Independent events — multiplication rule: Exercises

2.7.7 · D4 · Maths › Statistics & Probability — Intermediate › Independent events — multiplication rule

Yaad rakho woh ek tool jo is poori page par test ho raha hai: $P (A \cap B) = P (A) P (B) — valid only when A, B are ==independent==.$ Baaki sab cheezein (Complement Rule trick, General Multiplication Rule, Binomial Distribution) isi ke upar bani hain. Agar neeche koi symbol unfamiliar lage, toh parent note dobara dekho.

Ek chhoti si glossary taaki koi notation surprise na kare:

$P (A)$ ::: ek number ( $0$ aur $1$ ke beech) jo probability batata hai ki event $A$ hoga.
$A \cap B$ ::: " $A$ aur $B$ " — dono hote hain. $\cap$ symbol do events ka overlap hai.
$A \cup B$ ::: " $A$ ya $B$ " (ya dono) — kam se kam ek hota hai. $\cup$ symbol do events ka mila hua region hai.
$B^{c}$ ::: " $B$ nahin" — complement, woh sab kuch jahan $B$ nahin hota. Dekho Complement Rule.
$P (A ∣ B)$ ::: " $A$ ki probability given $B$ pehle ho chuka hai" — dekho Conditional Probability.
$\prod$ ::: ek bada "inhe sab ko milao" ka sign, jaise $\sum$ hota hai lekin products ke liye.

Level 1 — Recognition

Goal: decide karo ki events independent hain ya nahin, aur raw rule ek baar apply karo.

Recall Solution L1·Q1

Independence check. Coin feel nahin kar sakta die kya kar raha hai — koi physical link nahin hai. Toh yeh jaanna ki die $6$ par landa, coin ke heads aane ki chance ko $\frac{1}{2}$ par hi rehne deta hai. Yahi exact matlab hai $P (heads ∣ 6) = P (heads)$ ka, toh independent. ✓ Rule apply karo. $P (heads \cap 6) = P (heads) P (6) = \frac{1}{2} \cdot \frac{1}{6} = \frac{1}{12} \approx 0.0833.$

Recall Solution L1·Q2

$P (A) = \frac{4}{52} = \frac{1}{13}$ (chaar Kings).
$P (B) = \frac{13}{52} = \frac{1}{4}$ (terah Hearts).
$A \cap B =$ "King of Hearts", ek card: $P (A \cap B) = \frac{1}{52}$ . Test: $P (A) P (B) = \frac{1}{13} \cdot \frac{1}{4} = \frac{1}{52}$ . Yeh $P (A \cap B)$ ke barabar hai, toh $A$ aur $B$ independent hain. ✓ (Dhyan do: yahan independence deck ki balanced structure ka ek coincidence hai, koi physical "alag-alag cheez" wali story nahin — definition phir bhi yahi decide karti hai.)

Level 2 — Application

Goal: rule ko repeated trials par chain karo aur "at least one" complement use karo.

Recall Solution L2·Q1

Replace karne se bag bilkul waise hi wapas aa jaata hai jaise pehle tha, toh doosri draw ke odds untouched rehte hain → independent. $P (R_{1}) = \frac{4}{10}, P (R_{2}) = \frac{4}{10} .$ $P (R_{1} \cap R_{2}) = \frac{4}{10} \cdot \frac{4}{10} = \frac{16}{100} = 0.16.$

Recall Solution L2·Q2

"At least one miss" "no misses" ka ulta hai = "teeno made". Complement Rule use karo. $P (all made) = 0. 8^{3} = 0.512.$ $P (at least one miss) = 1 - 0.512 = 0.488.$ Complement kyun? "1 ya 2 ya 3 misses" count karna teen messy cases hain; "all made" ek single clean product hai.

Recall Solution L2·Q3

Har ek trigger karne mein fail karta hai $1 - 0.7 = 0.3$ probability se. Sab saath fail hote hain (failures ki independence): $P (none trigger) = 0. 3^{3} = 0.027.$ $P (at least one triggers) = 1 - 0.027 = 0.973.$

Level 3 — Analysis

Goal: independence ko dependence se alag karo, aur dekho answer kaise shift hota hai.

Neeche wali figure same bag se do baar draw dikhati hai. Left mein, marble replace ki jaati hai, toh draw 2 ke liye bag bilkul start jaisa hai — green label padhta hai $P (R_{2}) = 4/10 = 0.40$ . Right mein, ek red hata di gayi hai (dashed red outline "removed" mark ke saath dikhaya gaya hai), toh draw 2 sirf $9$ mein se $3$ red dekhta hai — orange label padhta hai $P (R_{2} ∣ R_{1}) = 3/9 = 0.33$ . Red count ka ghatna exactly wahi reason hai kyun without-replacement answer chhota aata hai.

Figure — Independent events — multiplication rule

Recall Solution L3·Q1

Pehli draw bag ko change kar deti hai, toh doosri draw conditional hai — independent nahin. General Multiplication Rule use karo: $P (R_{1} \cap R_{2}) = P (R_{1}) P (R_{2} ∣ R_{1}) .$

$P (R_{1}) = \frac{4}{10}$ .
Ek red hatane ke baad: $9$ mein se $3$ red bache, toh $P (R_{2} ∣ R_{1}) = \frac{3}{9} = \frac{1}{3}$ . $P (R_{1} \cap R_{2}) = \frac{4}{10} \cdot \frac{1}{3} = \frac{4}{30} = \frac{2}{15} \approx 0.1333.$ Comparison: with replacement ne $0.16$ diya; without replacement $0.133$ deta hai. Dependence ne chance ghata diya — yeh figure ke right panel se match karta hai, jahan ek red hatane se draw 2 ke liye red count patla ho jaata hai.

Recall Solution L3·Q2

Independence test: independence ke liye zaroori hota $P (jam ∣ rain) = P (jam) = 0.3$ . Lekin humein bataya gaya hai yeh $0.6 \neq = 0.3$ hai. Toh dependent — rain jam ki odds badhata hai. General rule use karo (conditional value, $0.3$ nahin): $P (rain \cap jam) = P (rain) P (jam ∣ rain) = 0.4 \cdot 0.6 = 0.24.$ (Blind-independence error yeh hoti — $0.4 \cdot 0.3 = 0.12$ multiply karna — jo sahi value ka almost aadha hai.)

Level 4 — Synthesis

Goal: independence ko complements, systems, aur multiple stages ke saath combine karo.

Neeche wali figure reliability ko parallel copies $n$ ki sankhya ke against plot karti hai, ek aise component ka jo $0.2$ probability se fail karta hai. Blue curve hai $P (unit works) = 1 - 0. 2^{n}$ ; har point apni value se labelled hai. Dotted gray line ek single copy ki reliability mark karti hai ( $0.8$ ); dashed red line $0.99$ target mark karti hai. Dhyan do ki blue curve sirf kuch copies ke saath hi target ke paar kaisi rocket kar jaati hai — yeh redundancy ke baare mein padhai ka point hai woh steep climb.

Recall Solution L4·Q1

"System works" = "pump works aur filter works". Independent parts multiply hote hain: $P (works) = 0.95 \cdot 0.9 = 0.855.$ $P (fails) = 1 - 0.855 = 0.145.$

Recall Solution L4·Q2

"At least one works" = "all fail" ka ulta. Har ek $0.2$ probability se fail hota hai: $P (all fail) = 0. 2^{3} = 0.008.$ $P (unit works) = 1 - 0.008 = 0.992.$ Design insight: ek component $0.8$ reliable tha; teen parallel mein $0.992$ tak pahunch jaate hain. Yahi redundancy ka poora point hai — upar ki figure mein $n = 3$ par blue curve dashed $0.99$ target ke bilkul upar baitha hai.

Recall Solution L4·Q3

"Exactly one" do disjoint cases mein split hota hai: (A solve karta hai, B fail karta hai) YA (A fail karta hai, B solve karta hai). Har case ke andar, do independent facts multiply hote hain; do cases ke across hum add karte hain (dono ek saath sach nahin ho sakte — Mutually Exclusive Events). $P (A only) = 0.6 \cdot (1 - 0.5) = 0.6 \cdot 0.5 = 0.30.$ $P (B only) = (1 - 0.6) \cdot 0.5 = 0.4 \cdot 0.5 = 0.20.$ $P (exactly one) = 0.30 + 0.20 = 0.50.$

Level 5 — Mastery

Goal: ulta sochna seekho, structure prove karo, aur bade machinery se connect karo.

Recall Solution L5·Q1

$P (at least one pass) = 1 - (1 - p)^{n} = 1 - 0. 5^{n} .$ Chahiye $1 - 0. 5^{n} \geq 0.99 \Rightarrow 0. 5^{n} \leq 0.01.$ Values test karo: $0. 5^{6} = 0.015625$ (bahut bada), $0. 5^{7} = 0.0078125 \leq 0.01$ ✓. Toh sabse chhota $n$ hai $7$ .

Recall Solution L5·Q2

Humein dikhana hai $P (A \cap B^{c}) = P (A) P (B^{c})$ . $A$ ko $B$ ke andar wale hisse aur $B$ ke bahar wale hisse mein todo (yeh disjoint hain aur saath mein pura $A$ banate hain): $P (A) = P (A \cap B) + P (A \cap B^{c}) .$ Toh $P (A \cap B^{c}) = P (A) - P (A \cap B)$ . Independence $P (A \cap B) = P (A) P (B)$ use karo: $P (A \cap B^{c}) = P (A) - P (A) P (B) = P (A) (1 - P (B)) = P (A) P (B^{c}) .$ Yahi exactly $A$ aur $B^{c}$ ke liye independence condition hai. $■$ (Isliye "machine works" aur "har part fail hota hai" dono ko same product logic se handle kiya ja sakta hai.)

Recall Solution L5·Q3

Pehle $P (A \cap B)$ addition rule se recover karo (is page ke top par bataya gaya hai) $P (A \cup B) = P (A) + P (B) - P (A \cap B)$ . Yaad karo $A \cup B$ matlab " $A$ ya $B$ (ya dono)": $0.7 = 0.5 + 0.4 - P (A \cap B) \Rightarrow P (A \cap B) = 0.2.$ Independence ke liye chahiye hoga $P (A) P (B) = 0.5 \cdot 0.4 = 0.20$ . Kyunki $P (A \cap B) = 0.20$ barabar hai $0.20$ ke, events independent hain. ✓

Recall Feynman recap: poori page ek saansh mein

Agar do cheezein aapas mein baat nahin karti, toh dono ke hone ki chance unki chances ka product hai — yahi ek move hai. Jab woh baat karte hain (ek marble hati, rain traffic ko feed karta hai), doosre factor ko uske given-value se badal do. "At least one" hamesha "one minus (sab fail)" se sabse clean aata hai. Aur "exactly one" matlab hai har case ke andar multiply karo, cases ke across add karo.

Connections

Independent events — multiplication rule — parent rule jo upar har problem test karti hai.
Conditional Probability — usi waqt chahiye jab independence toot jaati hai (L3, L5·Q2).
General Multiplication Rule — without-replacement / rain-jam solutions.
Complement Rule — har "at least one" answer ko power deta hai.
Mutually Exclusive Events — isliye disjoint cases L4·Q3 mein add hote hain.
Binomial Distribution — L5·Q1 "n independent trials" ka pehla taste hai.
Bayes' Theorem — natural next step jab conditionals stack up hoti hain.

Exercises — Independent events — multiplication rule