Question bank — Independent events — multiplication rule

False — yeh mutually exclusive events ko describe karta hai (Venn picture mein koi overlap nahi). Independent events usually saath ho sakte hain; actually agar dono possible hain, toh $P(A\cap B)=P(A)P(B)>0$, isliye kabhi kabhi saath hote hain.

True — yeh equation hi independence ki definition/test hai, aur yeh symmetric hai, isliye yeh $P(A\mid B)=P(A)$ aur $P(B\mid A)=P(B)$ bhi guarantee karta hai.

False — yeh dependence ki sabse strong form hai: agar $A$ hota hai, toh $B$ ka hona forced nahi hota, isliye $A$ ko jaanna $B$ ki chance ko $P(B)$ se $0$ tak le aata hai.

False — pehla card hataane se jo bacha rehta hai woh badal jaata hai, isliye $P(\text{2nd}\mid \text{1st})\ne P(\text{2nd})$; tumhe conditional probability ke saath General Multiplication Rule use karni padegi.

True — kyunki $P(A^c\cap B^c)=1-P(A\cup B)=1-(P(A)+P(B)-P(A)P(B))=(1-P(A))(1-P(B))=P(A^c)P(B^c)$, jo complements ke liye independence test bilkul match karta hai.

False — summing karne se overlap double-count hota hai aur yeh $1$ se bhi zyada ho sakta hai; independent events ke liye Complement Rule ke zariye $1-{\prod }_i(1-P(A_i))$ use karo.

False (degenerate cases ko chodke) — iske liye $P(A)=P(A{)}^2$ chahiye, jo $P(A)=0$ ya $P(A)=1$ force karta hai; koi bhi genuinely uncertain event ($0<P(A)<1$) apne aap se dependent hoti hai.

True — lekin zaroori caveat dhyan mein rakho: dono conditionals defined hone chahiye, yaani $P(A)>0$ aur $P(B)>0$; jab kisi ek ki probability $0$ ho toh conditional undefined hai aur yeh symmetric implication apply nahi hoti.

True — independence ek numerical property hai $P(A\cap B)=P(A)P(B)$, isliye bilkul same labelled events ek probability assignment mein independent ho sakte hain aur doosre mein dependent.

False — independence transitive nahi hoti; $A$–$B$ aur $B$–$C$ ke beech pairwise links tumhe $A$–$C$ ke baare mein kuch guaranteed nahi bataate.

False — mutual independence zyada strong hoti hai; iske liye $P(A\cap B\cap C)=P(A)P(B)P(C)$ bhi chahiye, jo fail ho sakta hai tab bhi jab sab pairs check out kar jaayein.

Error yeh hai ki do unconditional probabilities multiply ki gayi hain. Bag badal gaya, isliye doosra factor $P(R_2\mid R_1)=\frac{3}{9}$ hona chahiye, jo $\frac{4}{10}\cdot \frac{3}{9}\approx 0.133$ deta hai.

"Dono ek saath nahi ho sakte" ka matlab hai $P(A\cap B)=0$, jo exclusivity hai, independence nahi. Actually yeh unhe dependent banata hai, kyunki $A$ ka hona guarantee karta hai ki $B$ nahi hua.

Yeh gambler's fallacy hai: independence ka matlab hai past flips agle flip ki probability ko bilkul $\frac{1}{2}$ par hi chhodte hain; coin ki koi memory nahi hoti, isliye kuch bhi "due" nahi hota.

Independence yahan galat tool hai; $P(A\cap B)$ subtract kiye bina addition sirf mutually exclusive events ke liye valid hai. Independent events ke liye, $P(A\cup B)=P(A)+P(B)-P(A)P(B)$.

Overlaps add karne se double-counting hoti hai aur yeh sirf approximate hai. Clean method complement hai: $1-0.9^3=0.271$.

Yeh "no information transfer" ko "no overlap" se confuse karta hai. Upar diya Venn figure dekho: independent events typically overlap karti hain (unka intersection area $P(A)P(B)$ ke barabar hota hai); zero overlap unhe exclusive aur isliye dependent banata.

Doosra General Multiplication Rule hai — conditional probability ki definition se hamesha true. Pehla ek special case hai jo tab milta hai jab $P(A\mid B)$ ko $P(A)$ se replace kiya jaaye, ek swap jo independence allow karta hai.

"At least one" kaafi messy cases mein split hoti hai (exactly one, exactly two, …), lekin uska opposite — "koi nahi hota" — ek single clean product $\prod (1-P(A_i))$ hai, isliye $1-\prod (1-P(A_i))$ bahut aasaan hai.

Drawn item ko replace karne se bag apni original composition par reset ho jaata hai, isliye doosra draw identical odds face karta hai; pehle draw ka koi trace nahi rehta, jo exactly independence ka meaning hai.

Har probability at most $1$ hoti hai, isliye multiply karne se result chhota hota hai — aur hona bhi chahiye: dono events require karna sirf kisi ek se zyada mushkil hai, isliye "both" ko chhota chance milna chahiye.

Formula hamesha ek number produce karta hai, lekin woh number tab hi sahi hota hai jab independence hold kare. Check skip karne se galat answers milte hain (jaise replacement ke bina wale bag mein).

Independence ka matlab hai ek event doosre ke baare mein zero information carry karta hai; exclusivity ka matlab hai ek event maximum information carry karta hai — yeh bata deta hai ki doosra definitely nahi hua.

Haan — $P(A\cap B)\le P(A)=0$, isliye $P(A\cap B)=0=0\cdot P(B)=P(A)P(B)$; ek impossible event trivially har cheez se independent hai.

Haan — ek certain event satisfy karta hai $P(A\cap B)=P(B)=1\cdot P(B)=P(A)P(B)$, isliye $B$ ko jaanna $A$ ki (already certain) status kabhi nahi badalta.

Sirf degenerate cases mein — tumhe ek saath $P(A\cap B)=0$ aur $P(A\cap B)=P(A)P(B)$ chahiye, jo $P(A)=0$ ya $P(B)=0$ force karta hai; dono positive hone par yeh impossible hai.

Haan — physically alag devices ka koi shared mechanism nahi hai, isliye $P(\text{H}\cap 6)=\frac{1}{2}\cdot \frac{1}{6}=\frac{1}{12}$ product se match karta hai, independence confirm karta hai.

Sirf special numeric cases mein — sab kuch cover karna dependence force nahi karta, lekin tumhe phir bhi $P(A\cap B)=P(A)P(B)$ check karna hoga; e.g. $A$ aur $A^c$ jahan $0<P(A)<1$ hain woh independent nahi hain kyunki $P(A\cap A^c)=0\ne P(A)P(A^c)$.