Noesis
2.7.7Statistics & Probability — Intermediate

Independent events — multiplication rule

1,758 words8 min readdifficulty · medium3 backlinks
🇮🇳 Yeh note Hinglish mein padho →

WHAT are we talking about?

The two boxes above say the same thing. The first is the meaning (no information transfer); the second is the multiplication rule you compute with.

WHY does multiplying work? (Derive it from scratch)

We never just "state" the multiplication rule — we build it from the definition of conditional probability.

So the multiplication rule is not a new law — it is the general rule plus the independence assumption. That's the whole story.

Figure — Independent events — multiplication rule

Extending to many events

A handy consequence — "at least one" via the complement: P(at least one occurs)=1−∏i(1−P(Ai))P(\text{at least one occurs}) = 1 - \prod_{i}\bigl(1 - P(A_i)\bigr) Why? "At least one" is the opposite of "none happen." All fail together with probability ∏(1−P(Ai))\prod(1-P(A_i)) (using independence of the failures), so subtract from 1.

Worked examples

Steel-man your mistakes

Recall Feynman: explain it to a 12-year-old

Imagine flipping a coin and rolling a die at the same time. The coin has no idea what the die does — they don't talk to each other. So to get the chance of "heads AND a six," you just take the chance of heads (12\tfrac12) and the chance of a six (16\tfrac16) and multiply: 112\tfrac1{12}. Multiplying is like saying "half of the six-times... and a sixth of the heads-times." But if pulling a red marble means there's now one fewer red marble left, the marbles do talk — the second pull knows about the first — so you can't just multiply the same numbers; you must update.

Flashcards

What is the multiplication rule for independent events?
P(A∩B)=P(A) P(B)P(A \cap B) = P(A)\,P(B)
What condition must hold to use P(A∩B)=P(A)P(B)P(A\cap B)=P(A)P(B)?
AA and BB must be independent, i.e. P(A∣B)=P(A)P(A\mid B)=P(A).
Derive the independence multiplication rule from conditional probability.
P(A∣B)=P(A∩B)P(B)⇒P(A∩B)=P(A∣B)P(B)P(A\mid B)=\frac{P(A\cap B)}{P(B)} \Rightarrow P(A\cap B)=P(A\mid B)P(B); independence gives P(A∣B)=P(A)P(A\mid B)=P(A), so P(A∩B)=P(A)P(B)P(A\cap B)=P(A)P(B).
Are mutually exclusive events (both with positive prob) independent?
No — they are strongly dependent, since P(A∩B)=0≠P(A)P(B)P(A\cap B)=0 \ne P(A)P(B).
Formula for P(at least one of independent Ai)P(\text{at least one of independent } A_i)?
1−∏i(1−P(Ai))1 - \prod_i (1-P(A_i)).
Bag: 4 red, 6 blue, two draws WITH replacement, P(both red)?
410⋅410=0.16\frac{4}{10}\cdot\frac{4}{10}=0.16
Same bag WITHOUT replacement, P(both red)?
410⋅39=215≈0.133\frac{4}{10}\cdot\frac{3}{9}=\frac{2}{15}\approx0.133
Three parts each work with prob 0.9 independently; P(at least one fails)?
1−0.93=0.2711-0.9^3=0.271
What does P(A∣B)=P(A)P(A\mid B)=P(A) mean in words?
Knowing BB occurred doesn't change the probability of AA.

Connections

  • Conditional Probability — the multiplication rule is derived directly from it.
  • General Multiplication Rule — P(A∩B)=P(A)P(B∣A)P(A\cap B)=P(A)P(B\mid A), the dependent-case parent.
  • Mutually Exclusive Events — contrast: exclusive ≠ independent.
  • Complement Rule — powers the "at least one" trick.
  • Bayes' Theorem — built on conditional probability too.
  • Binomial Distribution — repeated independent trials multiplied together.

Concept Map

meaning

formalised as

rearrange

impose independence

substituted into

is the test for

apply repeatedly

complement of none

used in

requires justifying

Independent events

Knowing one tells nothing about other

P(A|B) = P(A)

Conditional prob def: P(A|B)=P(A∩B)/P(B)

General mult rule: P(A∩B)=P(A|B)P(B)

Mult rule: P(A∩B)=P(A)P(B)

n events: product of P(Ai)

At least one = 1 − ∏(1−P(Ai))

Example: two coins both heads

Hinglish (regional understanding)

Intuition Hinglish mein samjho

Dekho, do events independent tab hote hain jab ek ke hone se doosre ki probability bilkul nahi badalti. Jaise coin uchaalna aur dice fenkna — coin ko pata hi nahi ki dice pe kya aaya. Aise case mein "dono hone" ki chance nikaalne ke liye simply dono ki alag-alag probabilities ko multiply kar do: P(A∩B)=P(A) P(B)P(A\cap B)=P(A)\,P(B). Yaad rakho: "AND" aaye toh multiply, lekin sirf tabhi jab dono ek doosre ko mind na karein.

Ye multiplication rule aasman se nahi giri — ye conditional probability se banti hai. P(A∣B)=P(A∩B)P(B)P(A\mid B)=\frac{P(A\cap B)}{P(B)} ko rearrange karo toh P(A∩B)=P(A∣B) P(B)P(A\cap B)=P(A\mid B)\,P(B) milta hai (ye har case mein sach hai). Ab agar independent hain toh P(A∣B)=P(A)P(A\mid B)=P(A), bas isko daal do aur P(A)P(B)P(A)P(B) ban jaata hai. Matlab rule alag koi law nahi, sirf general rule + independence assumption hai.

Sabse bada trap: replacement. Bag mein marble wapas daal do toh bag same rehta hai → independent → same probability multiply. Par wapas na daalo toh bag change ho gaya → dependent → doosri draw ke liye updated probability (3/93/9 type) use karni padegi, warna answer galat aayega. Ek aur trap: mutually exclusive ko independent samajh lena — ye ekdum ulta hai, exclusive events toh strongly dependent hote hain kyunki ek hua toh doosra pakka nahi hoga.

Aur jab "at least one" pucha jaye, seedha add mat karo. Complement lo: 1−(koi bhi nahi hone ki probability)1 - (\text{koi bhi nahi hone ki probability}). Independent failures ko multiply karke "sab fail" nikaalo, phir 1 se ghata do. Ye trick exams mein bahut time bachaati hai.

Go deeper — visual, from zero

D1FoundationsEvery symbol from zero · 1,776 wordsD2Visual walkthroughThe derivation in pictures · 1,765 wordsD3Worked examplesEvery scenario covered · 2,471 wordsD4ExercisesGraded, full solutions · 2,204 wordsD5Question bankConcept traps, quick reveals · 1,418 words

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Connections

Conditional probabilityAI-ML · 1.3.2Mutually exclusive events — addition ruleMaths · 2.7.6Bayes' theorem — derivation and applicationsMaths · 2.7.9Binomial distribution — PMF, mean, varianceMaths · 2.7.13Complementary events — P(A') = 1 − P(A)Maths · 1.3.7Mutually exclusive events — addition ruleMaths · 2.7.6Bayes' theorem — derivation and applicationsMaths · 2.7.9