Question bank — Probability — classical, empirical, axiomatic (Kolmogorov axioms)

False — with $n$ the total outcomes and $m$ the favourable ones, the formula also needs the $n$ outcomes to be equally likely; a bent coin has $n=2$ but the halves are not symmetric, so $m/n$ is meaningless there. (In Set Theory — Union, Intersection, Complement, $A$ is just a subset of the box $S$; counting its elements only gives a probability when every element is equally weighted.)

True — for disjoint events axiom 3 gives $P(A\cup B)=P(A)+P(B)\le P(S)=1$, so exceeding 1 forces $A\cap B\ne \varnothing$ (the double-counted overlap is what pushes the sum above 1).

False — $P=0$ can attach to nonempty events. This happens for any exact point on a continuous dartboard, but also in a countably-infinite space with a non-uniform assignment: one can give a specific outcome weight 0 while it stays a possible result. The axioms permit $P(A)=0$ without $A=\varnothing$.

False — short runs are noisy; the Law of Large Numbers (relative frequency $f/N$ converging to the true value) only promises the settling as $N\to \infty$, so more trials means a tighter, more reliable estimate.

True — $A$ and its complement $A^c$ (the white region outside $A$'s blob) are disjoint with $A\cup A^c=S$, so axiom 3 gives $P(A)+P(A^c)=P(S)$, and axiom 2 makes the right side 1.

True — $A$'s blob sits entirely inside $B$'s; write $B=A\cup (B\cap A^c)$ as disjoint pieces, so $P(B)=P(A)+P(B\cap A^c)$, and since the extra piece has probability $\ge 0$ (axiom 1), $P(B)\ge P(A)$.

False — it is derived by splitting $A\cup B$ into the disjoint pieces $A$ and $B\cap A^c$ and applying axiom 3; nothing new is assumed (full steps in the "Why questions" below).

False for infinite spaces — as with the $P=0$ case, an event can have $P=1$ yet miss occasionally (its complement is a nonempty $P=0$ set); it is guaranteed on every trial only in finite sample spaces.

True — the axiom is stated for a whole sequence $A_1,A_2,\dots$, which is why the $\sigma$-algebra is required to be closed under countable unions: the union ${\bigcup }_i{A}_i$ must itself be an event for $P$ of it to be defined.

The King of Hearts is counted in both the 4 kings and the 13 hearts (it lives in the $A\cap B$ overlap), so you must subtract $P(K\cap H)=\frac{1}{52}$: the answer is $\frac{16}{52}=\frac{4}{13}$.

This is the gambler's fallacy; independent tosses have no memory, so $P(H)=0.5$ regardless of history. The Law of Large Numbers balances things by swamping early results with many new ones, not by compensating.

Only valid if all three are pairwise disjoint. With overlaps, adding the three singles counts each pairwise overlap twice, so subtract $P(A\cap B)$, $P(A\cap C)$, $P(B\cap C)$; but that subtraction removes the triple region $A\cap B\cap C$ three times after it was added three times (net zero), so add it back once: $P(A\cup B\cup C)=P(A)+P(B)+P(C)-P(A\cap B)-P(A\cap C)-P(B\cap C)+P(A\cap B\cap C)$.

The 11 sums are not equally likely, so classical counting on them is invalid. Count the 36 equally-likely ordered pairs instead: $P(7)=\frac{6}{36}=\frac{1}{6}$.

That silently applies classical symmetry; a tack has no symmetry between its two states, so you must measure the frequency (e.g. $290/500=0.58$), not assume equal halves.

Non-negativity does not force a positive intersection; disjoint events (blobs that do not touch) have $P(A\cap B)=0$ even when both events are individually likely.

The complement rule uses the whole event $A\cup B$, so it is $1-P(A\cup B)=1-P(A)-P(B)+P(A\cap B)$ — the overlap term survives.

Classical needs a symmetry argument and empirical needs infinite trials; the three axioms instead state the minimal rules any probability must obey, so both older methods become valid ways to assign numbers inside one rigorous framework.

The derivation assigns each of the $n$ outcomes weight $\frac{1}{n}$ so they sum to 1; if they are not equally likely, no single value $\frac{1}{n}$ is correct and $P(A)=m\cdot \frac{1}{n}$ collapses.

Because "and/or/not" map exactly to intersection/union/complement — e.g. "$A$ or $B$" is the union blob $A\cup B$ in the Venn box (see Set Theory — Union, Intersection, Complement) — turning logical reasoning into computable set algebra.

"At least one" splits into many cases, but its complement "none" is usually a single case; computing $1-P(\text{none})$ replaces a long sum with one term.

Split the union into non-overlapping pieces: $A\cup B=A\cup (B\cap A^c)$, disjoint, so axiom 3 gives $P(A\cup B)=P(A)+P(B\cap A^c)$. Separately $B=(B\cap A)\cup (B\cap A^c)$, disjoint, so $P(B\cap A^c)=P(B)-P(A\cap B)$. Substituting yields $P(A)+P(B)-P(A\cap B)$; geometrically, $P(A)+P(B)$ shades the overlap twice, so one copy is removed.

A single relative frequency $f/N$ is just an estimate; the LLN guarantees this estimate converges to a fixed true value as $N\to \infty$, which is what lets us call it a probability.

Because it follows: $S=S\cup \varnothing$ disjoint gives $P(S)=P(S)+P(\varnothing )$, so $P(\varnothing )=0$ — no separate assumption needed.

If two blobs overlap, their outcomes are shared, and adding $P(A)+P(B)$ double-counts that shared region; only disjoint blobs partition the union cleanly so the probabilities simply sum.

$P(S)=1$ by axiom 2 (normalisation) — the statement "some outcome in the box occurs" is certain.

Yes — they must overlap, since disjointness would force $\frac{3}{4}+\frac{3}{4}=1.5>1$; their intersection has probability at least $\frac{3}{4}+\frac{3}{4}-1=\frac{1}{2}$.

Generally no — Independence of Events means $P(A\cap B)=P(A)P(B)$ (the overlap is the product), while mutually exclusive means $P(A\cap B)=0$ (no overlap at all); both hold together only if one event already has probability 0.

Spin a pointer that lands uniformly on a circle. The event "the pointer does not land exactly at the 12 o'clock mark" has probability 1 (that single mark is a $P=0$ point), yet on some spin it could land there — so a $P=1$ event can "miss occasionally" because its exception is a nonempty $P=0$ set.

The primes are $\{2,3,5\}$, so $P=\frac{3}{6}=\frac{1}{2}$; classical applies because a symmetric die makes all six faces equally likely, giving $n=6$, $m=3$.

Only the empirical (frequentist) one — you measure $f/N$ over many trials, because classical counting requires the equal-likelihood you cannot justify.

The outcomes $\{H,TH,TTH,\dots \}$ form an infinite disjoint list; countable additivity lets us sum their individual probabilities $\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\cdots =1$, confirming $P(S)=1$ — impossible with only a two-event rule.