Worked examples — Conditional probability

This page is a complete case sweep. The parent note gave you the definition and three examples. Here we build a matrix of every situation conditional probability can throw at you — every extreme value, every degenerate input, every real-world twist — and then work one example per cell so you never meet a case you haven't already seen.

Before we start, one reminder in plain words. The symbol $P(A\mid B)$ is read "probability of $A$ given $B$". It means: stand inside the world where $B$ has already happened, and ask what fraction of that world also has $A$ in it. The bar $\mid$ means "given" — it does not mean division. The one formula everything below rests on is

$P(A\mid B)=\frac{P(A\cap B)}{P(B)}\text{(needs }P(B)>0\text{)}$

where $P(A\cap B)$ is the chance both happen (the overlap), and dividing by $P(B)$ renormalises so that the world of $B$ counts as the new "total".

The scenario matrix

Every conditional-probability problem you will ever see lands in one of these cells. The worked examples below are each tagged with the cell they cover.

The three "extreme value" ideas — an answer of exactly 0, exactly 1, an undefined input ($P(B)=0$), or a certain condition ($P(B)=1$) — are the ones people skip. We hit the whole spectrum $0\le P(B)\le 1$ head-on in C5, C6 and C10.

Look at the map below before reading on: it shows the whole universe as a rectangle, event $B$ as a shaded region, and $A\cap B$ as the darker overlap. Conditioning on $B$ = zooming into the $B$ region and treating it as the whole picture.

[!example] Example 1 — Cell C1: plain overlap

Statement. A deck-of-cards style setup: in a class of $30$ students, $12$ play chess ($B$), and of those $12$, exactly $3$ also play piano ($A$). A student is picked at random and turns out to be a chess player. What is the probability they play piano?

Forecast: guess a number before reading. (Is it $3/30$ or $3/12$?)

1. Identify the events. $A=$ plays piano, $B=$ plays chess. We are told the outcome is a chess player, so we condition on $B$. Why this step? Naming events stops us mixing up which is "given".
2. Read off the pieces. $P(B)=\frac{12}{30}$ and $P(A\cap B)=\frac{3}{30}$ (3 of the 30 do both). Why this step? The formula needs the overlap and the condition as fractions of the same original universe of 30.
3. Apply the formula. $P(A\mid B)=\frac{P(A\cap B)}{P(B)}=\frac{3/30}{12/30}=\frac{3}{12}=0.25$ Why this step? The $30$s cancel — we have zoomed into the $12$ chess players (the blue box of Figure 1).

Verify: Directly, "3 of the 12 chess players play piano" $=3/12=0.25$. Matches. The answer is a fraction of the shrunk universe (12), not the original (30) — that is the whole point.

[!example] Example 2 — Cell C2: flip the bar with Bayes

Statement. A factory: $70\%$ of bolts come from machine $M_1$, $30\%$ from $M_2$. $M_1$ produces $2\%$ defective, $M_2$ produces $5\%$ defective. A bolt is defective. What is the probability it came from $M_1$?

Forecast: more bolts come from $M_1$, but $M_2$ is defect-prone. Which wins?

1. Name events. $A=$ from $M_1$, $D=$ defective. Given: $P(A)=0.7$, $P(D\mid A)=0.02$, $P(D\mid \neg A)=0.05$. (Recall $\neg A$ = "not from $M_1$" = "from $M_2$".) Why this step? We know $P(D\mid A)$ but want $P(A\mid D)$ — the bar is pointing the wrong way, so we need Bayes.
2. Total probability of a defect (the denominator): $P(D)=P(D\mid A)P(A)+P(D\mid \neg A)P(\neg A)=0.02(0.7)+0.05(0.3)=0.014+0.015=0.029$ Why this step? A bolt can be defective via two pathways; we sum them.
3. Bayes: $P(A\mid D)=\frac{P(D\mid A)P(A)}{P(D)}=\frac{0.014}{0.029}\approx 0.4828$ Why this step? We already have the overlap $P(D\cap A)=P(D\mid A)P(A)=0.014$ and the new total universe $P(D)=0.029$; dividing gives the fraction of defective bolts that came from $M_1$ — exactly the master move of Figure 1.

Verify: The complement should be $P(\neg A\mid D)=\frac{0.015}{0.029}\approx 0.5172$, and $0.4828+0.5172=1.0$. ✓ Even though $M_1$ makes most bolts, the defective one is slightly more likely from $M_2$.

[!example] Example 3 — Cell C3: rare-prior twist

Statement. (Same shape as the parent's medical example, new numbers so you re-derive.) A security scanner flags $99\%$ of real threats ($P(\text{flag}\mid T)=0.99$) and falsely flags $2\%$ of safe bags ($P(\text{flag}\mid \neg T)=0.02$). Only $1$ in $1000$ bags is a real threat. A bag is flagged. Probability it is a real threat?

Forecast: the scanner is $99\%$ accurate — so surely a flag means near-certain threat? Guess.

1. Events. $T=$ threat, $P(T)=0.001$, $P(\neg T)=0.999$. ($\neg T$ = "safe bag".) Why this step? We must pin the tiny prior $P(T)=0.001$ down explicitly, because this is the number our intuition tends to forget and it will dominate the answer.
2. Denominator (total probability of a flag): $P(\text{flag})=0.99(0.001)+0.02(0.999)=0.00099+0.01998=0.02097$ Why this step? The $2\%$ false-positive rate acts on the huge safe majority.
3. Bayes: $P(T\mid \text{flag})=\frac{0.00099}{0.02097}\approx 0.0472\approx 4.72\%$ Why this step? The numerator $0.00099$ is the overlap "flagged and truly a threat"; the denominator $0.02097$ is the whole flagged universe. Dividing zooms into the flagged bags and asks what fraction are real threats — the Figure 1 move again.

Verify: Under $5\%$! Sanity: expected in $100,000$ bags → $100$ threats (99 flagged) and $99,900$ safe ($\approx 1998$ falsely flagged). Flagged threats $/$ all flagged $=99/(99+1998)=99/2097\approx 0.0472$. ✓ This is the base-rate neglect from the parent's Mistake 2 — the rare prior dominates.

[!example] Example 4 — Cell C4: independence (degenerate case)

Statement. Roll a fair six-sided die. $A=$ result is even $\{2,4,6\}$, $B=$ result is $\le 4$ $\{1,2,3,4\}$. Does knowing $B$ change $P(A)$?

Forecast: guess whether these are independent.

1. Unconditional. $P(A)=3/6=0.5$. Why this step? We need the before-conditioning value as a baseline to compare against, since independence is defined by "conditioning changes nothing".
2. Overlap. $A\cap B=\{2,4\}$ (the outcomes in both sets), so $P(A\cap B)=2/6$; and $P(B)=4/6$. Why this step? The formula needs the intersection and the condition; listing the outcomes explicitly avoids miscounting.
3. Conditional: $P(A\mid B)=\frac{2/6}{4/6}=\frac{2}{4}=0.5$ Why this step? We compare it to $P(A)$: they are equal, so learning $B$ told us nothing — in Figure 1 the green overlap fills the same fraction of the blue box as $A$ fills of the whole rectangle.

Verify (product test): Independent iff $P(A\cap B)=P(A)P(B)$. Check: $2/6=0.3\overline{3}$ and $P(A)P(B)=0.5\times \frac{2}{3}=0.3\overline{3}$. ✓ This is the degenerate conditioning: the world shrinks but the fraction of $A$ stays put.

[!example] Example 5 — Cell C5: the certainty limits (0 and 1)

Statement. One card drawn from a standard $52$-card deck. Let $B=$ "the card is a heart". Find (a) $P(\text{black card}\mid B)$ and (b) $P(\text{red card}\mid B)$.

Forecast: one answer will be exactly $0$, the other exactly $1$. Predict which.

1. Compute the condition $P(B)$. A standard deck has $13$ hearts out of $52$ cards, so $P(B)=\frac{13}{52}=\frac{1}{4}$. Why this step? The formula divides by $P(B)$, so we must pin it down explicitly before using it — never leave it implicit.
2. Mutually exclusive case (answer $0$). A heart is red, so "black" and "heart" cannot both happen: their intersection is empty, giving $P(\text{black}\cap B)=0$. $P(\text{black}\mid B)=\frac{0}{P(B)}=\frac{0}{1/4}=0$ Why this step? If the overlap is empty, conditioning cannot resurrect it — $0$ is a legitimate answer, not an error.
3. Subset case (answer $1$). Every heart is red, so $B\subseteq \text{red}$ (the heart-event sits entirely inside the red-event), meaning $\text{red}\cap B=B$. $P(\text{red}\mid B)=\frac{P(B)}{P(B)}=\frac{1/4}{1/4}=1$ Why this step? When $B$ lives entirely inside $A$ (that is what $B\subseteq A$ means), being in $B$ guarantees $A$.

Verify: These are the two extreme values of any probability: $0\le P(A\mid B)\le 1$. We hit both boundaries exactly, and both used $P(B)=1/4$. ✓

[!example] Example 6 — Cell C6: undefined input ($P(B)=0$)

Statement. A spinner points to a real number uniformly in $[0,1]$. Let $B=$ "the pointer lands exactly on $0.5$". What is $P(A\mid B)$ for $A=$ "the number is $\ge 0.5$"?

Forecast: it feels like the answer should be $1$. But watch the denominator.

1. Probability of the condition. For a continuous uniform pick, probability = length of the target interval. A single exact point like $0.5$ is an interval of length $0$, so $P(B)=0$. Why this step? The formula demands $P(B)>0$; here it fails, and we must see why it fails before we can talk about the fix.
2. Probability of the overlap. Since $0.5\ge 0.5$, the single point $0.5$ lies inside $A$, so $B\subseteq A$ and $A\cap B=B$. But $B$ itself has length $0$, so $P(A\cap B)=P(B)=0$ as well. Why this step? Students often expect $P(A\cap B)$ to be positive here; spelling out that the overlap is also a single zero-length point shows both numerator and denominator vanish.
3. Try the formula: $P(A\mid B)=\frac{P(A\cap B)}{P(B)}=\frac{0}{0}\text{— undefined.}$ Why this step? Dividing $0/0$ is meaningless: there is no $B$-world to shrink into (Figure 1's blue box has zero area), so "the fraction of it that is $A$" has no answer.

Verify: This is a genuine gap in the elementary definition. The professional fix is conditioning on densities / limits (measure-theoretic conditioning) used in Bayesian inference over continuous parameters — but the elementary rule correctly refuses to answer. Always check $P(B)>0$ first. ✓

[!example] Example 7 — Cell C7: three pathways (total probability)

Statement. Emails come from three sources: newsletters ($50\%$), personal ($30\%$), promotions ($20\%$). The word "sale" appears with probability $0.05$, $0.01$, $0.60$ respectively. An email contains "sale". Probability it is a promotion?

Forecast: promotions are only $20\%$ of mail but love the word "sale". Guess the posterior.

1. Events. $B_1,B_2,B_3=$ newsletter, personal, promotion; $W=$ contains "sale". Why this step? Total probability needs a partition — three mutually exclusive, exhaustive sources — so we name all three before summing.
2. Total probability (three-branch): $P(W)=0.05(0.5)+0.01(0.3)+0.60(0.2)=0.025+0.003+0.12=0.148$ Why this step? Every source is a pathway to seeing "sale"; sum all three.
3. Bayes for the promotion branch: $P(B_3\mid W)=\frac{0.60(0.2)}{0.148}=\frac{0.12}{0.148}\approx 0.8108$ Why this step? The numerator $0.12$ is the overlap "promotion and contains sale"; the denominator $0.148$ is the whole 'sale' universe from all three sources. Dividing zooms into the emails containing "sale" and asks what fraction are promotions — the Figure 1 move with three input branches.

Verify: The three posteriors must sum to $1$: $\frac{0.025}{0.148}+\frac{0.003}{0.148}+\frac{0.12}{0.148}=\frac{0.148}{0.148}=1$. ✓ This 3-branch structure generalises directly to the naive Bayes classifier.

[!example] Example 8 — Cell C8: chain of events

Statement. A box has $3$ red and $2$ green balls. Draw two without replacement. What is the probability both are red?

Forecast: it is not $\frac{3}{5}\times \frac{3}{5}$ — why not?

1. First draw. $P(R_1)=\frac{3}{5}$. Why this step? We anchor the first factor of the chain rule; at this moment the universe is the full box of $5$ balls (Figure 1's whole rectangle).
2. Second draw given the first. After removing one red, $2$ red remain out of $4$: $P(R_2\mid R_1)=\frac{2}{4}$. Why this step? The first draw changes the universe for the second — we zoom into the world where a red is already gone, exactly the shrink-into-$B$ move of Figure 1.
3. Chain rule (multiplication rule from the parent): $P(R_1\cap R_2)=P(R_1)P(R_2\mid R_1)=\frac{3}{5}\cdot \frac{2}{4}=\frac{6}{20}=0.3$ Why this step? To get both, we multiply "reach the shrunk world" ($P(R_1)$) by "$A$ happens inside it" ($P(R_2\mid R_1)$).

Verify: Count directly with the "$n$ choose $k$" notation from the top of this page: number of ways to pick 2 reds $=\left(\genfrac{}{}{0ex}{}{3}{2}\right)=3$; total ways to pick any 2 of 5 $=\left(\genfrac{}{}{0ex}{}{5}{2}\right)=10$; so $3/10=0.3$. ✓ The chain rule is what makes Hidden Markov Models tractable.

[!example] Example 9 — Cell C9: exam twist ("at least one" + complement)

Statement. A student attempts $3$ independent problems, each solved with probability $0.6$. Given that they solved at least one, what is the probability they solved all three?

Forecast: conditioning removes the "solved none" world — does that push the answer up or down?

1. Event $B=$ "at least one solved". Easier via complement: $P(\text{none})=(0.4{)}^3=0.064$, so $P(B)=1-0.064=0.936$. Why this step? "At least one" is painful directly; its complement "none" (the $\neg$ of "at least one") is one clean product (independence lets us multiply the three $0.4$s).
2. Event $A=$ "all three". $P(A)=(0.6{)}^3=0.216$. Since "all three" $\subseteq$ "at least one" (solving all three certainly means solving at least one), $A\cap B=A$. Why this step? Recognising the subset relation $A\subseteq B$ collapses the overlap $A\cap B$ to just $A$, so we do not need a separate calculation for the numerator.
3. Condition: $P(A\mid B)=\frac{P(A)}{P(B)}=\frac{0.216}{0.936}\approx 0.2308$ Why this step? We shrink out only the "solved none" outcome, so the surviving world is slightly smaller and $A$'s share rises from $0.216$ to $\approx 0.231$.

Verify: Conditioning should increase the value ($0.231>0.216$) because we deleted a region that could never contain $A$. ✓ And $0<0.231<1$, a valid probability.

[!example] Example 10 — Cell C10: the certain condition ($P(B)=1$)

Statement. Roll a fair die. $A=$ "result is a $6$", and $B=$ "result is between $1$ and $6$" (i.e. the die shows some face — a certainty). Find $P(A\mid B)$.

Forecast: conditioning on something guaranteed — does it change $P(A)$ at all?

1. Probability of the condition. $B$ is the whole sample space, so $P(B)=1$. Why this step? We must check $P(B)$ before dividing; here it sits at the upper end of the spectrum $0\le P(B)\le 1$, the mirror image of C6.
2. Overlap. Since $A\subseteq B$ (every $6$ is a valid face, so the six-event sits entirely inside the certain-event), $A\cap B=A$, so $P(A\cap B)=P(A)=\frac{1}{6}$. Why this step? The subset relation collapses the intersection to $A$, so the numerator is just $P(A)$.
3. Condition: $P(A\mid B)=\frac{P(A\cap B)}{P(B)}=\frac{1/6}{1}=\frac{1}{6}=P(A)$ Why this step? Dividing by $1$ changes nothing — a certain condition shrinks the universe to... the same universe. In Figure 1 the blue box would fill the entire white rectangle.

Verify: $P(A\mid B)=P(A)=1/6$, and any event is independent of a certain event. ✓ This closes the spectrum: $P(B)=0$ (undefined, C6) → $P(B)=1$ (no effect, C10).

[!recall]- Self-test

See also: Bayes-Theorem · Law-of-Total-Probability · Independence · Chain-Rule-of-Probability · Naive-Bayes-Classifier · Mutual-Information · Bayesian-Inference · Hidden-Markov-Models