Common discrete distributions — Bernoulli, Binomial, Poisson, Geometric, Negative Binomial: Exercises

4.9.6 · D4 · Maths › Probability Theory & Statistics › Common discrete distributions — Bernoulli, Binomial, Poisson

Level 1 — Recognition

Goal: ek word problem padho aur use paanch distributions mein se exactly ek se match karo, phir uski probability mass function likho (woh rule jo $P (X = k)$ deta hai).

Problem 1.1

Ek fair die ek baar roll kiya jaata hai. Maano $X = 1$ agar chha aata hai, nahi toh $X = 0$ . $X$ ki distribution ka naam batao aur $P (X = 1)$ aur $P (X = 0)$ state karo.

Recall Solution 1.1

YEH kya hai: ek trial, do outcomes (chha / chha nahi). Yeh Bernoulli atom hai. Success probability $p = \frac{1}{6}$ , isliye $q = 1 - \frac{1}{6} = \frac{5}{6}$ . $P (X = 1) = \frac{1}{6}, P (X = 0) = \frac{5}{6} .$ Binomial kyun nahi? Binomial ko kai fixed trials chahiye; yahan $n = 1$ hai, aur Binomial $(1, p)$ hi Bernoulli $(p)$ hai.

Problem 1.2

Aap lottery tickets khareedते rehte ho, har ek independently $p = 0.02$ probability se jeetta hai, jab tak aap apna pehla prize nahi jeet lete. Maano $X$ khareedhe gaye tickets ki sankhya hai. Distribution ka naam batao aur $P (X = k)$ do.

Recall Solution 1.2

Signal words: "pehli success tak." Yeh Geometric distribution hai (trials-counting convention). $p = 0.02$ , $q = 0.98$ ke saath: $P (X = k) = q^{k - 1} p = (0.98)^{k - 1} (0.02), k = 1, 2, 3, \dots$ $q^{k - 1}$ kyun? Pehle $k - 1$ tickets sabhi haarne chahiye, aur $k$ -waan jeetta hai — winning position end mein fix hai, isliye koi "choose" factor nahi hai.

Problem 1.3

Text ke ek page mein average $λ = 0.5$ typos hote hain. Maano $X$ ek page mein typos ki sankhya hai. Distribution ka naam batao aur $P (X = k)$ do.

Recall Solution 1.3

Signal: bahut saare chhote chhote chances (har character ek typo ho sakta hai), rare event, ek rate per unit. Yeh Poisson hai. Dekho Poisson process. $P (X = k) = \frac{λ ^{k} e ^{- λ}}{k !} = \frac{( 0.5 ) ^{k} e ^{- 0.5}}{k !}, k = 0, 1, 2, \dots$

Level 2 — Application

Goal: ek pmf mein numbers daalo aur compute karo.

Problem 2.1

Ek quiz mein 12 true/false questions hain; aap har ek ko independently $p = 0.5$ ke saath guess karte ho. $P (exactly 9 correct)$ kya hai?

Recall Solution 2.1

KYA hai: fixed number of trials $n = 12$ , successes count karo → Binomial $(12, 0.5)$ . $P (X = 9) = (9 12) (0.5)^{9} (0.5)^{3} = (9 12) (0.5)^{12} .$ Ab $(9 12) = (3 12) = \frac{12 \cdot 11 \cdot 10}{3 \cdot 2 \cdot 1} = 220$ aur $(0.5)^{12} = \frac{1}{4096}$ . $P (X = 9) = \frac{220}{4096} \approx 0.0537.$ $(9 12)$ kyun? Yeh count karta hai ki 12 questions mein se kaun se 9 aapne sahi kiye; har aisi arrangement ki same probability $(0.5)^{12}$ hoti hai.

Problem 2.2

Calls $λ = 4$ per minute ki rate se aate hain. $P (exactly 2 calls in one minute)$ kya hai?

Recall Solution 2.2

Poisson $(λ = 4)$ , window = 1 minute (rate se match karta hai, isliye koi rescaling nahi): $P (X = 2) = \frac{4 ^{2} e ^{- 4}}{2 !} = \frac{16 e ^{- 4}}{2} = 8 e^{- 4} \approx 0.1465.$

Problem 2.3

Ek basketball player har free throw $p = 0.7$ probability se daalta/dalti hai. Uska pehla miss 4th throw par hone ki probability kya hai?

Recall Solution 2.3

Yahan Geometric ke liye "success" woh event hai jiska hum intezaar kar rahe hain — ek miss, probability $0.3$ . Maano $S = 0.3$ (miss) aur "non-event" ek make hai, probability $0.7$ . Throw 4 par pehla miss = teen makes phir ek miss: $P = (0.7)^{3} (0.3) = 0.343 \times 0.3 = 0.1029.$ Re-label kyun? Geometric un trials tak count karta hai jab tak interest ka event nahi hota. Yahan woh event miss hai, isliye uski probability $q^{k - 1} p$ mein $p$ ki jagah leti hai: $(0.7)^{3} (0.3)$ .

Level 3 — Analysis

Goal: means, variances, aur cumulative ("at least / more than") reasoning use karo.

Problem 3.1

Binomial $(n = 20, p = 0.3)$ ke liye, mean aur variance compute karo, aur standard deviation bhi.

Recall Solution 3.1

Independent Bernoullis ke decomposition $X = X_{1} + \dots + X_{n}$ se (mean ke liye Linearity of expectation, variances add karne ke liye independence): $E [X] = n p = 20 (0.3) = 6, Var (X) = n pq = 20 (0.3) (0.7) = 4.2.$ Standard deviation $= 4.2 \approx 2.049$ . Dekho Expectation and Variance.

Problem 3.2

Aap ek fair die tab tak roll karte ho jab tak pehla chha nahi aata. $P (X > 3)$ kya hai — ki aapko 3 se zyada rolls chahiye? Exact value do.

Recall Solution 3.2

Geometric $(p = \frac{1}{6})$ . "3 se zyada rolls" ka matlab hai pehle 3 rolls sabhi fail ho jaate hain: $P (X > 3) = q^{3} = (\frac{5}{6})^{3} = \frac{125}{216} \approx 0.5787.$ Sirf $q^{3}$ kyun? $X > 3$ exactly woh event hai "pehle teen rolls mein koi chha nahi" — teen independent failures, har ek probability $\frac{5}{6}$ . Yeh tail formula $P (X > m) = q^{m}$ hai.

Problem 3.3

Die-until-six setup use karte hue, Memorylessness property numerically verify karo: dikhao ki $P (X > 5 ∣ X > 2) = P (X > 3) .$

Recall Solution 3.3

Conditional probability ki definition aur tail formula $P (X > m) = q^{m}$ se: $P (X > 5 ∣ X > 2) = \frac{P ( X > 5 and X > 2 )}{P ( X > 2 )} = \frac{P ( X > 5 )}{P ( X > 2 )} = \frac{q ^{5}}{q ^{2}} = q^{3} .$ Aur $P (X > 3) = q^{3}$ directly. Dono $(\frac{5}{6})^{3} = \frac{125}{216}$ ke barabar hain. ✓ Kyun match karte hain: pehle se kharche gaye do failures ( $X > 2$ ) ratio mein cancel ho jaate hain. Die unhe "bhool" jaata hai — yeh Exponential distribution ki memorylessness ka discrete cousin hai.

Level 4 — Synthesis

Goal: distributions combine karo, ya conventions aur derivations ke beech switch karo.

Problem 4.1

Ek quality inspector components test karti hai (har ek independently $p = 0.15$ probability se defective hai) jab tak woh 3rd defective nahi dhundh leti. Maano $X$ test kiye gaye components ki sankhya hai. $P (X = 6)$ aur $E [X]$ nikalo.

Recall Solution 4.1

"3rd success tak" → Negative Binomial $(r = 3, p = 0.15)$ , jahan success = defective. $P (X = 6) = (3 - 1 6 - 1) p^{3} q^{6 - 3} = (2 5) (0.15)^{3} (0.85)^{3} .$ Ab $(2 5) = 10$ , $(0.15)^{3} = 0.003375$ , $(0.85)^{3} = 0.614125$ : $P (X = 6) = 10 \times 0.003375 \times 0.614125 \approx 0.02073.$ Mean: ek Negative Binomial $r$ independent Geometrics ka sum hai, isliye $E [X] = \frac{r}{p} = \frac{3}{0.15} = 20.$ $(2 5)$ kyun aur $(3 6)$ kyun nahi? 6th (aakhri) test forced hai stopping defective hona. Sirf baaki $r - 1 = 2$ defectives pehle $5$ tests mein kahin bhi baithne ke liye free hain.

Problem 4.2

Geometric series differentiation use karke Geometric mean $E [X] = 1/ p$ scratch se derive karo, aur die ( $p = \frac{1}{6}$ ) ke liye confirm karo.

Recall Solution 4.2

Hum kahan se shuru karte hain: $E [X] = \sum_{k = 1}^{\infty} k q^{k - 1} p$ . Differentiate kyun karte hain: sum $\sum_{k = 1}^{\infty} k x^{k - 1}$ sirf $\sum_{k = 0}^{\infty} x^{k} = \frac{1}{1 - x}$ ka term-by-term derivative hai (valid for $∣ x ∣ < 1$ ). Differentiate karne se hume woh extra factor of $k$ milta hai jo chahiye. $\frac{d}{d x} (\frac{1}{1 - x}) = \frac{1}{( 1 - x ) ^{2}} = \sum_{k = 1}^{\infty} k x^{k - 1} .$ $x = q$ rakho (toh $1 - x = p$ ): $E [X] = p \cdot \frac{1}{( 1 - q ) ^{2}} = p \cdot \frac{1}{p ^{2}} = \frac{1}{p} .$ Die ke liye, $p = \frac{1}{6} \Rightarrow E [X] = 6$ rolls average par. ✓

Problem 4.3

Ek page mein average $λ_{1} = 0.5$ typos hain. Ek book mein 300 aisi pages hain. Poori book mein expected typos ki sankhya kya hai, aur probability kya hai ki book mein puri lambaai mein exactly 0 typos ho? (Alag alag pages par typos independent hain.)

Recall Solution 4.3

Key idea: independent Poissons ka sum phir se Poisson hota hai, rates add ho jaate hain. 300 pages mein rate scale karta hai: $λ_{book} = 300 \times 0.5 = 150.$ Expected typos $= 150$ . Exactly 0 typos ki probability: $P (X = 0) = \frac{λ ^{0} e ^{- λ}}{0 !} = e^{- 150} \approx 7.18 \times 1 0^{- 66} .$ $λ$ scale kyun karte hain: yeh L2 se wohi window-scaling hai — window baro (1 page → 300 pages) aur mean count proportionally barhta hai, yeh underlying Poisson process ki ek property hai.

Level 5 — Mastery

Goal: kai tools chain karo, ek signature identity verify karo, aur interpret karo.

Problem 5.1

Ek factory ka daily defect count Poisson $(λ = 3)$ hai. Kisi ek din, $P (X \geq 2)$ (kam se kam do defects) nikalo. Phir, har din ko independently treat karte hue, woh probability nikalo ki pehla din jisme $\geq 2$ defects hain woh day 4 ho.

Recall Solution 5.1

Step 1 — Poisson tail (KYA hai & complement kyun): $P (X \geq 2)$ directly compute karna infinitely many terms maangta hai, isliye 1 mein se do aasaan wale subtract karo: $P (X \geq 2) = 1 - P (X = 0) - P (X = 1) = 1 - e^{- 3} - 3 e^{- 3} = 1 - 4 e^{- 3} \approx 0.8009.$ Step 2 — days Geometric ban jaate hain: ek din ko "success" bolo agar usmein $\geq 2$ defects hain, toh per-day success probability $s = 1 - 4 e^{- 3} \approx 0.8009$ hai. Day 4 par pehli success Geometric hai: $P = (1 - s)^{3} s = (4 e^{- 3})^{3} (1 - 4 e^{- 3}) \approx (0.1991)^{3} (0.8009) \approx 0.006318.$ Do layers kyun: din ke andar count Poisson hai; days mein waiting Geometric hai jo har din ek yes/no event par built hai — paanch distributions ko nest karne ka ek clean example.

Problem 5.2

$λ = 3$ ke liye Poisson signature fact numerically confirm karo: $k = 0, \dots, 40$ par truncated definition se mean aur variance compute karo aur dikhao ki dono $3$ ke barabar hain high accuracy ke saath.

Recall Solution 5.2

KYA hai: $E [X] = \sum_{k} k \frac{λ ^{k} e ^{- λ}}{k !}$ aur $Var (X) = E [X^{2}] - (E [X])^{2}$ . Parent derivation $E [X] = λ = 3$ aur $Var (X) = λ = 3$ exactly deta hai. Sums ko $k = 40$ par truncate karna (usse aage ki tail $λ = 3$ ke liye astronomically small hai) dono ke liye $3.000000$ reproduce karta hai. Signature: mean $=$ variance Poisson ki fingerprint hai; agar kisi dataset ka variance uske mean se kaafi zyada ho, toh Poisson galat model hai (yeh "over-dispersed" hai). Yeh equality Binomial limit se inherit hoti hai jahan $n pq = n p (1 - p) \to λ$ jab $p \to 0$ .

Problem 5.3

Maano $X \sim Binomial (n = 5, p = 0.4)$ . MGF $M_{X} (t) = (q + p e^{t})^{n}$ use karke $E [X]$ recover karo, aur check karo ki woh $n p$ ke barabar hai.

Recall Solution 5.3

MGF kyun: $M_{X} (t)$ ke derivatives $t = 0$ par moments generate karte hain — $M_{X}^{'} (0) = E [X]$ — ek infinite sum ko calculus mein badal dete hain. $M_{X} (t) = (q + p e^{t})^{n} \Rightarrow M_{X}^{'} (t) = n (q + p e^{t})^{n - 1} p e^{t} .$ $t = 0$ par, $e^{0} = 1$ aur $q + p = 1$ : $M_{X}^{'} (0) = n \cdot 1^{n - 1} \cdot p = n p = 5 (0.4) = 2.$ Yeh $E [X] = n p = 2$ se match karta hai. ✓

Recall Ek-line self-quiz

Geometric $P (X > m)$ shortcut? ::: $q^{m}$ — pehle $m$ trials sabhi fail ho gaye. Negative Binomial "choose" factor? ::: $(r - 1 k - 1)$ — aakhri trial forced stopping success hai. Poisson mean vs variance? ::: Dono $λ$ ke barabar hain. Poisson ko bade window ke liye scale karna? ::: $λ$ ko window ratio se multiply karo.