Independent events — multiplication rule: Worked examples

2.7.7 · D3 · Maths › Statistics & Probability — Intermediate › Independent events — multiplication rule

Scenario matrix

Is topic ka har problem inhi cells mein se exactly ek mein aata hai. Sahi move cell se cell par alag hota hai.

Cell	Ise woh cell kya banata hai	Sahi move	Example mein
A. Genuinely independent	Physically alag trials; ek doosre ko touch nahi kar sakta	$P (A) P (B)$ multiply karo	Ex 1
B. With replacement	Setup draws ke beech reset hoti hai	Multiply karo — independence restore ho jaati hai	Ex 2
C. Without replacement	Pehla event setup badal deta hai	$P (A) P (B ∣ A)$ use karo, na ki product	Ex 3
D. "At least one" (complement)	Kai trials, $\geq 1$ success/failure chahiye	$1 - \prod (1 - p_{i})$	Ex 4
E. Degenerate: certain / impossible event	Koi $P = 1$ ya $P = 0$ ho	Multiply karo — lekin dekho yeh kya force karta hai	Ex 5
F. Mutually exclusive trap	Events saath nahi ho sakte ( $P (A \cap B) = 0$ )	Kabhi multiply mat karo — ye dependent hain	Ex 6
G. Real-world word problem	Independence story mein chupi hai	Independence test karo, phir multiply karo	Ex 7
H. Exam twist: solve backwards	Joint answer diya hai, missing $P$ nikalni hai	Rule rearrange karo	Ex 8
I. Limiting behaviour	"Almost sure hone ke liye kitne trials?"	$1 - (1 - p)^{n} \to 1$ push karo	Ex 9

Neeche ki picture poori page ek board par hai — ek decision tree jo kisi bhi "AND" problem ko uski cell mein route karta hai. Har example jo aage aata hai woh is tree ka ek leaf hai, isliye jab bhi lost feel karo yahan wapas aao.

Figure — Independent events — multiplication rule

Cell A — Genuinely independent

Neeche ka grid woh verification hai: 12 equal cells, aur multiply sirf ek pale-yellow cell pick karta hai. Highlighted cell count karo — woh ratio $1/12$ answer hai, compute nahi dekha gaya.

Cell B — With replacement (independence restore hoti hai)

Cell C — Without replacement (the trap)

Neeche ke do side-by-side trees yeh contrast unmissable banate hain: blue (replaced) tree apni doosri branch par $4/10$ rakhta hai, pink (not replaced) tree $3/9$ par aa jaata hai. Same pehli branch, alag doosri branch — woh ek badla hua number $0.16$ aur $0.133$ ke beech ka poora fark hai.

Cell C⁺ — Do se zyada events: the chain rule

Cell D — "At least one" complement ke through

Cell E — Degenerate: a certain or impossible event

Worked example Ex 5 ·

$1$ aur $0$ se multiply karna Ek drawer light probability $1$ se on hoti hai (certain hai) jab bhi drawer kholo, aur independently andar ek bulb probability $0$ se already dead hai (brand-new, kabhi dead nahi). Maano $A = light on$ ( $P (A) = 1$ ), $B = bulb dead$ ( $P (B) = 0$ ). $P (A \cap B)$ nikalo.

Forecast: agar ek event impossible hai toh "dono hona" kya hona chahiye?

$P (A) = 1, P (B) = 0$ . Yeh step kyun? Given — yeh probability ke degenerate endpoints hain.
Yahan hum $P (A ∣ B) = P (A)$ par lean nahi kar sakte, kyunki $P (B) = 0$ conditional $P (A ∣ B) = \frac{P ( A \cap B )}{P ( B )}$ ko $\frac{0}{0}$ bana deta hai — literally undefined. Yeh step kyun? Bar-formula ko positive denominator chahiye; ek impossible condition ke saath koi "woh world nahi jahan $B$ hua" shrink karne ke liye.
Toh seedha argue karo: kyunki $A \cap B \subseteq B$ , hame milta hai $0 \leq P (A \cap B) \leq P (B) = 0$ , jo $P (A \cap B) = 0$ force karta hai. Aur product $P (A) P (B) = 1 \times 0 = 0$ agree karta hai — isliye multiplication rule phir bhi hold karta hai, chahe conditional route unavailable ho. Yeh step kyun? Hum answer monotonicity se prove karte hain, undefined quantity se nahi.

Verify: "dono hona" impossible part ke chance se zyada nahi ho sakta, isliye $P (A \cap B) = 0$ . ✓ Doosre endpoint par sanity check: ek certain event $A$ ( $P (A) = 1$ ) satisfy karta hai $P (A \cap C) = 1 \cdot P (C) = P (C)$ kisi bhi independent $C$ ke liye — yahan $P (C) > 0$ har conditional well-defined rakhta hai, aur certainty koi constraint nahi laata, bilkul sahi.

Cell F — Mutually exclusive, independent NAHI hai

Worked example Ex 6 · Classic confusion, quantified

Ek fair die ka ek roll. Maano $A = roll a 2$ aur $B = roll a 5$ . Kya $P (A \cap B) = P (A) P (B)$ hai?

Forecast: "alag outcomes" wale words independence ki taraf tempt karte hain — lekin check karo.

$P (A) = \frac{1}{6}, P (B) = \frac{1}{6}$ . Yeh step kyun? Ek face each.
$A$ aur $B$ dono ek roll par nahi ho sakte → $P (A \cap B) = 0$ . Yeh step kyun? Ek die ek face dikhata hai; $2$ aur $5$ Mutually Exclusive Events hain.
Rule test karo: $P (A) P (B) = \frac{1}{6} \cdot \frac{1}{6} = \frac{1}{36} \neq = 0 = P (A \cap B) .$ Yeh step kyun? Product rule fail hota hai → yeh events dependent hain, independent nahi.
Intuition: yeh jaanna ki tumne $2$ roll kiya $5$ ko impossible bana deta hai — yeh information transfer hai, independence ka opposite.

Verify: $\frac{1}{36} \neq = 0$ , isliye jo equality independence define karti hai woh violate ho gayi. ✓ Exclusive (dono possible) $\Rightarrow$ dependent — hamesha.

Cell G — Real-world word problem (independence chupi hai)

Worked example Ex 7 · Do commuters, ek storm

Kisi bhi subah, Asha apni bus probability $0.8$ se pakad leti hai aur, ek alag route par, Ben apni bus probability $0.75$ se pakad leta hai. $P (both catch their bus)$ nikalo — phir trap ke liye dobara padho.

Forecast: kya multiply apply hota hai, ya koi hidden link hai?

Independence test karo. Alag routes, alag buses → normally ek doosre ko affect nahi karta → independent treat karo. Yeh step kyun? Word problems mein multiply karne se pehle tumhe ek shared cause (storm, shared bus) dhoondna hota hai.
$P (both) = 0.8 \times 0.75 = 0.6.$ Yeh step kyun? Independent AND → product.
The twist: agar ek akela storm dono buses delay kar sakta, toh misses correlated honge aur product galat hoga. Problem separate routes batata hai bina common cause ke → assumption valid rehti hai.

Verify: $0.6$ zero aur "surely both" cap $min (0.8, 0.75) = 0.75$ ke beech hai, aur $0.8 \cdot 0.75$ ke equal hai. Units pure probabilities hain (dimensionless), $0 \leq 0.6 \leq 1$ . ✓

Cell H — Exam twist: rule backwards solve karo

Cell I — Limiting behaviour: "almost sure hone ke liye kitne?"

Worked example Ex 9 · Six ke liye roll karte raho

Ek fair die ka har roll $6$ dikhata hai probability $p = \frac{1}{6}$ se, rolls independent hain. Kitne rolls $n$ se $P (at least one 6) \geq 0.9$ ho jaata hai? $n \to \infty$ par kya hota hai?

Forecast: compute karne se pehle rolls ki number guess karo.

$P (no 6 in n rolls) = (\frac{5}{6})^{n}$ . Yeh step kyun? Independent failures multiply hote hain — Binomial Distribution ka "all-fail" corner.
$P (at least one 6) = 1 - (\frac{5}{6})^{n}$ . Yeh step kyun? Phir se Complement Rule — "at least one" matlab "not none."
Chahiye $1 - (\frac{5}{6})^{n} \geq 0.9 ⟺ (\frac{5}{6})^{n} \leq 0.1$ . Log lo: $n \geq \frac{ln 0.1}{ln ( 5/6 )} \approx \frac{- 2.3026}{- 0.1823} \approx 12.63$ , isliye $n = 13$ . Yeh step kyun? Log repeated product ko ek solvable line mein badal deta hai.
Limit: jaise $n \to \infty$ , $(\frac{5}{6})^{n} \to 0$ , isliye $P (at least one 6) \to 1$ . Yeh step kyun? $1$ se neeche ka base growing power par raise kiya jaaye toh vanish ho jaata hai — tum eventually almost certain ho jaate ho, lekin finite rolls mein exactly $1$ kabhi nahi.

Verify: $1 - (5/6)^{12} \approx 0.888 < 0.9$ lekin $1 - (5/6)^{13} \approx 0.907 \geq 0.9$ , isliye $13$ pehla roll count hai jo kaam karta hai. ✓

Neeche ka curve yeh climb dikhata hai: dekho blue staircase kahan pehli baar pink $0.9$ line ke upar jaati hai (woh $n = 13$ hai), aur dekho yeh $1$ par yellow ceiling ki taraf flatten hoti hai lekin kabhi touch nahi karti.

Recall Which cell am I in? (quick triage)

Kya pehle event ne doosre ka setup badala? ::: Agar haan → Cell C ( $P (B ∣ A)$ use karo); agar nahi → multiply karo. Problem "at least one" kehti hai — kya karna hai? ::: Complement: $1 - \prod (1 - p_{i})$ (Cells D, I). Ek single trial ke do outcomes saath nahi ho sakte — independent hain? ::: Nahi — positive prob ke saath mutually exclusive ⇒ dependent (Cell F). $P (A \cap B)$ aur $P (A)$ dono independent given hain, $P (B)$ nikalo? ::: Divide karo: $P (B) = P (A \cap B) / P (A)$ (Cell H). Teen dependent stages — kaunsa rule? ::: Chain rule $P (A) P (B ∣ A) P (C ∣ A \cap B)$ (Cell C⁺).

Connections

General Multiplication Rule — Cell C / chain-rule engine: $P (A \cap B) = P (A) P (B ∣ A)$ .
Conditional Probability — $P (B ∣ A)$ supply karta hai aur is poori page par use hone wale " $∣$ " bar ka meaning deta hai.
Complement Rule — har "at least one" cell ko power deta hai (D, I).
Mutually Exclusive Events — independence se Cell F ka contrast.
Binomial Distribution — repeated independent trials, Cell I ka natural ghar.
Bayes' Theorem — Cell H ke peeche ka deeper reversal engine.