Common discrete distributions — Bernoulli, Binomial, Poisson, Geometric, Negative Binomial: Worked examples

4.9.6 · D3 · Maths › Probability Theory & Statistics › Common discrete distributions — Bernoulli, Binomial, Poisson

Scenario matrix

Is topic ka har question in cells mein se kisi ek mein aata hai. Neeche ke worked examples par cell label laga hai, aur milake ye sab cover karte hain.

Cell	Kya special hai	Kaunsa example
A. Binomial, ordinary	successes count karo, seedha plug-in	Ex 1
B. Degenerate $p$	$p = 0$ ya $p = 1$ — variance collapse ho jaata hai	Ex 2
C. Boundary $k = 0$	"kuch nahi hua" — aksar "at least one" ka easy leg	Ex 3
D. Geometric, ordinary	pehli success ka wait + tail $P (X > k)$	Ex 4 (figure)
E. Geometric memorylessness	conditional "given tum already wait kar chuke ho"	Ex 5 (figure)
F. Negative Binomial	$r$ -th success ka wait, $(r - 1 k - 1)$ wala trap	Ex 6
G. Poisson, ordinary + window scaling	rate per unit diya gaya, alag window pe poochha gaya	Ex 7
H. Limiting: Binomial $\to$ Poisson	large $n$ , small $p$ — dikhao ki dono agree karte hain	Ex 8
I. Word problem, mixed	distribution khud choose karo	Ex 9
J. Exam twist	mean/variance signature, ya "at least" flip	Ex 10

Ex 1 — Cell A · Binomial, ordinary

Worked example Free throws

Ek player har free throw independently $p = 0.7$ probability se sink karti hai. Wo $n = 5$ shots leti hai. $P (exactly 3 made)$ kya hai?

Forecast: padhne se pehle guess karo — kya ye $0.3$ se upar hai ya neeche? (Uski average $n p = 3.5$ makes hai, toh exactly $3$ kaafi likely hona chahiye.)

Distribution identify karo. Fixed number of trials $n = 5$ , successes ka count → Binomial $(5, 0.7)$ . Why this step? "Fixed $n$ , count successes" Binomial ka signature hai (parent §2).
$P (X = k) = (k n) p^{k} q^{n - k}$ mein plug karo $k = 3$ , $q = 1 - 0.7 = 0.3$ ke saath: $P (X = 3) = (3 5) (0.7)^{3} (0.3)^{2} .$ Why this step? $(3 5) = 10$ choose karta hai ki 5 shots mein se kaunse 3 andar jaate hain; unke liye $(0.7)^{3}$ , aur do misses ke liye $(0.3)^{2}$ .
Compute karo: $(3 5) = 10$ , $(0.7)^{3} = 0.343$ , $(0.3)^{2} = 0.09$ , toh $P = 10 \cdot 0.343 \cdot 0.09 = 0.3087.$

Verify (normalization, haath se). Ek pmf ko $1$ mein sum hona chahiye, toh chalo actually $k = 0..5$ ke saath $p = 0.7, q = 0.3$ ke saath saare chhe terms add karte hain:

P (0) P (1) P (2) P (3) P (4) P (5) = 1 \cdot 0. 3^{5} = 0.00243, = 5 \cdot 0.7 \cdot 0. 3^{4} = 0.02835, = 10 \cdot 0. 7^{2} \cdot 0. 3^{3} = 0.13230, = 10 \cdot 0. 7^{3} \cdot 0. 3^{2} = 0.30870, = 5 \cdot 0. 7^{4} \cdot 0.3 = 0.36015, = 1 \cdot 0. 7^{5} = 0.16807.

Add karo: $0.00243 + 0.02835 + 0.13230 + 0.30870 + 0.36015 + 0.16807 = 1.00000$ . ✓ Hamara $P (3) = 0.3087$ exactly upar ka middle term hai, aur ye forecast threshold $0.3$ se upar bhi hai.

Verify (mode $k = 4$ kyun hai — ratio test poora). Consecutive terms tab tak badhte hain jab tak ratio $1$ se zyada ho: $\frac{P ( k )}{P ( k - 1 )} = \frac{( k n ) p ^{k} q ^{n - k}}{( k - 1 n ) p ^{k - 1} q ^{n - k + 1}} = \frac{n - k + 1}{k} \cdot \frac{p}{q} .$ Why this shape? $k - 1$ se $k$ jaane par ek aur $p$ multiply hota hai, ek $q$ divide hota hai, aur choose-count factor $\frac{n - k + 1}{k}$ se change hota hai. Is ratio ko $\geq 1$ set karo aur $k$ solve karo: $(n - k + 1) p \geq k q \Rightarrow k \leq (n + 1) p$ . Toh terms $k = ⌊(n + 1) p ⌋ = ⌊ 6 \cdot 0.7 ⌋ = 4$ tak badhti hain aur baad mein ghatti hain — mode $4$ hai. Table se, $P (4) = 0.36015$ sabse bada hai, jo confirm karta hai ki $k = 3$ peak se thoda left mein hai. ✓

Ex 2 — Cell B · Degenerate $p$

Worked example The certain and the impossible

Binomial $(4, p)$ lo. Poora distribution evaluate karo jab (i) $p = 1$ aur (ii) $p = 0$ ho. Dono cases mein mean aur variance kya hain?

Forecast: agar har trial hamesha succeed ho, toh count kitna spread out ho sakta hai?

Case $p = 1$ : $q = 0$ . Toh $P (X = k) = (k 4) 1^{k} 0^{4 - k}$ . Factor $0^{4 - k}$ zero hai jab tak $4 - k = 0$ na ho, yaani $k = 4$ . Why this step? $0$ kisi bhi positive power pe zero hota hai, jo $k < 4$ wale har term ko kill karta hai. Surviving term $k = 4$ mein exponent $0^{0}$ hai, jise hum $1$ maante hain. $0^{0} = 1$ kyun hai yahan? Discrete-distribution formulas mein factor $q^{n - k}$ ek count-of-failures bookkeeping term hai: jab zero failures hon toh use product mein kuch contribute nahi karna chahiye, yaani $1$ se multiply karna. Combinatorial convention $0^{0} = 1$ adopt karne se single formula $(k n) p^{k} q^{n - k}$ boundaries $k = 0$ aur $k = n$ par bina special-casing ke sahi rehta hai. (Ye ek convention hai, computed limit nahi — isliye choose kiya gaya taaki pmf $1$ mein sum ho.) $P (X = 4) = 1, everything else 0.$
Mean & variance, $p = 1$ : $E [X] = n p = 4$ , $Var = n pq = 4 \cdot 1 \cdot 0 = 0$ . Why this step? Variance spread measure karta hai. Jo variable hamesha $4$ ho uski koi spread nahi hoti — $Var = 0$ . (Dekho Expectation and Variance.)
Case $p = 0$ : symmetric mirror. Ab $q = 1$ hai aur $p^{k}$ har term ko except $k = 0$ kill karta hai (same $0^{0} = 1$ convention $p^{0}$ factor ke liye use karte hue). $P (X = 0) = 1$ , mean $= 0$ , variance $= n pq = 0$ . Why this step? Kabhi succeed na karna matlab count certainly $0$ hai; phir se ek point mass, zero spread.

Verify: $Var = n pq = n p (1 - p)$ , $p$ mein ek downward parabola hai, dono ends $p = 0, 1$ par zero aur $p = \frac{1}{2}$ par maximum. Dono degenerate cases sahi se variance $0$ dete hain. Units: probabilities dimensionless hain, $1$ mein sum hoti hain. ✓

Ex 3 — Cell C · Boundary $k = 0$ ("at least one" trick)

Worked example At least one six

Ek fair die $4$ baar roll karo. $P (at least one six)$ kya hai?

Forecast: har roll mein $1/6$ chance hai; chaar rolls pe, kya "at least one six" aadhe se zyada ya kam hai?

Pehle complement lo. "At least one" sum karna tedious hai ( $k = 1, 2, 3, 4$ ). Iska complement ek aasaan event hai "koi six nahi," yaani $k = 0$ . Why this step? $P (\geq 1) = 1 - P (0)$ chaar terms ko ek mein convert kar deta hai. Isliye $k = 0$ boundary cell matter karta hai.
Binomial $(4, 1/6)$ se $P (0)$ : $k = 0$ ke saath, $(0 4) = 1$ aur $p^{0} = 1$ , toh $P (X = 0) = (5/6)^{4} .$ Why this step? Zero successes matlab chaaon rolls fail; har fail $q = 5/6$ se hota hai, multiply karo (independence).
Subtract karo: $(5/6)^{4} = 625/1296 \approx 0.4823$ , toh $P (\geq 1) = 1 - 0.4823 = 0.5177.$

Verify: Aadhe se thoda zyada — us folk rule se match karta hai ki ~ $4$ rolls pe six ke liye even-money shot milta hai. Cross-check: $1 - (5/6)^{4} = 671/1296$ . ✓

Ex 4 — Cell D · Geometric, ordinary + tail

Figure — Common discrete distributions — Bernoulli, Binomial, Poisson, Geometric, Negative Binomial

Worked example Pehle six ka wait

Ek fair die tab tak roll karo jab tak pehla six na aaye. $X$ = jis roll number par six aata hai. (a) $P (X = 3)$ , (b) $P (X > 3)$ , (c) $E [X]$ nikalo.

Figure mein kya dikhta hai: $k = 1$ se $8$ tak $P (X = k)$ ka bar chart. Bars geometrically shrink hote hain (har ek apne pehle wale ka $5/6$ hota hai). Dono leftmost bars ( $k = 1, 2$ ) red hain — ye roll 3 se pehle ki forced failures hain. $k = 3$ bar green hai — target event $P (X = 3)$ . $k = 4$ se aage sab kuch ek pale shaded band mein hai jis par $P (X > 3) = q^{3}$ lika hai, yaani teen rolls ke baad abhi tak six nahi aaya iska probability.

Forecast: $p = 1/6$ ke saath, roughly kitne rolls mein pehla six milega?

Distribution. "Pehli success tak rolls" → Geometric $(1/6)$ , toh $q = 5/6$ (upar definition box). Why this step? Koi "choose" nahi aata: success force karke last hoti hai, toh arrange karne kuch nahi hai.
(a) $P (X = 3) = q^{k - 1} p$ $k = 3$ ke saath. Figure mein ye green bar hai; uske left ke dono red bars rolls 1–2 par required failures hain. $P (X = 3) = (5/6)^{2} (1/6) = 25/216 \approx 0.1157.$ Why this step? Do failures phir ek success — timeline ke saath independent pieces multiply karo.
(b) Tail $P (X > 3)$ . $X > 3$ matlab pehle teen rolls sab failures hain (success abhi tak nahi hua). $P (X > 3) = q^{3} = (5/6)^{3} = 125/216 \approx 0.5787.$ Why this step? Figure mein shaded band = "3 rolls ke baad bhi wait kar rahe ho" = teen failures in a row.
(c) Mean $E [X] = 1/ p = 6$ . Why this step? Upar in-line intuition box se — $1/6$ chance per roll ⇒ $6$ rolls wait karo.

Verify: $P (X = 3) + P (X > 3)$ ko $P (X \geq 3) = q^{2} = (5/6)^{2} = 25/36 \approx 0.6944$ ke barabar hona chahiye; sach mein $0.1157 + 0.5787 = 0.6944$ . ✓ Mean $6$ forecast se match karta hai.

Ex 5 — Cell E · Geometric memorylessness

Worked example Jab tum already wait kar chuke ho

Same die Ex 4 wala ( $p = 1/6$ ). Tumne $5$ baar roll kiya aur koi six nahi aaya. Probability kya hai ki tumhe $3$ se zyada additional rolls chahienge? Ise fresh start se compare karo.

Figure mein kya dikhta hai: top strip mein paanch red boxes hain jo "F" (tumhari paanch past failures) mark karti hain, unke baad dashed yellow line hai jis par "NOW" lika hai. NOW ke neeche aur daaye, blue bars ka fresh set nikalta hai — ek identical Geometric distribution — jaise abhi clock reset hua ho. Ek arrow annotate karta hai ki "P(more than 3 extra) $= q^{3} = P (X > 3)$ ": left ke paanch failures simply calculation se drop out ho jaate hain.

Forecast: paanch failures ki string se kya six "due" ho jaata hai? Apne gut pe trust karo, phir dekho ye galat ho jaaye.

Conditional likho. Hum $P (X > 5 + 3 ∣ X > 5)$ chahte hain. Why this step? "Given already past 5" mein " $3$ se zyada additional" exactly Memorylessness pattern $P (X > m + n ∣ X > m)$ hai jahan $m = 5, n = 3$ .
Conditional probability ki definition: $P (X > 8 ∣ X > 5) = \frac{P ( X > 8 and X > 5 )}{P ( X > 5 )} = \frac{P ( X > 8 )}{P ( X > 5 )} .$ Why this step? ${X > 8}$ already ${X > 5}$ ke andar hai, toh joint event sirf ${X > 8}$ hai.
Ex 4 ke definition box se tail $P (X > k) = q^{k}$ use karo: $= \frac{q ^{8}}{q ^{5}} = q^{3} = (5/6)^{3} \approx 0.5787.$ Why this step? $q^{5}$ cancel ho jaata hai — paanch wasted rolls gayab ho jaate hain. Figure mein, origin ko "NOW" pe slide karne par ek identical fresh geometric milti hai.
Compare karo: ek brand-new player $P (X > 3)$ poochhe toh usse bhi $q^{3} \approx 0.5787$ (Ex 4b) milta hai. Bilkul same. Why this step? Yahi memorylessness hai: die ke paas tumhari suffering ka koi record nahi.

Verify: $q^{8} / q^{5} = q^{3}$ exactly, toh conditional unconditional $P (X > 3)$ ke barabar hai. Die "due" nahi hai. ✓

Ex 6 — Cell F · Negative Binomial ( $(r - 1 k - 1)$ wala trap)

Worked example Saatwen trial par teesri success

Ek drug trial: har patient independently $p = 0.4$ ke saath respond karta hai. $X$ = jis trial number par 3rd responder aata hai. $P (X = 7)$ aur $E [X]$ nikalo.

Forecast: 3rd success ka average trial — $7$ se bada ya chhota?

Distribution. " $r$ -th success tak trials" → Negative Binomial $(r, p)$ , $r = 3$ , $p = 0.4$ , $q = 0.6$ , pmf $P (X = k) = (r - 1 k - 1) p^{r} q^{k - r}$ $k \geq r$ ke liye. Why this step? Aur edge case notice karo: $r = 1$ par ye Geometric mein reduce ho jaata hai, kyunki $(0 k - 1) = 1$ chhod deta hai $P (X = k) = p q^{k - 1}$ — exactly Ex 4 wala pmf. Negative Binomial hai "Geometric, but $r$ -th success tak wait karo."
Key insight — last trial forced hai. Trial $7$ zaroor ek success hona chahiye (ye 3rd hai, stopping event). Toh pehle $6$ trials mein hum baaki $2$ successes freely rakhte hain: $(2 6)$ , $(3 7)$ nahi. Why this step? Ye parent ki mistake (a) hai. Last slot fix karne se ek degree of freedom remove ho jaata hai.
Plug in karo, $k = 7$ : $P (X = 7) = (2 6) (0.4)^{3} (0.6)^{4} .$ Why this step? Teen responders ke liye $p^{3}$ , chaar non-responders ke liye $q^{4}$ , do early successes ke $(2 6) = 15$ arrangements.
Compute karo: $15 \cdot 0.064 \cdot 0.1296 = 0.124416$ .
Mean $E [X] = r / p = 3/0.4 = 7.5$ . Why this step? Neg-Bin, $r$ independent Geometrics stack kiya hua hai; har ek $1/ p = 2.5$ trials wait karta hai, toh $3 \times 2.5 = 7.5$ .

Verify: $E [X] = 7.5$ , toh $P (X = 7)$ ka mean se thoda neeche hona aur ek solid peak-region value ( $\approx 0.124$ ) hona consistent hai. Trap check: agar humne (galti se) $(3 7) = 35$ use kiya hota toh $0.29$ milta — obviously ek single point mass ke liye bahut bada. ✓

Ex 7 — Cell G · Poisson with window scaling

Worked example Do ghante mein emails

Emails ek Poisson process mein $λ_{0} = 4$ per hour rate par aate hain. $P (exactly 5 emails in the next 2 hours)$ kya hai?

Forecast: 2 ghanton mein tum $8$ expect karte ho; kya exactly $5$ common hai ya rare?

$λ$ ko window ke liye scale karo. Rate per hour hai, lekin question $2$ ghanton ka hai, toh $λ = λ_{0} \cdot t = 4 \cdot 2 = 8.$ Why this step? Ye parent ki mistake (c) hai — parameter wo expected count hai jo tumse pooche gaye interval pe ho, raw rate nahi.
$P (X = k) = \frac{λ ^{k} e ^{- λ}}{k !}$ mein plug karo, $k = 5$ , $λ = 8$ : $P (X = 5) = \frac{8 ^{5} e ^{- 8}}{5 !} = \frac{32768 e ^{- 8}}{120} .$ Why this step? Seedha Poisson pmf; $λ = 8$ 2-ghante window ka mean hai.
Compute karo, ek multiplication at a time. Pehle $e^{- 8} \approx 3.3546 \times 1 0^{- 4}$ . Phir numerator: $32768 \times 3.3546 \times 1 0^{- 4} \approx 10.9924.$ Aakhir mein $5! = 120$ se divide karo: $P (X = 5) \approx \frac{10.9924}{120} \approx 0.0916.$

Verify: Poisson ke liye mean $=$ variance $= 8$ , toh standard deviation $\approx 2.83$ . Value $5$ mean $8$ se roughly ek SD neeche hai, toh $\approx 0.09$ probability (peak $\approx 0.14$ se $k = 7, 8$ par neeche) sahi lagti hai. Units: count dimensionless hai. ✓

Ex 8 — Cell H · Limiting Binomial → Poisson

Worked example Rare defects, do tareekon se

Ek factory $n = 1000$ chips banati hai; har chip independently $p = 0.002$ ke saath defective hai. $P (exactly 3 defective)$ do tareekon se estimate karo aur dikhao ki dono agree karte hain.

Forecast: expected defects $n p = 2$ ke saath, kya exactly $3$ peak ke paas hai?

Exact Binomial $(1000, 0.002)$ : $P (X = 3) = (3 1000) (0.002)^{3} (0.998)^{997} \approx 0.18047.$ Why this step? Ye honest count model hai; lekin 4-digit $n$ wala binomial coefficient haath se unwieldy hai.
Poisson approximation. Kyunki $n$ bada hai aur $p$ tiny hai, $λ = n p = 1000 \cdot 0.002 = 2$ set karo: $P (X = 3) \approx \frac{2 ^{3} e ^{- 2}}{3 !} = \frac{8 e ^{- 2}}{6} \approx 0.18045.$ Why this step? Parent §5 ne prove kiya ki Binomial $(n, λ / n) \to$ Poisson $(λ)$ via $(1 - λ / n)^{n} \to e^{- λ}$ (dekho Limit e^x as (1+x/n)^n). Yahan $n = 1000$ itna "large" hai ki dono limits kick in kar gaye.
Compare karo: $0.18047$ vs $0.18045$ — chaar decimals tak agree karte hain. Why this step? Confirm karta hai ki limit sirf theory nahi hai: rare events over many trials ke liye messy binomial ek one-parameter Poisson mein collapse ho jaata hai.

Verify: Relative error $0.02%$ se kam hai. Dono models ka mean $\approx 2$ hai; Poisson variance $= 2$ , Binomial variance $= n pq = 2 \cdot 0.998 = 1.996$ — lagbhag equal kyunki $p \to 0$ se $q \to 1$ ho jaata hai. ✓

Ex 9 — Cell I · Word problem (distribution tum choose karo)

Worked example Ek page par typos

Ek editor average $0.5$ typos per page find karta hai. Ek chapter mein $10$ pages hain. (a) Kisi ek page par koi typo nahi hone ka chance kya hai? (b) Poore chapter mein expected number of typos aur uska variance kya hai?

Forecast: zyaadatar pages clean hain ya zyaadatar pages mein typos hain?

Model choose karo. Typos rare events hain jo ek continuum (page) par sprinkle hote hain → Poisson, rate $λ_{page} = 0.5$ . Why this step? Koi fixed number of "trials" nahi, sirf ek rate over an extent — ye Poisson signature hai, Binomial nahi.
(a) Ek page par $P (0)$ , $k = 0$ , $λ = 0.5$ : $P (X = 0) = \frac{0. 5 ^{0} e ^{- 0.5}}{0 !} = e^{- 0.5} \approx 0.6065.$ Why this step? $k = 0$ boundary phir se — $λ^{0} = 1$ aur $0! = 1$ sirf $e^{- λ}$ chhod dete hain. Toh lagbhag $61%$ pages clean hain.
(b) Poora chapter. $λ$ ko $10$ pages tak scale karo: $λ_{chap} = 0.5 \cdot 10 = 5$ . Phir $E [X] = 5, Var (X) = 5.$ Why this step? Chapter total $10$ independent per-page Poisson counts ka sum hai. Poisson family ka ek key structural fact ye hai ki independent Poissons add hote hain: agar $Y_{i} \sim$ Poisson $(λ_{i})$ independently, toh $\sum Y_{i} \sim$ Poisson $(\sum λ_{i})$ . (Ek clean proof unke probability generating functions $e^{λ_{i} (s - 1)}$ multiply karta hai, jinka product $e^{(\sum λ_{i}) (s - 1)}$ hai — phir se ek Poisson pgf.) Toh chapter count genuinely Poisson $(5)$ hai, sirf mean $5$ wali koi cheez nahi; isliye uska variance bhi $5$ hai (mean $=$ variance Poisson signature hai). Linearity of expectation akele mean $5$ deta lekin poori distribution nahi bata sakta tha — uske liye hume parameter-additivity chahiye thi.

Verify: Lagbhag $61%$ pages clean hain — "half a typo per page" ke liye plausible. Kyunki chapter count exactly Poisson $(5)$ hai, mean aur variance dono $5$ hain. ✓

Ex 10 — Cell J · Exam twist (mean = variance signature + "at least")

Worked example Kis distribution ka mean = variance hota hai?

Counts ka ek data set hai jiska sample mean $\overset{x}{ˉ} = 3.0$ aur sample variance $s^{2} = 3.0$ hai. Ek exam tumse poochha hai ki (a) natural model ka naam batao, phir (b) use karke $P (X \geq 2)$ compute karo.

Forecast: kya " $\geq 2$ " ke liye complement trick chahiye?

Signature padho. Mean $=$ variance Poisson ka fingerprint hai (parent §5). Toh Poisson $(λ = 3)$ model karo. Why this step? Binomial ka variance $n pq < n p$ hai (kyunki $q < 1$ ); Geometric ka variance $q / p^{2} \neq = 1/ p$ generally hai. Sirf Poisson hi mean $=$ variance force karta hai, aur data exactly yahi dikhata hai ( $3.0 = 3.0$ ).
"At least 2" via complement. $P (X \geq 2) = 1 - P (X = 0) - P (X = 1) .$ Why this step? $k = 2, 3, 4, \dots$ sum karna infinite hai; do subtractions finite hain. Ye Ex 3 ke $k = 0$ boundary cell ka "at least" flip hai.
Do easy terms compute karo, $λ = 3$ : $P (0) = \frac{3 ^{0} e ^{- 3}}{0 !} = e^{- 3}, P (1) = \frac{3 ^{1} e ^{- 3}}{1 !} = 3 e^{- 3} .$ Why this step? $k = 0, 1$ par direct Poisson pmf; $3^{0} = 1$ , $0! = 1! = 1$ .
Combine karo aur $e^{- 3}$ factor karo: $P (X \geq 2) = 1 - e^{- 3} - 3 e^{- 3} = 1 - 4 e^{- 3} .$
Number: $e^{- 3} \approx 0.049787$ , toh $4 e^{- 3} \approx 0.199148$ aur $P (X \geq 2) = 1 - 0.199148 \approx 0.8009.$

Verify: Mean $3$ hai, toh at least $2$ counts dekhna bahut likely hona chahiye — $\approx 0.80$ fit karta hai. Cross-check ki probabilities balance hoti hain: $P (0) + P (1) = 4 e^{- 3} \approx 0.1991$ , aur $0.1991 + 0.8009 = 1.0000$ . ✓

Recall Quick self-test (answers cover karo)

"4 rolls mein at least one six" ke liye complement trick deta hai ::: $1 - (5/6)^{4} = 671/1296 \approx 0.5177$ Neg-Bin mein $(r k)$ nahi $(r - 1 k - 1)$ kyun aata hai? ::: $k$ -wa (aakhri) trial forced hota hai stopping success hone ke liye, toh sirf $r - 1$ successes $k - 1$ trials mein free hain Negative Binomial jab $r = 1$ ho toh reduce ho jaata hai ::: Geometric distribution mein Poisson rate $4$ /hr over $2$ hr mein $λ =$ ::: $8$ (rate ko window length se scale karo) Memoryless conditional $P (X > 8 ∣ X > 5)$ Geometric ke liye equal hai ::: $q^{3} = P (X > 3)$ — past cancel ho jaata hai Wo distribution jiska mean $=$ variance ho ::: Poisson $10$ independent Poissons ka sum Poisson kyun rehta hai? ::: unke parameters add hote hain (pgfs ka product phir se ek Poisson pgf hai), sirf means nahi