Intuition The big picture (WHY these five exist)Almost every "count" question in probability is one of five flavours of the same single trial : a coin flip with success probability p p p
Bernoulli — flip the coin once . (the atom)Binomial — flip n n n fixed times, count successes.Geometric — flip until the first success, count trials.Negative Binomial — flip until the r r r Poisson — what Binomial becomes when n → ∞ n\to\infty n → ∞ p → 0 p\to0 p → 0 n p = λ np=\lambda n p = λ
Learn the Bernoulli atom + the counting argument and the other four fall out for free. That is the 80/20 of this whole topic.
( p ) (p) ( p ) X ∈ { 0 , 1 } X\in\{0,1\} X ∈ { 0 , 1 } P ( X = 1 ) = p , P ( X = 0 ) = 1 − p = q . P(X=1)=p,\qquad P(X=0)=1-p=q. P ( X = 1 ) = p , P ( X = 0 ) = 1 − p = q . trial with two outcomes: "success" (1) and "failure" (0).
If I do n n n any specific sequence with k k k p k q n − k p^k q^{n-k} p k q n − k k k k ( n k ) \binom{n}{k} ( k n ) which trials succeeded. Add them up.
( n , p ) (n,p) ( n , p ) X = X= X = n n n ( p ) (p) ( p ) P ( X = k ) = ( n k ) p k q n − k , k = 0 , 1 , … , n . P(X=k)=\binom{n}{k}p^k q^{\,n-k},\qquad k=0,1,\dots,n. P ( X = k ) = ( k n ) p k q n − k , k = 0 , 1 , … , n .
Worked example Faulty bulbs10 bulbs, each defective with p = 0.1 p=0.1 p = 0.1 P ( X = 2 ) = ( 10 2 ) ( 0.1 ) 2 ( 0.9 ) 8 . P(X=2)=\binom{10}{2}(0.1)^2(0.9)^8. P ( X = 2 ) = ( 2 10 ) ( 0.1 ) 2 ( 0.9 ) 8 . Why this step? ( 10 2 ) = 45 \binom{10}{2}=45 ( 2 10 ) = 45 ( 0.1 ) 2 (0.1)^2 ( 0.1 ) 2 ( 0.9 ) 8 (0.9)^8 ( 0.9 ) 8 = 45 ⋅ 0.01 ⋅ 0.4305 ≈ 0.194. =45\cdot0.01\cdot0.4305\approx 0.194. = 45 ⋅ 0.01 ⋅ 0.4305 ≈ 0.194.
q k − 1 p q^{k-1}p q k − 1 p k k k k − 1 k-1 k − 1 all fail (q k − 1 q^{k-1} q k − 1 k k k p p p
( p ) (p) ( p ) X = X= X = P ( X = k ) = q k − 1 p , k = 1 , 2 , 3 , … P(X=k)=q^{\,k-1}p,\qquad k=1,2,3,\dots P ( X = k ) = q k − 1 p , k = 1 , 2 , 3 , …
Intuition Memorylessness — the unique fingerprintGeometric is the only discrete distribution with P ( X > m + n ∣ X > m ) = P ( X > n ) P(X>m+n\mid X>m)=P(X>n) P ( X > m + n ∣ X > m ) = P ( X > n )
Definition Negative Binomial( r , p ) (r,p) ( r , p ) X = X= X = r r r P ( X = k ) = ( k − 1 r − 1 ) p r q k − r , k = r , r + 1 , … P(X=k)=\binom{k-1}{r-1}p^r q^{\,k-r},\qquad k=r,r+1,\dots P ( X = k ) = ( r − 1 k − 1 ) p r q k − r , k = r , r + 1 , …
( k − 1 r − 1 ) \binom{k-1}{r-1} ( r − 1 k − 1 ) ( k r ) \binom{k}{r} ( r k ) k k k must be a success (it's the r r r stopping event). So we only freely arrange the remaining r − 1 r-1 r − 1 k − 1 k-1 k − 1 ( k − 1 r − 1 ) \binom{k-1}{r-1} ( r − 1 k − 1 ) p r p^r p r r r r q k − r q^{k-r} q k − r r = 1 r=1 r = 1
Intuition WHERE it comes fromTake Binomial( n , p ) (n,p) ( n , p ) n → ∞ n\to\infty n → ∞ p → 0 p\to0 p → 0 expected count n p = λ np=\lambda n p = λ
Worked example Call centreAverage λ = 3 \lambda=3 λ = 3 P ( X = 5 ) = 3 5 e − 3 5 ! = 243 e − 3 120 ≈ 0.101. P(X=5)=\frac{3^5 e^{-3}}{5!}=\frac{243\,e^{-3}}{120}\approx 0.101. P ( X = 5 ) = 5 ! 3 5 e − 3 = 120 243 e − 3 ≈ 0.101. Why λ = 3 \lambda=3 λ = 3 The minute is the unit over which the rate is given, so λ \lambda λ
Dist.
P ( X = k ) P(X=k) P ( X = k ) Mean
Variance
Counts
Bernoulli( p ) (p) ( p )
p k q 1 − k p^k q^{1-k} p k q 1 − k k ∈ { 0 , 1 } k\in\{0,1\} k ∈ { 0 , 1 } p p p p q pq pq 1 trial
Binomial( n , p ) (n,p) ( n , p )
( n k ) p k q n − k \binom{n}{k}p^kq^{n-k} ( k n ) p k q n − k n p np n p n p q npq n pq successes in n n n
Geometric( p ) (p) ( p )
q k − 1 p q^{k-1}p q k − 1 p 1 / p 1/p 1/ p q / p 2 q/p^2 q / p 2 trials to 1st success
Neg.Bin.( r , p ) (r,p) ( r , p )
( k − 1 r − 1 ) p r q k − r \binom{k-1}{r-1}p^rq^{k-r} ( r − 1 k − 1 ) p r q k − r r / p r/p r / p r q / p 2 rq/p^2 r q / p 2 trials to r r r
Poisson( λ ) (\lambda) ( λ )
λ k e − λ / k ! \lambda^k e^{-\lambda}/k! λ k e − λ / k ! λ \lambda λ λ \lambda λ rare events / unit
Common mistake Steel-manned common errors(a) Using ( k r ) \binom{k}{r} ( r k ) Why it feels right: "I'm choosing r r r k k k The fix: the last trial is forced to be the stopping success, so only r − 1 r-1 r − 1 k − 1 k-1 k − 1 ( k − 1 r − 1 ) \binom{k-1}{r-1} ( r − 1 k − 1 ) (b) Adding variances without independence. Why it feels right: means add freely, so variances "should" too. Fix: Var ( X + Y ) = Var X + Var Y + 2 Cov ( X , Y ) \operatorname{Var}(X+Y)=\operatorname{Var}X+\operatorname{Var}Y+2\operatorname{Cov}(X,Y) Var ( X + Y ) = Var X + Var Y + 2 Cov ( X , Y ) only when independent.
(c) Poisson with a mismatched window. Why it feels right: you grab the rate number blindly. Fix: scale λ \lambda λ λ = 6 \lambda=6 λ = 6 (d) Off-by-one in Geometric. Two conventions exist: "number of trials" (k ≥ 1 k\ge1 k ≥ 1 1 / p 1/p 1/ p k ≥ 0 k\ge0 k ≥ 0 q / p q/p q / p
Recall Feynman: explain to a 12-year-oldImagine flipping a slightly-bent coin that lands heads (=win) sometimes.
Bernoulli: one flip — did you win? Yes/no.Binomial: flip exactly 10 times — how many wins?Geometric: keep flipping until your first win — how long did it take?Negative Binomial: keep flipping until your 5th win.Poisson: instead of flips, count how many shooting stars you see in an hour when each second the chance is tiny but the night is long.
Same coin, different questions . That's all five distributions.
Mnemonic BIG-PN ordering by "when do I stop?"B ernoulli (1 flip) → B inomial (stop at fixed n n n G eometric (stop at 1st success) → N egBin (stop at r r r P oisson (never stop counting — continuous time, rare hits).
Memory line: "Be In Good Numbers, Please" — Bernoulli, Binomial, Geometric, NegBin, Poisson.
Bernoulli mean and variance E [ X ] = p E[X]=p E [ X ] = p ,
Var = p q \operatorname{Var}=pq Var = pq (using
X 2 = X X^2=X X 2 = X ).
Why does the Binomial PMF have a ( n k ) \binom{n}{k} ( k n ) Each specific sequence of
k k k successes has prob
p k q n − k p^kq^{n-k} p k q n − k ; there are
( n k ) \binom{n}{k} ( k n ) such sequences, so we sum them.
Binomial mean and variance n p np n p and
n p q npq n pq (sum of
n n n i.i.d. Bernoullis).
Geometric PMF (trials convention) P ( X = k ) = q k − 1 p P(X=k)=q^{k-1}p P ( X = k ) = q k − 1 p ,
k ≥ 1 k\ge1 k ≥ 1 .
Geometric mean 1 / p 1/p 1/ p (derived from
∑ k x k − 1 = 1 / ( 1 − x ) 2 \sum kx^{k-1}=1/(1-x)^2 ∑ k x k − 1 = 1/ ( 1 − x ) 2 ).
Which distribution is memoryless? The Geometric (discrete) — only one with
P ( X > m + n ∣ X > m ) = P ( X > n ) P(X>m+n\mid X>m)=P(X>n) P ( X > m + n ∣ X > m ) = P ( X > n ) .
Negative Binomial PMF ( k − 1 r − 1 ) p r q k − r \binom{k-1}{r-1}p^rq^{k-r} ( r − 1 k − 1 ) p r q k − r ,
k ≥ r k\ge r k ≥ r .
Why ( k − 1 r − 1 ) \binom{k-1}{r-1} ( r − 1 k − 1 ) ( k r ) \binom{k}{r} ( r k ) The
k k k -th trial is forced to be the
r r r -th success (stopping event); only
r − 1 r-1 r − 1 successes arrange among first
k − 1 k-1 k − 1 trials.
NegBin mean and variance r / p r/p r / p and
r q / p 2 rq/p^2 r q / p 2 (sum of
r r r Geometrics).
Poisson PMF λ k e − λ / k ! \lambda^k e^{-\lambda}/k! λ k e − λ / k ! ,
k ≥ 0 k\ge0 k ≥ 0 .
Poisson as a limit of what? Binomial
( n , p ) (n,p) ( n , p ) with
n → ∞ n\to\infty n → ∞ ,
p → 0 p\to0 p → 0 ,
n p = λ np=\lambda n p = λ fixed.
Poisson mean and variance Key Poisson signature mean = variance.
Geometric is NegBin with which r r r When can you add variances? Only when the variables are independent (covariance term vanishes).
Bernoulli trial — the shared atom of all fiveExpectation and Variance — every formula here is derived from these definitionsLinearity of expectation — gives Binomial/NegBin means instantlyGeometric series and its derivative — engine behind Geometric meanLimit e^x as (1+x/n)^n — engine behind the Poisson derivationPoisson process — continuous-time origin of the Poisson distributionMemorylessness — links Geometric (discrete) and Exponential distribution (continuous)Moment generating functions — alternative unified route to all these moments
Bernoulli p - single trial
Intuition Hinglish mein samjhoDekho, in paanchon distributions ko alag-alag yaad karne ki zaroorat nahi hai — sab ek hi cheez par based hain: ek "coin flip" jiska success probability p p p Bernoulli kehte hain — sirf ek baar flip, win ya lose. Bas yahi atom hai, baaki sab isi se bante hain.
Ab sawaal badalta hai: agar main coin ko fixed n n n flip karun aur wins ginun, woh Binomial hai — isme ( n k ) \binom{n}{k} ( k n ) pehla win na aaye, woh Geometric hai (average wait 1 / p 1/p 1/ p r r r Negative Binomial hai (Geometric ka r r r
Poisson thoda alag feel hota hai par actually woh Binomial ka hi limit hai: jab n n n p p p n p = λ np=\lambda n p = λ λ k e − λ / k ! \lambda^k e^{-\lambda}/k! λ k e − λ / k ! λ \lambda λ
Why ye important hai? Real-world counting problems — defective items, calls per minute, attempts till success — sab inhi mein fit ho jaate hain. Ek galti se bacho: Poisson mein λ \lambda λ ( k − 1 r − 1 ) \binom{k-1}{r-1} ( r − 1 k − 1 ) ( k r ) \binom{k}{r} ( r k )
