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

(a) Ek fair die ka matlab hai ki saatey chhekhne equally likely hain, toh hum count karte hain — yeh classical hai. (b) Favourable outcomes = $S$ mein $3$ ke multiples = $\{3,6\}$, toh $m=2$. Total $n=6$. $P(A)=\frac{m}{n}=\frac{2}{6}=\frac{1}{3}.$

Pin mein koi symmetry nahi hoti — "point-up" aur "point-down" equally likely nahi hain, toh count karne ke liye equal outcomes ka koi $n$ nahi hai. Hume measure karna padega: yeh empirical hai. $P(\text{point-up})\approx \frac{f}{N}=\frac{118}{200}=\mathrm{0.59.}$

Axiom 1, Non-negativity: $P(A)\ge 0$ har event ke liye. Probability kabhi negative nahi ho sakti, toh $-0.2$ illegal hai. (Yeh No–Sure–Split mein "No" hai.)

Saare $52$ cards equally likely hain → classical. Overlap $F\cap D$ = face cards jo spades bhi hain = $\{$J♠, Q♠, K♠$\}$, toh $|F\cap D|=3$. $P(F)=\frac{12}{52},P(D)=\frac{13}{52},P(F\cap D)=\frac{3}{52}.$ Addition rule (double-counted overlap ko subtract karo): $P(F\cup D)=\frac{12}{52}+\frac{13}{52}-\frac{3}{52}=\frac{22}{52}=\frac{11}{26}.$

"At least one head" ke bahut saare cases hain; iska complement "koi head nahi" (sab tails, TTTT) mein exactly ek case hai. Complement rule use karo. $2^4=16$ sequences mein se har ek equally likely hai, aur TTTT unhi mein se ek hai: $P(\text{no head})=\frac{1}{16}.$ $P(\text{at least one head})=1-\frac{1}{16}=\frac{15}{16}.$

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

Overlap ke liye addition rule rearrange karo: $P(A\cap B)=P(A)+P(B)-P(A\cup B)=0.5+0.4-0.7=\mathrm{0.2.}$ Ab $A^c\cap B^c$ ka matlab hai "na $A$, na $B$" $=(A\cup B{)}^c$ (De Morgan). Complement rule use karo: $P(A^c\cap B^c)=1-P(A\cup B)=1-0.7=\mathrm{0.3.}$

$m/n$ kyun nahi? Faces equally likely nahi hain, toh counting fail hoti hai — hum seedha axioms use karte hain. (a) Normalisation (axiom 2, $P(S)=1$) aur chhe disjoint singletons par additivity: $c(1+2+3+4+5+6)=1\Rightarrow 21c=1\Rightarrow c=\frac{1}{21}.$ (b) Even faces $\{2,4,6\}$ disjoint hain, toh add karo: $P(\text{even})=c(2+4+6)=\frac{12}{21}=\frac{4}{7}.$

(a) Empirical: $P(H)\approx \frac{620}{1000}=\mathrm{0.62.}$ (b) Independent events ke liye parent ki Independence of Events deti hai $P(H\cap H)=P(H)P(H)$: $P(HH)\approx 0.62\times 0.62=\mathrm{0.3844.}$

"At least one" ka direct counting messy hai; iska complement "kisi bhi roll mein koi six nahi" clean hai, aur rolls independent hain toh probabilities multiply hoti hain. $P(\text{no six on one roll})=\frac{5}{6}.$ $P(\text{no six in four rolls})={\left(\frac{5}{6}\right)}^4=\frac{625}{1296}.$ $P(\text{at least one six})=1-\frac{625}{1296}=\frac{671}{1296}\approx \mathrm{0.5177.}$

Event "red" ko split karo ki hum kaun sa urn use kar rahe hain (ek disjoint split — Law of Total Probability, Conditional Probability & Bayes' Theorem par based): $P(\text{red})=P(X)P(\text{red}\mid X)+P(Y)P(\text{red}\mid Y).$ Yahaan $P(X)=P(Y)=\frac{1}{2}$, $P(\text{red}\mid X)=\frac{3}{5}$, $P(\text{red}\mid Y)=\frac{1}{5}$: $P(\text{red})=\frac{1}{2}\cdot \frac{3}{5}+\frac{1}{2}\cdot \frac{1}{5}=\frac{3}{10}+\frac{1}{10}=\frac{4}{10}=\frac{2}{5}.$

Saare committees equally likely hain → classical, aur hum Permutations & Combinations se count karte hain. Total committees: $\left(\genfrac{}{}{0ex}{}{10}{3}\right)=120.$ Complement "koi woman nahi" = saare $3$ $6$ mardon mein se: $\left(\genfrac{}{}{0ex}{}{6}{3}\right)=20,P(\text{no woman})=\frac{20}{120}=\frac{1}{6}.$ $P(\text{at least one woman})=1-\frac{1}{6}=\frac{5}{6}.$

Exact addition rule se shuru karo (parent mein axioms se derive kiya gaya): $P(A\cup B)=P(A)+P(B)-P(A\cap B).$ Axiom 1 se, $P(A\cap B)\ge 0$. Ek non-negative number subtract karne se total sirf ghatt sakta hai, toh: $P(A\cup B)=P(A)+P(B)-\underset{\ge 0}{\underbrace{P(A\cap B)}}\le P(A)+P(B).$ Equality exactly tab hoti hai jab $P(A\cap B)=0$ — khaas taur par jab $A$ aur $B$ disjoint hon (tab axiom 3 seedha addition deta hai). ∎

Agar $A\subseteq B$ hai, toh $B$ ko do disjoint pieces mein split karo: $A$ ke andar wala hissa, aur baaki. $B=A\cup (B\cap A^c),A\cap (B\cap A^c)=\varnothing .$ Axiom 3 se (disjoint add hote hain): $P(B)=P(A)+P(B\cap A^c).$ Kyunki $P(B\cap A^c)\ge 0$ hai (axiom 1), hume milta hai $P(B)\ge P(A)$. ∎ (Isliye bade events ki probability kabhi chhoti nahi hoti — ek picture: $A$ ek chhota region hai jo poori tarah $B$ ke andar baitha hai.)

(a) $A\cup A^c=S$, toh $P(A\cup A^c)=P(S)=1$ (axiom 2). Hamesha $1$, chahe $A$ kuch bhi ho. (b) Monotonicity se, $A\cap B\subseteq B$ toh $P(A\cap B)\le P(B)$. Aur $A\cap B\supseteq$ ... zyaada precisely: $B=(A\cap B)\cup (A^c\cap B)$ disjoint hai, toh $P(B)=P(A\cap B)+P(A^c\cap B)$. Ab $A^c\cap B\subseteq A^c$ aur $P(A^c)=1-P(A)=0$, toh monotonicity se $P(A^c\cap B)=0$. Isliye $P(A\cap B)=P(B).$ (c) Ek continuous space mein ek exact point ke liye, $P(\{0.5\})=0$. Phir bhi dart wahaan land kar sakta hai — toh yeh genuinely probability $0$ wala event hai jo impossible nahi hai. Axioms $P=0$ ko permit karte hain bina $A=\varnothing$ ke (parent ki teesri "steel-manned" mistake dekho).

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; $1$; $P(A\cap B)=P(B)$; $0$ lekin possible.