Conditional probability — P(A - B) = P(A∩B) - P(A): Worked examples

The scenario matrix

Every conditional-probability question you meet falls into one of these case classes. We will hit each cell at least once.

#	Case class	What makes it tricky	Worked in
C1	Basic overlap (equally likely, count in a shrunken universe)	forgetting to restrict to $B$	Ex 1
C2	Flip the bar — $P(A	B) $v s$ P(B	A)$
C3	Chaining without replacement (multiplication rule)	universe shrinks each step	Ex 3
C4	Independence check — is $P(A	B)=P(A)$?	overlap can look dependent
C5	Degenerate: $P (B) = 0$ (impossible condition)	formula divides by zero	Ex 5
C6	Degenerate: $A \subseteq B$ or $B \subseteq A$ (nested events)	answer is $0$ or $1$	Ex 5
C7	Condition on a complement $P(A	B^c)$	you may not touch the numerator's $A$
C8	Real-world word problem (base rates / false positives)	reversing the condition gives the wrong answer	Ex 7
C9	Limiting behaviour — condition shrinks toward a single outcome	probabilities snap to $0$ or $1$	Ex 8
C10	Condition on one cell of a partition (disjoint $B_{1}, \dots, B_{n}$ )	the parts must be disjoint and cover all	Ex 9
C11	Exam twist — condition on a union	must not double-count the overlap	Ex 10

Prerequisite pictures we lean on: Sample space and events (the universe), Tree diagrams (chaining), Multiplication rule of probability, Independence of events, Bayes' Theorem, Law of total probability.

C1 — Basic overlap

Worked example Ex 1 — Fair die, restrict to the even world

Roll a fair six-sided die. Let $A = {$ score is a prime $} = {2, 3, 5}$ and $B = {$ score is even $} = {2, 4, 6}$ . Find $P (A ∣ B)$ .

Forecast: guess a fraction now — how many of the even numbers are prime? Write your guess.

Step 1 — draw the universe and shade $B$ .

Why this step? Conditional probability means "throw away everything outside

B

." The picture makes the new (smaller) universe literal: only

{2, 4, 6}

survive.

Step 2 — find the overlap $A \cap B$ . The primes that are also even: only $2$ . So $A \cap B = {2}$ and $P (A \cap B) = \frac{1}{6}$ . Why this step? The numerator counts outcomes in both — the amber cell in the figure.

Step 3 — find $P (B)$ . $B = {2, 4, 6}$ has $3$ outcomes, so $P (B) = \frac{3}{6} = \frac{1}{2}$ . Why this step? $B$ is the denominator: it is the size of the world we now live in.

Step 4 — divide. $P (A ∣ B) = \frac{P ( A \cap B )}{P ( B )} = \frac{1/6}{3/6} = \frac{1}{3} .$ Why this step? This is the definition itself — overlap over shrunken universe. Dividing rescales the small amber slice by the size of the world we now live in, turning a raw count into a probability that sums to $1$ inside $B$ .

Verify: Restrict to ${2, 4, 6}$ directly and count primes inside: only ${2}$ , i.e. $1$ of $3$ . Same $\frac{1}{3}$ . ✔

C2 — Flip the bar

Worked example Ex 2 — Same events, opposite condition

Using the exact same $A = {2, 3, 5}$ , $B = {2, 4, 6}$ from Ex 1, now find $P (B ∣ A)$ and compare.

Forecast: will it equal $\frac{1}{3}$ again? Guess yes/no.

Step 1 — keep the numerator, it is symmetric. $A \cap B = {2}$ still, so $P (A \cap B) = \frac{1}{6}$ . Why this step? "Both $A$ and $B$ " does not care about order — that is why the numerator is the same as Ex 1. This is exactly the trap.

Step 2 — change the denominator to the new condition $A$ . Now the given event is $A = {2, 3, 5}$ , so we divide by $P (A) = \frac{3}{6} = \frac{1}{2}$ . Why this step? "Bottom = the Bar's Buddy." The condition sits right of the bar; here that is $A$ .

Step 3 — divide. $P (B ∣ A) = \frac{1/6}{3/6} = \frac{1}{3} .$ Why this step? Same definition as Ex 1, but the world is now $A$ , not $B$ . We rescale the overlap by $P (A)$ so that probabilities inside the new universe $A$ sum to $1$ — a different rescaling gives a different meaning.

Step 4 — read the coincidence carefully. Here $P (A ∣ B) = P (B ∣ A) = \frac{1}{3}$ only because $P (A) = P (B) = \frac{1}{2}$ . Change one event's size and the two split apart — see Ex 7. Why this step? To kill the false lesson "they're always equal." They are equal iff $P (A) = P (B)$ .

Verify: Bayes' Theorem says $P (B ∣ A) = P (A ∣ B) \frac{P ( B )}{P ( A )} = \frac{1}{3} \cdot \frac{1/2}{1/2} = \frac{1}{3}$ . ✔

C3 — Chaining without replacement

Worked example Ex 3 — Three aces in a row

Draw $3$ cards from a standard $52$ -card deck, no replacement. Find $P (all three are aces)$ .

Forecast: bigger or smaller than $(\frac{4}{52})^{3}$ ? Guess.

Step 1 — build the chain with a tree.

Why this step? Tree diagrams show the universe shrinking at each node — aces left and cards left both drop by one after a success.

Step 2 — first draw. $P (A_{1}) = \frac{4}{52}$ . Why this step? No information yet; $4$ aces among $52$ cards is the raw base rate.

Step 3 — second draw given first was an ace. $P (A_{2} ∣ A_{1}) = \frac{3}{51}$ . Why this step? Universe shrank: conditioning on "first was an ace" leaves $3$ aces in $51$ cards.

Step 4 — third draw given the first two were aces. $P (A_{3} ∣ A_{1} \cap A_{2}) = \frac{2}{50}$ . Why this step? Two aces are gone, $50$ cards remain — the condition tightens the universe again.

Step 5 — multiply along the branch. $P = \frac{4}{52} \cdot \frac{3}{51} \cdot \frac{2}{50} = \frac{24}{132600} = \frac{1}{5525} .$ Why this step? The general Multiplication rule of probability chains conditionals: $P (A_{1} \cap A_{2} \cap A_{3}) = P (A_{1}) P (A_{2} ∣ A_{1}) P (A_{3} ∣ A_{1} \cap A_{2})$ .

Verify: $\frac{1}{5525} \approx 0.000181$ , and $(\frac{4}{52})^{3} \approx 0.000455$ — the true value is smaller because removing aces makes each later ace rarer. ✔

C4 — Independence check

Worked example Ex 4 — Does knowing

$B$ change $A$ ? A card is drawn from $52$ . $A = {$ the card is a King $}$ , $B = {$ the card is a Heart $}$ . Is $A$ independent of $B$ ? Find $P (A ∣ B)$ .

Forecast: does knowing "it's a Heart" change the chance it's a King?

Step 1 — the overlap. $A \cap B = {$ King of Hearts $}$ , one card, so $P (A \cap B) = \frac{1}{52}$ . Why this step? Exactly one card is both a King and a Heart — that single card is the numerator.

Step 2 — the condition. $P (B) = \frac{13}{52} = \frac{1}{4}$ ( $13$ Hearts). Why this step? $B$ is the given event, so its size is the denominator.

Step 3 — divide. $P (A ∣ B) = \frac{1/52}{13/52} = \frac{1}{13} .$ Why this step? Definition — we rescale the single overlap card by the $13$ -card Heart world to get the fraction of Hearts that are Kings.

Step 4 — compare to $P (A)$ . $P (A) = \frac{4}{52} = \frac{1}{13}$ . Since $P (A ∣ B) = P (A)$ , the events are independent — knowing it's a Heart tells you nothing about Kinghood. Why this step? Independence of events is precisely the case $P (A ∣ B) = P (A)$ .

Verify: independence also means $P (A \cap B) = P (A) P (B)$ : check $\frac{1}{4} \cdot \frac{1}{13} = \frac{1}{52}$ . ✔

C5 & C6 — Degenerate inputs

Worked example Ex 5 — When the machine breaks or trivialises

Same fair die. Answer three edge questions.

(a) $P (B) = 0$ — impossible condition. Let $B = {$ score is $7}$ . Find $P (A ∣ B)$ for any $A$ . Step 1. $P (B) = 0$ because $7$ is not a face. Why this step? $B$ is empty — no outcome realises it. Step 2. $P (A ∣ B) = \frac{P ( A \cap B )}{0}$ — undefined. Why this step: the formula's fine print is $P (B) > 0$ ; dividing by zero has no value, so "given the impossible" has no answer.

(b) Nested $A \subseteq B$ . Let $A = {2}$ , $B = {2, 4, 6}$ . Find $P (A ∣ B)$ . Step 1. $A \cap B = A = {2}$ , so $P (A \cap B) = \frac{1}{6}$ . Why this step? Since $A$ lives entirely inside $B$ , the overlap is all of $A$ . Step 2. $P (A ∣ B) = \frac{1/6}{3/6} = \frac{1}{3}$ . Why this step? We divide the overlap by $P (B)$ exactly as always; because $A \subseteq B$ the numerator equals $P (A)$ , so the formula reduces to the clean shape $P (A ∣ B) = \frac{P ( A )}{P ( B )}$ .

(c) Reverse nesting $B \subseteq A$ . Let $A = {2, 4, 6}$ , $B = {4}$ . Find $P (A ∣ B)$ . Step 1. $A \cap B = B = {4}$ , $P (A \cap B) = \frac{1}{6}$ . Why this step? $B$ sits inside $A$ , so the overlap is all of $B$ . Step 2. $P (A ∣ B) = \frac{1/6}{1/6} = 1$ . Why this step? We still divide by $P (B)$ , but now every outcome of the condition already lies in $A$ ; the overlap equals the whole condition, so the ratio is forced to $1$ — certainty.

Verify: (b) $\frac{1}{3}$ and (c) $1$ ; and $P (B ∣ B) = 1$ always, matching the rescaling logic in the parent. ✔

C7 — Condition on a complement

Worked example Ex 6 — Given "NOT even", how likely is prime?

Same fair die, $A = {$ prime $} = {2, 3, 5}$ , $B = {$ even $} = {2, 4, 6}$ . Now condition on $B^{c}$ (the die is odd). Find $P (A ∣ B^{c})$ .

Forecast: does $P (A ∣ B) + P (A ∣ B^{c})$ add to $1$ ? Guess yes/no before computing.

Step 1 — build the complement. $B^{c} = {1, 3, 5}$ (the odd faces), so $P (B^{c}) = \frac{3}{6} = \frac{1}{2}$ . Why this step? Conditioning on $B^{c}$ means the new universe is "everything not in $B$ ." We may change the condition to its complement, but the event $A$ on the left of the bar stays untouched.

Step 2 — overlap of $A$ with the new universe. $A \cap B^{c} = {2, 3, 5} \cap {1, 3, 5} = {3, 5}$ , so $P (A \cap B^{c}) = \frac{2}{6}$ . Why this step? The numerator counts primes that are also odd — the primes surviving in the odd world.

Step 3 — divide. $P (A ∣ B^{c}) = \frac{2/6}{3/6} = \frac{2}{3} .$ Why this step? Definition — overlap over the shrunken (odd) universe.

Step 4 — check the two false "complement" laws. From Ex 1, $P (A ∣ B) = \frac{1}{3}$ . Then $P (A ∣ B) + P (A ∣ B^{c}) = \frac{1}{3} + \frac{2}{3} = 1$ — here it happens to equal $1$ , but only by coincidence (both conditions have size $3$ and split the primes evenly). The law that is always true is on the event: $P (A ∣ B) + P (A^{c} ∣ B) = 1$ , changing $A$ , never the condition. Why this step? The parent's mistake box warns that $P (A ∣ B) + P (A ∣ B^{c}) = 1$ is not a law; this example shows it can hold by accident, so you must not rely on it.

Verify: the always-true law: $P (A^{c} ∣ B) = \frac{P ( A ^{c} \cap B )}{P ( B )}$ with $A^{c} \cap B = {4, 6}$ gives $\frac{2/6}{3/6} = \frac{2}{3}$ , and $\frac{1}{3} + \frac{2}{3} = 1$ . ✔

C8 — Real-world word problem

Worked example Ex 7 — Rare disease, good test, surprising answer

A disease affects $2%$ of a population. A test gives a positive result for $95%$ of sick people (true positive) and for $5%$ of healthy people (false positive). You test positive. Find $P (sick ∣ +)$ .

Forecast: guess a number. Most people guess "about $95%$ ." Watch what happens.

Step 1 — take a concrete crowd of $10000$ . Sick $= 2%$ of $10000 = 200$ . Healthy $= 9800$ . Why this step? Whole numbers make the shrunken universe visible — this is Law of total probability in disguise.

Step 2 — split each group by test result (build the $2 \times 2$ tree).

Sick and

+

95%

200 = 190

. Healthy and

+

5%

9800 = 490

. Why this step? The numerator we need is "sick and positive"; the denominator is "everyone positive."

Step 3 — restrict to the new universe: everyone positive. Total positives $= 190 + 490 = 680$ . Why this step? We condition on $+$ , so $+$ is the world now.

Step 4 — the sick fraction of that world. $P (sick ∣ +) = \frac{190}{680} = \frac{19}{68} \approx 0.279.$ Why this step? Definition — favourable (sick $+$ ) over total (all $+$ ).

Step 5 — the moral. Only about $28%$ , not $95%$ ! Because the healthy group is so huge, its $5%$ of false positives ( $490$ ) swamps the sick group's true positives ( $190$ ). This is the Bayes' Theorem flip: $P (sick ∣ +) \neq = P (+ ∣ sick)$ . Why this step? To fix in memory that reversing the bar (base-rate neglect) is the classic real-world error.

Verify: Bayes directly: $\frac{0.02 \cdot 0.95}{0.02 \cdot 0.95 + 0.98 \cdot 0.05} = \frac{0.019}{0.068} \approx 0.279$ . ✔

C9 — Limiting behaviour

Worked example Ex 8 — Squeezing the condition to a single point

Roll a fair die. Fix $A = {6}$ . Watch $P (A ∣ B_{k})$ as the condition $B_{k}$ shrinks: $B_{1} = {4, 5, 6}$ , then $B_{2} = {5, 6}$ , then $B_{3} = {6}$ .

Forecast: as $B$ closes in on ${6}$ , where does the answer head?

Step 1 — $B_{1} = {4, 5, 6}$ . $A \cap B_{1} = {6}$ , so $P (A ∣ B_{1}) = \frac{1/6}{3/6} = \frac{1}{3}$ . Why this step? One favourable outcome out of three in the universe.

Step 2 — $B_{2} = {5, 6}$ . $P (A ∣ B_{2}) = \frac{1/6}{2/6} = \frac{1}{2}$ . Why this step? Universe halved, but the favourable outcome $6$ still survives — the fraction rises.

Step 3 — $B_{3} = {6}$ . $P (A ∣ B_{3}) = \frac{1/6}{1/6} = 1$ . Why this step? The condition now is the outcome — certainty.

Step 4 — read the limit. As the condition tightens around the single outcome that makes $A$ true, $P (A ∣ B_{k}) \to 1$ . Had we squeezed toward ${4, 5}$ (missing $6$ ) instead, the overlap would vanish and $P (A ∣ B_{k}) \to 0$ . Why this step? Limiting a condition either guarantees or forbids $A$ — the answer snaps to $1$ or $0$ , never stalls in between at the endpoint.

Verify: the sequence is $\frac{1}{3}, \frac{1}{2}, 1$ — strictly increasing to $1$ . ✔

C10 — Condition on one cell of a partition

Worked example Ex 9 — Balls in three coloured urns (a true partition)

A factory sends parts through three disjoint machines $M_{1}, M_{2}, M_{3}$ that together handle all parts: $M_{1}$ makes $50%$ , $M_{2}$ makes $30%$ , $M_{3}$ makes $20%$ . A part is defective with probability $0.02, 0.04, 0.10$ on $M_{1}, M_{2}, M_{3}$ respectively. Find $P (defective ∣ M_{2})$ , and then the overall $P (defective)$ .

Forecast: is $P (defective ∣ M_{2})$ just $0.04$ , or do the machine shares matter?

Step 1 — check that $M_{1}, M_{2}, M_{3}$ form a partition. They are disjoint (a part comes from exactly one machine) and cover everything ( $0.5 + 0.3 + 0.2 = 1$ ). Why this step? Law of total probability only works when the conditions are disjoint parts that tile the whole sample space — this is what "partition" means.

Step 2 — condition on the single cell $M_{2}$ . $P (defective ∣ M_{2}) = 0.04$ is given directly — it is already the fraction of defectives within the $M_{2}$ world. Why this step? Conditioning on one cell means the universe is just that cell; the per-cell defect rate is exactly the conditional probability, no rescaling needed.

Step 3 — recombine the cells for the overall rate. $P (defective) = \sum_{i} P (M_{i}) P (defective ∣ M_{i}) = 0.5 (0.02) + 0.3 (0.04) + 0.2 (0.10) .$ Why this step? Law of total probability weights each cell's conditional by how big that cell is, then sums — because the cells partition the space.

Step 4 — compute. $P (defective) = 0.010 + 0.012 + 0.020 = 0.042.$ Why this step? Arithmetic finishes the total-probability sum.

Verify: $0.5 \cdot 0.02 + 0.3 \cdot 0.04 + 0.2 \cdot 0.10 = 0.042$ , and conditioning on $M_{2}$ alone returns $0.04$ . ✔

C11 — Exam twist: condition on a union

Worked example Ex 10 — Condition on "

$B$ or $C$ " Fair die. $A = {$ prime $} = {2, 3, 5}$ . Condition on $B \cup C$ where $B = {$ even $} = {2, 4, 6}$ and $C = {$ score $\leq 2} = {1, 2}$ . Find $P (A ∣ B \cup C)$ .

Forecast: the condition is now a union — will the denominator be $P (B) + P (C)$ ? Careful.

Step 1 — build the union carefully (inclusion–exclusion). $B \cup C = {2, 4, 6} \cup {1, 2} = {1, 2, 4, 6}$ . Note $2$ is counted once, not twice. Why this step? $P (B \cup C) = P (B) + P (C) - P (B \cap C) = \frac{3}{6} + \frac{2}{6} - \frac{1}{6} = \frac{4}{6}$ . Naïvely adding $P (B) + P (C) = \frac{5}{6}$ would be wrong.

Step 2 — overlap of $A$ with the union. $A \cap (B \cup C) = {2, 3, 5} \cap {1, 2, 4, 6} = {2}$ , so $P = \frac{1}{6}$ . Why this step? Only $2$ is both prime and in the conditioning set.

Step 3 — divide. $P (A ∣ B \cup C) = \frac{1/6}{4/6} = \frac{1}{4} .$ Why this step? Definition, with the correct union denominator — the shrunken universe is ${1, 2, 4, 6}$ .

Verify: restrict to ${1, 2, 4, 6}$ and count primes: only ${2}$ — that's $1$ of $4$ . Same $\frac{1}{4}$ . ✔

Recall Quick self-test on the matrix

Which cell does "given two aces already drawn, chance the next is an ace" hit? ::: C3 — chaining without replacement. Why is $P (A ∣ {7})$ undefined on a die? ::: The condition has $P (B) = 0$ ; dividing by zero is undefined (C5). If $A \subseteq B$ , simplify $P (A ∣ B)$ . ::: $P (A) / P (B)$ (C6). When conditioning on $B^{c}$ , which quantity must you NOT change? ::: The event $A$ on the left of the bar; only the condition becomes $B^{c}$ (C7). In the disease problem, why isn't the answer $95%$ ? ::: False positives from the huge healthy group dominate (C8 / Bayes flip). What two properties make $M_{1}, M_{2}, M_{3}$ a partition? ::: Disjoint (no overlap) and exhaustive (probabilities sum to 1) (C10). Denominator for conditioning on $B \cup C$ ? ::: $P (B \cup C) = P (B) + P (C) - P (B \cap C)$ , not $P (B) + P (C)$ (C11).

Connections

Multiplication rule of probability — the engine behind Ex 3's chaining.
Tree diagrams — the picture for Ex 3 and Ex 7.
Independence of events — the test in Ex 4.
Bayes' Theorem — the correct flip in Ex 2 and Ex 7.
Law of total probability — the partition recombination in Ex 9.
Sample space and events — the universe we keep shrinking.