P(A|B) = “double-shaded” (A ∩ B) ÷ “all of B“. The given event becomes the new whole.
Conditional probabilityP(A|B) = P(A ∩ B)P(B)in the formula booklet ✓
🧠 Memory aid — “double shading over all shading”
On a Venn diagram, P(A|B) = double-shaded ÷ all of the B-shading. The condition B is the bottom (the new total); the overlap A ∩ B is the top. With raw counts: out of the B group, what fraction are also A?
Counting method (often fastest)
When you have actual numbers, you don’t even need the formula — just restrict to the given group and count.
🧭 Recipe — P(A|B) from counts
Find the B group — how many are in B? That’s your denominator.
Of those, how many are also in A? That’s your numerator.
Divide: P(A|B) = n(A ∩ B)n(B).
Without replacement is conditional in disguise: after taking one item, the total drops by 1 (and the count of that type drops too), so the next probability is conditioned on the first pick.
🤔 Why does P(A|B) usually differ from P(B|A)?
They have different denominators. P(A|B) divides the overlap by P(B); P(B|A) divides the same overlap by P(A). Same top, different bottom → different answers (unless P(A) = P(B)). So “A given B” and “B given A” are not interchangeable — read the wording carefully.
Worked examples
WE 1
Basic probability from a class
In a class of 30: 19 have a dog, 17 have a cat, 11 have both. A student is chosen at random. Find P(has a dog).
P(D) = n(D)/n(U)= 19/30P(D) = 19/30full sample space here is all 30 students.
WE 2
Conditional — dog given cat
Using the same class, find P(has a dog given they have a cat).
restrict to the cat group17 have a cat (new denominator)of those, how many also have a dog?11 have bothP(D|C) = 11/17“given cat” → divide by the 17 cat-owners, not 30.
WE 3
Conditional — cat given dog
For the same class, find P(has a cat given they have a dog).
restrict to the dog group19 have a dog (new denominator)of those, how many also have a cat?11 have bothP(C|D) = 11/19note P(C|D) ≠ P(D|C) — different denominators.
WE 4
Use the formula
P(A ∩ B) = 0.24 and P(B) = 0.4. Find P(A|B).
P(A|B) = P(A ∩ B) / P(B)= 0.24 / 0.4P(A|B) = 0.6straight substitution into the formula.
WE 5
Without replacement
A bag has 10 balls, 6 red. Two are taken without replacement. Given the first is red, find P(second is red).
one red already removedballs left: 9, reds left: 5probability next is red= 5/9P(2nd red | 1st red) = 5/9the total and red count both drop — that’s the conditioning.
💡 Top tips
“Given” shrinks the sample space — the condition becomes your new total (denominator).
With counts, just restrict and divide: out of the B group, what fraction is also A?
Formula: P(A|B) = P(A ∩ B) ÷ P(B).
Without replacement → conditional: reduce the totals after each pick.
P(A|B) ≠ P(B|A) in general — read which is the condition.
If independent, the condition does nothing: P(A|B) = P(A).
⚠ Common mistakes
Dividing by the whole total instead of the given group. The denominator is P(B), not 1.
Swapping P(A|B) and P(B|A). They have different denominators.
Forgetting to reduce totals in without-replacement problems.
Putting the wrong event on top. The numerator is the overlap A ∩ B.
Assuming independence so P(A|B) = P(A) without checking.
Using n(A) instead of n(A ∩ B) for the numerator.
Next up — Venn Diagrams. You’ve used Venn regions to picture P(A|B); next you’ll fill them in properly. Working from the centre out (intersections first), you’ll lay out frequencies or probabilities, read off ∩, ∪, ′ and conditional values directly, and use algebra when a region is unknown.
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