Worked examples — Independence and mutual exclusivity

This page is a stress test. The parent Independence and mutual exclusivity note built the ideas; here we throw every kind of situation at those ideas and grind out the answer by hand, so you never meet a case you haven't already seen.

Before we start, three tiny reminders in plain words so nothing is used before it's defined:

The whole topic is two yes/no questions about two regions on that dartboard:

1. Mutually exclusive? → Do the regions touch at all? $P(A\cap B)=0$ means no overlap.
2. Independent? → Does the overlap have exactly the "no-information" size? $P(A\cap B)=P(A)P(B)$.

Look at the two dartboards above. On the left the regions are pushed apart — zero overlap — that is mutually exclusive. On the right they overlap by exactly the product of their sizes — that is independent. These two pictures are the whole chapter.

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The scenario matrix

Every problem this topic can throw lands in one of these cells. Our job below is to fill every cell with a worked example.

#	Case class	What makes it tricky	Example that hits it
C1	Disjoint, both positive	classic "can't both happen"	Ex 1 (die faces)
C2	Independent, overlap positive	both can happen, no info	Ex 2 (two coins)
C3	Zero-probability event	$P(B)=0$, degenerate	Ex 3 (impossible event)
C4	Certain event	$P(B)=1$, limiting case	Ex 4 (sure event)
C5	Dependent but NOT disjoint	the confusing middle	Ex 5 (rain / cloudy)
C6	Same experiment, sign flip	sampling without replacement	Ex 6 (two cards)
C7	Three-way / conditional independence	independence appears only after conditioning	Ex 7 (flu, cough, fever)
C8	Real-world word problem	translate English → sets	Ex 8 (spam filter)
C9	Exam twist	independent given $C$, dependent unconditionally	Ex 9 (two dice, sum)

The two "extremes" of the dartboard (C1 = no overlap, C2 = product overlap) plus the "degenerate" edges (C3 zero, C4 one) plus the "middle spectrum" (C5) cover every possible overlap size. C6–C9 test whether you can spot which cell a real problem sits in.

The amber bar above is a number line of overlap sizes. Every event pair sits somewhere on it. Mutual exclusivity is the far-left point (overlap $=0$); independence is the single amber tick at $P(A)P(B)$; everything else is "just dependent." Keep this line in your head.

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Ex 1 — C1: Disjoint, both positive

Forecast: guess now — can they be both? Can knowing one leave the other unchanged?

1. $P(A)=\frac{1}{6}$, $P(B)=\frac{1}{6}$. Why this step? One favourable face out of six equally likely faces.
2. Both at once would mean the die shows $2$ and $5$ on a single roll — impossible. So $P(A\cap B)=0$. Why this step? The overlap region is empty; on our dartboard the two patches don't touch.
3. Mutually exclusive? $P(A\cap B)=0$ → Yes.
4. Independent? Test $P(A)P(B)=\frac{1}{6}\cdot \frac{1}{6}=\frac{1}{36}$. Compare to $P(A\cap B)=0$. Since $\frac{1}{36}\ne 0$ → No.

Verify: union should be $P(A)+P(B)=\frac{1}{3}$; also equal $P(A\cup B)=\frac{2}{6}=\frac{1}{3}$. ✓ Consistent — disjoint events add cleanly.

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Ex 2 — C2: Independent, overlap positive

Forecast: both heads possible? Does the first flip change the second?

1. $P(A)=\frac{1}{2}$, $P(B)=\frac{1}{2}$. Why this step? Each flip is fair.
2. The four equally likely outcomes are HH, HT, TH, TT. Both heads = HH, one outcome. $P(A\cap B)=\frac{1}{4}$. Why this step? We count the overlap directly instead of assuming — this is how you confirm independence rather than take it on faith.
3. Independent? $P(A)P(B)=\frac{1}{2}\cdot \frac{1}{2}=\frac{1}{4}=P(A\cap B)$ → Yes.
4. Mutually exclusive? $P(A\cap B)=\frac{1}{4}\ne 0$ → No (both can happen).

Verify: conditional check $P(A\mid B)=\frac{P(A\cap B)}{P(B)}=\frac{1/4}{1/2}=\frac{1}{2}=P(A)$. ✓ Knowing the second flip tells you nothing about the first.

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Ex 3 — C3: The zero-probability (degenerate) case

Forecast: an event that cannot occur — is it independent of everything? Mutually exclusive with everything?

1. $P(B)=0$ (no face shows $7$). $P(A)=\frac{3}{6}=\frac{1}{2}$. Why this step? Establish the degenerate input explicitly — this is the edge cell.
2. $P(A\cap B)=0$ because $B$ never happens, so $A\cap B$ never happens. Why this step? If one region is empty, any overlap with it is empty too.
3. Mutually exclusive? $P(A\cap B)=0$ → Yes (trivially — $B$ excludes everything).
4. Independent? Test $P(A)P(B)=\frac{1}{2}\cdot 0=0=P(A\cap B)$ → Yes, also independent.

Verify: the equivalent conditional $P(A\mid B)$ is undefined (dividing by $P(B)=0$), which is exactly why we test with the product form $P(A\cap B)=P(A)P(B)$ here. The product form gives $0=0$. ✓

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Ex 4 — C4: The certain event (limiting case $P(B)=1$)

Forecast: an event that always happens — does knowing it change anything?

1. $P(B)=1$, $P(A)=\frac{1}{6}$. Why this step? $B$ is the whole dartboard.
2. $A\cap B$ = "roll a $3$ and roll something $1$–$6$" = just "roll a $3$". So $P(A\cap B)=\frac{1}{6}$. Why this step? Intersecting with the whole sample space changes nothing.
3. Independent? $P(A)P(B)=\frac{1}{6}\cdot 1=\frac{1}{6}=P(A\cap B)$ → Yes.
4. Mutually exclusive? $P(A\cap B)=\frac{1}{6}\ne 0$ → No.

Verify: $P(A\mid B)=\frac{1/6}{1}=\frac{1}{6}=P(A)$. ✓ A certain event carries no information, so it is independent of everything — the mirror image of Ex 3.

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Ex 5 — C5: Dependent but NOT mutually exclusive (the confusing middle)

Forecast: these two clearly happen together, yet they're related — which cell?

1. $P(A\cap B)=0.2\ne 0$ → not mutually exclusive. Both happen (rainy and cloudy). Why this step? Rule out the far-left point of our overlap number line.
2. Independence test: $P(\text{rain})P(\text{cloudy})=0.2\times 0.4=0.08$. Why this step? This is the "no-information" overlap size — the amber tick on the line.
3. Actual overlap $0.2\ne 0.08$ → Not independent (dependent). Why this step? The real overlap is bigger than the product → positively correlated.
4. Feel it: $P(\text{rain}\mid \text{cloudy})=\frac{0.2}{0.4}=0.5$, but $P(\text{rain})=0.2$. Clouds push rain from $20\%$ up to $50\%$.

Verify: dependence direction check — $P(\text{rain}\mid \text{cloudy})=0.5>0.2=P(\text{rain})$, so knowing "cloudy" raises rain's chance. Positive dependence, not exclusion. This is the cell where beginners wrongly shout "mutually exclusive." ✓

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Ex 6 — C6: Same setup, sign flip (sampling without replacement)

Forecast: with replacement the two draws were independent (parent note). Remove replacement — does independence survive?

1. $P(A)=\frac{4}{52}=\frac{1}{13}$. Why this step? Four aces, fifty-two cards.
2. $P(A\cap B)$: first ace then second ace $=\frac{4}{52}\cdot \frac{3}{51}=\frac{12}{2652}=\frac{1}{221}$. Why this step? After pulling one ace, only $3$ aces remain among $51$ cards — the deck changed.
3. $P(B)$ overall $=\frac{4}{52}=\frac{1}{13}$ (by symmetry, the second draw is an ace with the same marginal chance). Why this step? Before we look at the first card, every position is equally likely to be an ace.
4. Independence test: $P(A)P(B)=\frac{1}{13}\cdot \frac{1}{13}=\frac{1}{169}$. Compare $P(A\cap B)=\frac{1}{221}$. Since $\frac{1}{221}\ne \frac{1}{169}$ → Not independent.

Verify: $\frac{1}{221}=0.004525\dots$ vs $\frac{1}{169}=0.005917\dots$; overlap is smaller than the product → knowing "first was an ace" lowers the second's chance ($3/51<4/52$). Negative dependence — the opposite direction from Ex 5. ✓

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Ex 7 — C7: Conditional independence (independence appears only after conditioning)

Forecast: cough and fever look linked in the crowd — but are they linked once we fix the flu status?

1. Given flu: $P(\text{cough}\mid \text{flu})=0.8$, $P(\text{fever}\mid \text{flu})=0.6$, and $P(\text{cough}\cap \text{fever}\mid \text{flu})=0.48=0.8\times 0.6$ → conditionally independent given flu. ✓ Why this step? This is the exact definition $P(A\cap B\mid C)=P(A\mid C)P(B\mid C)$.
2. Unconditionally, marginal cough $=\frac{80+10}{200}=0.45$; marginal fever $=\frac{60+5}{200}=0.325$. Why this step? Pool both groups to see the raw, un-conditioned picture.
3. Unconditional joint: $P(\text{cough}\cap \text{fever})=\frac{0.48\cdot 100+0.005\cdot 100}{200}=\frac{48+0.5}{200}=0.2425$. Why this step? Add the "both symptoms" counts from each group, divide by total.
4. Test unconditional independence: $0.45\times 0.325=0.14625\ne 0.2425$ → NOT independent unconditionally.

Verify: conditional product $0.8\times 0.6=0.48$ matches the given joint exactly ($0=|0.48-0.48|$). And $0.2425>0.14625$: in the raw crowd, cough and fever do co-occur more than chance — because flu is a hidden common cause. Conditioning on flu removes the link. This is precisely what Naive Bayes and graphical models rely on. ✓

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Ex 8 — C8: Real-world word problem (translate English → sets)

Forecast: if "free" were independent of spam, the filter would be useless — guess the answer.

1. Let $S=$ spam, $F=$ contains "free". Given: $P(F\mid S)=0.6$, $P(F\mid \neg S)=0.1$, $P(S)=0.3$, $P(\neg S)=0.7$. Why this step? Turn each English phrase into a labelled probability.
2. Marginal $P(F)$ by the law of total probability: $P(F)=P(F\mid S)P(S)+P(F\mid \neg S)P(\neg S)=0.6\cdot 0.3+0.1\cdot 0.7=0.18+0.07=0.25$. Why this step? We need $P(F)$ to run the independence test; split it over the two spam states.
3. Independence would require $P(F\mid S)=P(F)$, i.e. $0.6\stackrel{?}{=}0.25$. False → dependent. Why this step? Independence means "knowing spam doesn't move the chance of 'free'." Here it moves it from $0.25$ to $0.6$.
4. Bonus (Bayes, from Bayes' theorem): $P(S\mid F)=\frac{P(F\mid S)P(S)}{P(F)}=\frac{0.18}{0.25}=0.72$.

Verify: $P(S\mid F)=0.72>0.3=P(S)$ — seeing "free" more than doubles the spam probability, confirming strong dependence. A filter needs dependent features; independence here would mean the word carries zero signal. ✓

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Ex 9 — C9: Exam twist (independent in pieces, dependent through a sum)

Forecast: $A,B$ are the poster child for independence. Is $A$ still independent of the derived event $C$?

1. $P(A)=\frac{1}{6}$, $P(B)=\frac{1}{6}$, and $P(A\cap B)=\frac{1}{36}=\frac{1}{6}\cdot \frac{1}{6}$ → $A,B$ independent. Why this step? Separate physical dice — the standard independent pair.
2. Sum-to-$8$ outcomes: $(2,6),(3,5),(4,4),(5,3),(6,2)$ → $P(C)=\frac{5}{36}$. Why this step? List every way to hit the target sum; leave none out.
3. $A\cap C$ = first die $4$ and sum $8$ = only $(4,4)$ → $P(A\cap C)=\frac{1}{36}$. Why this step? Fixing the first die to $4$ forces the second to $4$ for the sum to work.
4. Test: $P(A)P(C)=\frac{1}{6}\cdot \frac{5}{36}=\frac{5}{216}\ne \frac{1}{36}=\frac{6}{216}$ → ==$A$ and $C$ NOT independent==.

Verify: $P(C\mid A)=\frac{1/36}{1/6}=\frac{1}{6}$ but $P(C)=\frac{5}{36}\approx 0.139$; knowing the first die is $4$ lowers the sum-8 chance to $\frac{1}{6}\approx 0.167$… wait, it raises it. Indeed $\frac{1}{6}=0.1\overline{6}>0.139$: information gain either way, so dependent. The trap: independence of $A,B$ does not propagate to functions like $C=A+B$. ✓

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Recall Rapid self-test

Overlap $=0$ means what property? ::: Mutually exclusive (disjoint). Overlap $=P(A)P(B)$ means what property? ::: Independent. Can a zero-probability event be BOTH disjoint and independent of $A$? ::: Yes — Ex 3; the "opposites" rule needs both probabilities positive. Two dice are independent — must "first die = 4" be independent of "sum = 8"? ::: No — Ex 9; functions of independent variables can be dependent. Sampling two cards without replacement: independent? ::: No — Ex 6; the first draw changes the deck.

See also the parent Independence and mutual exclusivity and prerequisites 1.3.01-Basic-probability-concepts, 1.3.02-Conditional-probability.