Binomial distribution — PMF, mean, variance: Foundations

2.7.13 · D1 · Maths › Statistics & Probability — Intermediate › Binomial distribution — PMF, mean, variance

Is page pe assume kiya gaya hai ki tumne koi bhi notation pehle nahi dekha. Hum har symbol ko ground up se build karenge, ek picture at a time, taaki jab tum parent note dubara padho toh kuch bhi mystery na lage.

0. Scene: baar-baar yes/no trials

Ek row of light bulbs imagine karo. Tum har switch flip karte ho. Har bulb ya toh jal jaata hai (hum use success, ✓ kehte hain) ya band rehta hai (ek failure, ✗). Bas itna hi — poora topic ✓ aur ✗ ki rows ke baare mein hai.

Figure s01 (neche) bilkul yahi draw karta hai: chhe labelled bulbs ek row mein, green ✓ boxes for successes aur gray ✗ boxes for failures, taaki tum dekh sako ki "ek experiment = $n$ trials ki ek row" ka matlab kya hai. Baad ke har figure mein yahi green-✓ / gray-✗ colour code reuse hota hai. Yeh BINS ka B hai — Binary — visible form mein: sirf do colours dikhte hain.

Figure — Binomial distribution — PMF, mean, variance

Yeh vocabulary kyun chahiye? Kyunki parent note baar-baar "success/fail" kehta hai — aur agar tum ✓/✗ ki woh row visualize nahi karte, toh baad ki koi bhi counting sense nahi banegi.

1. Symbol $p$ — ek single success ka chance

Picture: $p$ ko ek bar ka fraction socho jo success colour se painted hai. Ek bar jo $0.7$ blue hai matlab har trial mein $70%$ chance hai ✓ ka.

$[0, 1]$ mein number kyun? Kyunki probability certainty ka share hai. Zero share = impossible; full share ( $1$ ) = guaranteed; beech mein kuch bhi partial chance hai. Kuch bhi certain se zyada certain nahi ho sakta, isliye yeh $1$ se kabhi zyada nahi hoti.

Kyunki $p$ har trial pe same hota hai, yeh BINS ka S hai — Same probability. Agar $p$ har trial pe drift karta, toh neat formulas toot jaate.

Parent note har jagah $(1 - p)$ likhta hai aur kabhi-kabhi use $q$ se short karta hai. Dono same cheez hain — bar ka un-painted part.

$p + (1 - p) = 1 ⟺ p + q = 1.$

2. Symbol $n$ — kitne trials

Picture literally bulbs ki row ki length hai. Das bulbs → $n = 10$ .

Pehle se fixed kyun hona chahiye? Kyunki agar tum "jab tak mann lage tab tak" flip karte rehte, toh count tumhare mood pe depend karta, chance pe nahi — aur neat formulas toot jaate. $n$ fix karna BINS ka N hai — Number fixed.

3. Symbol $X$ aur $k$ — count, aur ek value jo wo le sakta hai

Toh " $X = k$ " English sentence ka shorthand hai: "successes ka count $k$ ke equal nikla." Aur " $P (X = k)$ " padhte hain "probability ki exactly $k$ successes milein."

Figure s02 (neche) ek five-trial row dikhata hai jisme teen ✓ hain; ek blue arrow kehta hai "checks count karo," orange result $X = 3$ pe land karta hai, yaani value $k = 3$ . Iska pedagogical kaam hai: $X$ rule ko $k$ value se visually alag karna, taaki notation $P (X = k)$ abstract lagna band ho jaaye.

$X$ ko $k$ se alag kyun karo? $X$ rule hai ("✓ count karo"); $k$ ek particular answer hai jo hum plug in karte hain. Dono ko alag rakhne se hum ek PMF likh sakte hain aur use har possible answer $k = 0$ se $k = n$ tak reuse kar sakte hain.

4. Independence — isliye hum multiply kar sakte hain

Picture aur rule: independent events ke liye, "yeh AND woh AND woh" ki probability separate probabilities ka product hoti hai.

$P (✓ then ✓ then ✓) = p \times p \times p = p^{3} .$

Yeh itna important kyun hai? Parent note ka pehla PMF step $p$ aur $q$ ko saath multiply karta hai. Woh multiplication tabhi legal hai jab trials independent hon — yeh BINS ka I hai, Independent. Single-trial atom ke liye Bernoulli distribution dekho jis pe yeh sab built hai.

5. Powers — $p^{k}$ aur $q^{n - k}$ seedhe shabdon mein

Toh jab parent $p^{k}$ likhta hai, matlab hai: " $k$ trials succeed hue, toh $p$ ko $k$ baar khud se multiply karo." Aur $q^{n - k} = (1 - p)^{n - k}$ matlab: "baaki $n - k$ trials fail hue, toh $q$ ko utni baar multiply karo."

$n - k$ kyun? Agar $n$ trials mein se $k$ succeed hue, toh $n - k$ (baaki sab) fail hue. Dono exponents hamesha $n$ mein add hote hain: $k + (n - k) = n$ , ek factor per trial.

$k successes p^{k} \times n - k failures (1 - p)^{n - k} = probability of ONE particular row of ✓ / \times .$

6. Factorial $n!$ — arrangements count karna

Picture: $n!$ woh number count karta hai jo $n$ distinct objects ko order mein line up karne ke kitne tarike hain. Pehla slot: $n$ choices; agla slot: $n - 1$ baaki; aur aise $1$ tak.

Yeh kyun chahiye? Kyunki $(k n)$ build karne ke liye humen ek tool chahiye jo arrangements count kare — aur $n!$ exactly wahi tool hai.

7. Binomial coefficient $(k n)$ — "kitne tarike?"

Yeh poore topic ka star symbol hai, isliye hum ise slowly build karte hain.

Figure s03 (neche) $n = 4, k = 2$ ke liye char boxes mein do ✓ ke saare $6$ patterns draw karta hai — ek grid mein laid out taaki tum literally unhe count kar sako, aur confirm kar sako ki formula wahi $6$ return karta hai. Iska kaam: $(k n)$ ko "distinct ✓-placements ki number" ke roop mein concrete banana, kisi bhi formula se pehle.

Figure dekho: $n = 4$ slots aur $k = 2$ successes ke saath, char boxes mein do ✓ ke exactly $6$ distinct patterns hain. Humen ek aisa formula chahiye jo bina har pattern draw kiye woh $6$ de.

YEH FORMULA KYUN (upar ke pieces se build karke):

$n!$ sab $n$ slots ko order karta hai — lekin yeh over-count karta hai, kyunki humen chosen $k$ mein se order ki parwah nahi, na hi baaki $n - k$ mein se.
$k!$ se divide karo successes ke beech ordering mitaane ke liye.
$(n - k)!$ se divide karo failures ke beech ordering mitaane ke liye.

Figure se check karo ( $n = 4, k = 2$ ): $(2 4) = \frac{4 !}{2 ! 2 !} = \frac{24}{2 \cdot 2} = \frac{24}{4} = 6. ✓$

Topic ko $(k n)$ kyun chahiye? Kyunki bahut saari alag-alag rows same count $k$ deti hain, aur har ek ki same probability $p^{k} q^{n - k}$ hoti hai. Toh "exactly $k$ successes" ka sahi chance hai (woh probability) $\times$ (kitni rows), aur $(k n)$ doosra factor supply karta hai. Yeh exactly parent note ka Binomial Theorem ingredient hai.

8. Sum symbol $\sum$ — sab cases mein add karna

Picture: $k = 0, 1, \dots, n$ ke liye probability bars line up karo aur unhe end-to-end glue karo.

Topic ko yeh kyun chahiye: check karne ke liye ki PMF honest hai, tum har possible count ki probabilities add karte ho. Kyunki koi na koi count zaroor hoga, unhe exactly $1$ total hona chahiye: $\sum_{k = 0}^{n} (k n) p^{k} (1 - p)^{n - k} = (p + (1 - p))^{n} = 1^{n} = 1.$

9. Expectation $E [X]$ — long-run average, aur yeh $n p$ kyun hai

Picture: yeh probability bars ka balance point hai — jahan poori distribution ek ungali ki tip pe balance ho jaaye.

Ab $E [X] = n p$ zero se derive karo. Poore count ko har trial ke liye ek tiny counter mein break karo. Maano $X_{i} = 1$ agar trial $i$ succeed kare aur $X_{i} = 0$ agar fail kare. Toh total count simply inhi chhote counters ka sum hai: $X = X_{1} + X_{2} + \dots + X_{n} .$

Yeh step kyun? Har success ke liye $1$ add karna literally successes count karne ke barabar hai.

Ek single counter ka average easy hai — value times probability, add karke: $E [X_{i}] = success 1 \cdot p + failure 0 \cdot (1 - p) = p .$

Intuitive picture: average pe har trial ek success ka $p$ fraction contribute karta hai (ek fair coin $0.5$ per toss contribute karta hai).

Ab $n$ trials ko linearity of expectation se jodo — independence ki zarurat nahi: $E [X] = E [X_{1}] + \dots + E [X_{n}] = n copies p + p + \dots + p = n p .$

$E [X] = n p$

Toh mean bas "har trial kitna pitch in karta hai ( $p$ ), times kitne trials hain ( $n$ )" hai.

10. Variance $Var (X)$ aur $σ$ — kitna spread out, aur yeh $n pq$ kyun hai

Picture: bars ka narrow cluster = small variance (outcomes average ke paas bunch karte hain); wide spread = large variance.

$Var (X) = n p (1 - p)$ step by step derive karo. Section 9 ke same chhote counters $X_{i}$ use karo.

Step 1 — ek counter ki wiggle. Ek handy shortcut: kyunki $X_{i}$ sirf $0$ ya $1$ hota hai, square karne se kuch nahi badalta ( $0^{2} = 0, 1^{2} = 1$ ), isliye $E [X_{i}^{2}] = E [X_{i}] = p$ . Phir $Var (X_{i}) = E [X_{i}^{2}] - (E [X_{i}])^{2} = p - p^{2} = p (1 - p) = pq .$

Intuitive picture: $pq$ tab sabse bada hota hai jab $p = 0.5$ (ek fair coin sabse zyada unpredictable hai — maximum wiggle) aur $0$ ho jaata hai jab $p = 0$ ya $p = 1$ (ek certain trial mein koi wiggle nahi hoti).

Step 2 — wiggles add karo. Yahan — mean ke unlike — humen independence chahiye (BINS ka I). Sirf jab trials interfere nahi karte tab unki wiggles bina kisi cross-terms (covariances) ke add hoti hain; yeh exactly Variance and covariance rule hai. $n$ identical wiggles add karo: $Var (X) = Var (X_{1}) + \dots + Var (X_{n}) = n copies pq + \dots + pq = n pq = n p (1 - p) .$

$Var (X) = n p (1 - p), σ = n p (1 - p) .$

Topic ko yeh kyun chahiye: sirf average jaanna kaafi nahi — " $2$ defects expected" tab kaafi zyada useful hota hai jab tum yeh bhi jaano ki "give or take $σ$ ."

Prerequisite map

Neche ka diagram dikhata hai ki plain-word ingredients upar kaise feed karte hain: atomic trial $p$ aur $q$ mein split hota hai; $p$ , independence aur powers per-row probability build karte hain; $n!$ coefficient build karta hai; saath milke woh PMF assemble karte hain, jo phir mean aur variance deta hai. Ise bottom-up padho jaise "isse pehle mujhe kya samajhna chahiye?"

Equipment checklist

Right side cover karo aur reveal karne se pehle har ek ka jawab do.

BINS ke charon letters ka kya matlab hai?

Binary (do outcomes), Independent trials, Number

n

fixed, Same probability

p

p

ka kya matlab hai aur woh kaunsi range le sakta hai?

Single trial pe success ki probability;

[0, 1]

mein ek number.

q

ek line mein kya hai?

Failure probability

q = 1 - p

; saath mein

p + q = 1

n

shuru karne se pehle fixed kyun hona chahiye?

Taaki count is baat pe depend na kare ki tum kab ruke; yeh BINS mein "N" (Number fixed) hai.

X

aur

k

mein kya fark hai?

X

successes ka random count hai;

k

ek specific value hai jiske baare mein hum poochte hain, jaise

P (X = k)

mein.

"PMF" ka kya matlab hai aur kya hota hai?

Probability mass function — woh rule jo har value

k

ke liye

P (X = k)

deta hai.

Probabilities multiply karne ki permission kab milti hai?

Sirf jab events independent hon — ek outcome doosre ke baare mein kuch nahi batata.

p^{k}

ka words mein kya matlab hai?

p

ko khud se

k

baar multiply karo — woh chance ki

k

chosen trials sab succeed karein.

0!

aur

a^{0}

kya hain?

Dono

1

ke equal hain (kuch bhi arrange karne ka ek tarika; zero copies multiply karna

1

chhodta hai).

(k n)

kya count karta hai, aur iska formula kya hai?

Woh tarike kitne hain ki

n

slots mein se kaunse

k

succeed hote hain;

\frac{n !}{k ! ( n - k )!}

PMF mein

(k n)

kyun include hota hai?

Bahut saari equally-likely rows

k

successes deti hain; coefficient unhe sab count karta hai.

Linearity of expectation kya hai aur iska kyun matlab rakhte hain?

E [A + B] = E [A] + E [B]

kisi bhi quantities ke liye; yeh

E [X] = n p

independence ke bina deta hai.

E [X] = n p

ek line mein derive karo.

X = \sum X_{i}

with

E [X_{i}] = p

, toh linearity se

E [X] = n p

Var (X) = n p (1 - p)

ek line mein derive karo.

Har

X_{i}

ka variance

p (1 - p)

hai; independence se woh add hokar

n p (1 - p)

bante hain.

Kaun se

p

pe per-trial wiggle

pq

sabse bada hota hai?

p = 0.5

pe;

p = 0

p = 1

pe yeh

0

hota hai.

Connections

Bernoulli distribution — single trial ( $n = 1$ ) jis se yahan har symbol build hua hai.
Binomial Theorem — $(p + q)^{n}$ expand karta hai, $(k n)$ supply karta hai aur prove karta hai ki PMF ka sum $1$ hota hai.
Linearity of expectation — woh tool jo per-trial averages ko $E [X] = n p$ mein turn karta hai.
Variance and covariance — kyun independence per-trial wiggles ko $n p (1 - p)$ mein add hone deta hai.
Poisson distribution — jahan binomial jaata hai jab $n$ bahut bada ho jaata hai aur $p$ chhota hota jaata hai.
Normal approximation to binomial — woh smooth bell shape jiske paas bars large $n$ ke liye approach karte hain.
Hypergeometric distribution — without-replacement cousin (dependent trials independence tod dete hain).