Exercises — Probability — classical, empirical, axiomatic (Kolmogorov axioms)

(a) A fair die means all six faces are equally likely, so we count — this is classical. (b) Favourable outcomes = multiples of $3$ in $S$ = $\{3,6\}$, so $m=2$. Total $n=6$. $P(A)=\frac{m}{n}=\frac{2}{6}=\frac{1}{3}.$

A pin has no symmetry — "point-up" and "point-down" are not equally likely, so there is no $n$ of equal outcomes to count. We must measure: this is empirical. $P(\text{point-up})\approx \frac{f}{N}=\frac{118}{200}=\mathrm{0.59.}$

Axiom 1, Non-negativity: $P(A)\ge 0$ for every event. A probability can never be negative, so $-0.2$ is illegal. (This is the "No" in No–Sure–Split.)

All $52$ cards are equally likely → classical. The overlap $F\cap D$ = face cards that are spades = $\{$J♠, Q♠, K♠$\}$, so $|F\cap D|=3$. $P(F)=\frac{12}{52},P(D)=\frac{13}{52},P(F\cap D)=\frac{3}{52}.$ Addition rule (subtract the double-counted overlap): $P(F\cup D)=\frac{12}{52}+\frac{13}{52}-\frac{3}{52}=\frac{22}{52}=\frac{11}{26}.$

"At least one head" has many cases; its complement "no heads" (all tails, TTTT) has exactly one. Use the complement rule. Each of the $2^4=16$ sequences is equally likely, and TTTT is one of them: $P(\text{no head})=\frac{1}{16}.$ $P(\text{at least one head})=1-\frac{1}{16}=\frac{15}{16}.$

Ordered pairs keep outcomes equally likely → $n=36$. Sums $\le 4$: sum $2$: $(1,1)$; sum $3$: $(1,2),(2,1)$; sum $4$: $(1,3),(2,2),(3,1)$. That is $m=1+2+3=6$. $P(\text{sum}\le 4)=\frac{6}{36}=\frac{1}{6}.$

Rearrange the addition rule to solve for the overlap: $P(A\cap B)=P(A)+P(B)-P(A\cup B)=0.5+0.4-0.7=\mathrm{0.2.}$ Now $A^c\cap B^c$ means "neither $A$ nor $B$" $=(A\cup B{)}^c$ (De Morgan). Use the complement rule: $P(A^c\cap B^c)=1-P(A\cup B)=1-0.7=\mathrm{0.3.}$

Why not $m/n$? Faces are not equally likely, so counting fails — we use the axioms directly. (a) Normalisation (axiom 2, $P(S)=1$) and additivity over the six disjoint singletons: $c(1+2+3+4+5+6)=1\Rightarrow 21c=1\Rightarrow c=\frac{1}{21}.$ (b) Even faces $\{2,4,6\}$ are disjoint, so add: $P(\text{even})=c(2+4+6)=\frac{12}{21}=\frac{4}{7}.$

(a) Empirical: $P(H)\approx \frac{620}{1000}=\mathrm{0.62.}$ (b) For independent events the parent's Independence of Events gives $P(H\cap H)=P(H)P(H)$: $P(HH)\approx 0.62\times 0.62=\mathrm{0.3844.}$

Direct counting of "at least one" is messy; its complement "no six in any roll" is clean, and rolls are independent so probabilities multiply. $P(\text{no six on one roll})=\frac{5}{6}.$ $P(\text{no six in four rolls})={\left(\frac{5}{6}\right)}^4=\frac{625}{1296}.$ $P(\text{at least one six})=1-\frac{625}{1296}=\frac{671}{1296}\approx \mathrm{0.5177.}$

Split the event "red" by which urn we used (a disjoint split — Law of Total Probability, built on Conditional Probability & Bayes' Theorem): $P(\text{red})=P(X)P(\text{red}\mid X)+P(Y)P(\text{red}\mid Y).$ Here $P(X)=P(Y)=\frac{1}{2}$, $P(\text{red}\mid X)=\frac{3}{5}$, $P(\text{red}\mid Y)=\frac{1}{5}$: $P(\text{red})=\frac{1}{2}\cdot \frac{3}{5}+\frac{1}{2}\cdot \frac{1}{5}=\frac{3}{10}+\frac{1}{10}=\frac{4}{10}=\frac{2}{5}.$

All committees equally likely → classical, and we count with Permutations & Combinations. Total committees: $\left(\genfrac{}{}{0ex}{}{10}{3}\right)=120.$ Complement "no woman" = all $3$ from the $6$ men: $\left(\genfrac{}{}{0ex}{}{6}{3}\right)=20,P(\text{no woman})=\frac{20}{120}=\frac{1}{6}.$ $P(\text{at least one woman})=1-\frac{1}{6}=\frac{5}{6}.$

Start from the exact addition rule (derived in the parent from the axioms): $P(A\cup B)=P(A)+P(B)-P(A\cap B).$ By axiom 1, $P(A\cap B)\ge 0$. Subtracting a non-negative number can only decrease the total, so: $P(A\cup B)=P(A)+P(B)-\underset{\ge 0}{\underbrace{P(A\cap B)}}\le P(A)+P(B).$ Equality holds exactly when $P(A\cap B)=0$ — in particular when $A$ and $B$ are disjoint (then axiom 3 gives plain addition). ∎

If $A\subseteq B$, split $B$ into two disjoint pieces: the part inside $A$, and the rest. $B=A\cup (B\cap A^c),A\cap (B\cap A^c)=\varnothing .$ By axiom 3 (disjoint add): $P(B)=P(A)+P(B\cap A^c).$ Since $P(B\cap A^c)\ge 0$ (axiom 1), we get $P(B)\ge P(A)$. ∎ (This is why bigger events never have smaller probability — a picture: $A$ is a smaller region sitting entirely inside $B$.)

(a) $A\cup A^c=S$, so $P(A\cup A^c)=P(S)=1$ (axiom 2). Always $1$, whatever $A$ is. (b) From monotonicity, $A\cap B\subseteq B$ so $P(A\cap B)\le P(B)$. Also $A\cap B\supseteq$ ... more precisely: $B=(A\cap B)\cup (A^c\cap B)$ disjoint, so $P(B)=P(A\cap B)+P(A^c\cap B)$. Now $A^c\cap B\subseteq A^c$ and $P(A^c)=1-P(A)=0$, so by monotonicity $P(A^c\cap B)=0$. Hence $P(A\cap B)=P(B).$ (c) For a single exact point in a continuous space, $P(\{0.5\})=0$. Yet the dart can land there — so this is a genuine event of probability $0$ that is not impossible. The axioms permit $P=0$ without $A=\varnothing$ (see the parent's third "steel-manned" mistake).

L1: $\frac{1}{3}$, $0.59$, axiom 1. — L2: $\frac{11}{26}$, $\frac{15}{16}$, $\frac{1}{6}$. — L3: $P(A\cap B)=0.2$, $P(A^c\cap B^c)=0.3$; $c=\frac{1}{21}$, even $=\frac{4}{7}$; $0.62$, $0.3844$. — L4: $\frac{671}{1296}$, $\frac{2}{5}$, $\frac{5}{6}$. — L5: proofs; $1$; $P(A\cap B)=P(B)$; $0$ but possible.