Binomial distribution — PMF, mean, variance: Worked examples

2.7.13 · D3 · Maths › Statistics & Probability — Intermediate › Binomial distribution — PMF, mean, variance

Scenario matrix

Har binomial question in cells mein se ek hai. Neeche har worked example us cell ke saath tagged hai jisme woh fit baith ta hai.

Cell	Kya poochha jaata hai	Tool
A. Exact	$P (X = k)$ ek value of $k$ ke liye	PMF ek baar
B. At least one	$P (X \geq 1)$	complement $1 - (1 - p)^{n}$
C. At most / range	$P (X \leq m)$ ya $P (a \leq X \leq b)$	kuch PMF terms sum karo
D. Summary stats	mean, variance, standard deviation, "typical range"	$n p$ , $n p (1 - p)$
E. Extreme $p$	$p = 0$ ya $p = 1$ (degenerate)	distribution ek point par collapse ho jaati hai
F. Extreme $k$	$k = 0$ ya $k = n$ (the tails)	$(0 n) = (n n) = 1$
G. Not binomial	dependent trials / changing $p$	pehchano aur model switch karo
H. Poisson limit	bahut bada $n$ , bahut chhota $p$ , $n p$ moderate	$λ = n p$ approximation
I. Normal limit	large $n$ , middling $p$	$N (n p, n p (1 - p))$ + continuity

Figure — Binomial distribution — PMF, mean, variance

Neeche har worked example exactly woh cell naam leta hai jisme woh land karta hai, taaki tum dekh sako ki tree ka poora exercise ho raha hai.

Example 1 — Cell A (exact $k$ )

Worked example Exactly 3 sixes in 6 rolls of a fair die

Ek fair die $n = 6$ baar roll ki jaati hai. "Success" $=$ six aana, isliye $p = \frac{1}{6}$ . $P (X = 3)$ nikalo.

Forecast: andaaza lagao — kya yeh $10%$ se zyada hai ya kam? (Sixes rare hote hain, to teen milna unlikely lagta hai.)

Check BINS. Chhe fixed rolls, har ek six/not-six, independent, same $p = \frac{1}{6}$ . Binomial. Yeh step kyun? PMF apply karne se pehle chaaon conditions confirm karo (Binary, Independent, Number fixed, Same $p$ ) — cell G exactly isliye exist karta hai kyunki log yeh step skip karte hain.
PMF pick karo. Hum ek exact count $k = 3$ chahte hain, to yeh cell A hai: PMF ek baar apply karo. Yeh step kyun? Cell A ek single value of $k$ maangta hai, aur PMF $(k n) p^{k} (1 - p)^{n - k}$ exactly woh tool hai jo ek specific count ki probability return karta hai — koi summing nahi chahiye. $P (X = 3) = (3 6) (\frac{1}{6})^{3} (\frac{5}{6})^{3} .$
Orderings count karo. $(3 6) = \frac{6 !}{3 ! 3 !} = 20$ . Yeh step kyun? Teen sixes chhe rolls mein $20$ alag-alag positions mein baith sakti hain; har ordering ki same probability hai, isliye hum multiply karte hain.
Pieces multiply karo. $P (X = 3) = 20 \cdot \frac{1}{216} \cdot \frac{125}{216} = \frac{2500}{46656} \approx 0.0536.$ Yeh step kyun? Hum orderings ka count ( $20$ ) ko ek ordering ki shared probability $(\frac{1}{6})^{3} (\frac{5}{6})^{3} = \frac{125}{46656}$ ke saath combine karte hain, exactly jaisa PMF prescribe karta hai.

Verify: $\approx 5.4%$ — $10%$ se neeche, forecast se match karta hai. Units: ek probability, dimensionless, aur $0 \leq 0.0536 \leq 1$ . ✓

Is distribution ki poori shape Figure 2 ke histogram mein hai, jisme $k = 3$ bar highlighted hai.

Example 2 — Cell B (at least one)

Worked example At least one bullseye

Ek archer har independent shot mein bullseye hit karta hai probability $p = 0.2$ se. Woh $n = 10$ shots lete hai. $P (at least one bullseye)$ nikalo.

Forecast: 10 tries at 20% each ke saath, "at least one" $50%$ ke kareeb lagta hai ya $90%$ ke?

Cell B pehchano. "At least one" ka matlab hai $X \geq 1$ , yaani $k = 1, 2, \dots, 10$ — nau terms. Yeh step kyun? Nau PMF values sum karna slow hai; complement ek line mein ho jaata hai (yeh parent note se 80/20 move hai).
Complement ki taraf jaao. $P (X \geq 1) = 1 - P (X = 0) .$ Yeh step kyun? At least one na milne ka ek hi tarika hai — exactly zero milna — jo ek asaan sa term hai.
$P (X = 0)$ compute karo. $(0 10) = 1$ , to $P (X = 0) = (0.8)^{10} \approx 0.1074.$ Yeh step kyun? $k = 0$ par sirf ek arrangement hai (sab misses), isliye coefficient $1$ hai aur sirf das failure probabilities ka product bachta hai.
Subtract karo. $P (X \geq 1) = 1 - 0.1074 = 0.8926$ . Yeh step kyun? Total probability $1$ hai, isliye jo "zero bullseyes" nahi hai woh "at least one bullseye" zaroor hai.

Verify: $\approx 0.893$ , comfortably $90%$ ke kareeb — forecast se match karta hai. Sanity: jaise $n$ badhta hai, $(0.8)^{n} \to 0$ isliye yeh probability $1$ ki taraf rise karti hai, exactly jaisa intuition maangta hai. ✓

Example 3 — Cell C (at most / range)

Worked example At most 2 defects

Ek production line $p = 0.05$ ke saath defects produce karta hai. $n = 8$ ke sample mein $P (X \leq 2)$ nikalo.

Forecast: defects rare hain, isliye "at most 2" bahut likely hona chahiye — $95%$ se upar?

Cell C pehchano. "At most 2" $= k \in {0, 1, 2}$ — ek chhota sum. Yeh step kyun? $P (X \leq m)$ ek single PMF value nahi hai; yeh ek cumulative probability hai, isliye hume har allowed count add karna hoga. Yahan sirf teen terms qualify karti hain, isliye direct addition kisi bhi complement se quick hai.
Term $k = 0$ . $(0 8) (0.95)^{8} = (0.95)^{8} \approx 0.6634$ . Yeh step kyun? Ek arrangement (sab good), isliye coefficient $1$ hai aur 8 items mein se har ek non-defective hona chahiye.
Term $k = 1$ . $(1 8) (0.05) (0.95)^{7} = 8 \cdot 0.05 \cdot 0.6983 \approx 0.2793$ . Yeh step kyun? $(1 8) = 8$ — single defect 8 items mein se kisi mein bhi ho sakta hai, aur har aisa arrangement ek $0.05$ aur saat $0.95$ factors carry karta hai.
Term $k = 2$ . $(2 8) (0.05)^{2} (0.95)^{6} = 28 \cdot 0.0025 \cdot 0.7351 \approx 0.0514$ . Yeh step kyun? $(2 8) = 28$ count karta hai ki do defects kitne positions ke pairs mein ho sakte hain, har ek do $0.05$ aur chhe $0.95$ factors ke saath.
Add karo. $0.6634 + 0.2793 + 0.0514 \approx 0.9941$ . Yeh step kyun? Teen counts $k = 0, 1, 2$ mutually exclusive outcomes hain, isliye "at most 2" ki probability unka sum hai.

Verify: $\approx 0.994$ — $95%$ se upar, forecast se match karta hai. Cross-check: $P (X \geq 3) = 1 - 0.9941 = 0.0059$ , ek chhoti si tail, exactly jo "rare defects" predict karta hai. ✓

Example 4 — Cell D (summary statistics)

Worked example Correct answers ka typical range

Ek student $n = 40$ four-option multiple-choice questions par guess karta hai, isliye $p = 0.25$ . Mean, variance, standard deviation, aur correct answers ka "typical" range do.

Forecast: 40 questions par 25% ke saath, average mein roughly kitne correct? Aur kya luck se $20$ correct possible hai?

Mean (cell D). $E [X] = n p = 40 (0.25) = 10$ . Yeh step kyun? Linearity of expectation: har question average mein $p$ correct contribute karta hai, aur 40 questions hain.
Variance. $Var (X) = n p (1 - p) = 40 (0.25) (0.75) = 7.5$ . Yeh step kyun? Trials independent hain, isliye variances add hote hain — har Bernoulli $p (1 - p)$ contribute karta hai.
Standard deviation. $σ = 7.5 \approx 2.739$ , upar wali definition use karke.
Typical range. $E [X] \pm 2 σ = 10 \pm 5.48$ , yaani roughly $5$ se $15$ correct. Yeh step kyun? Bell-ish distribution ka zyaadaatar mass mean ke do standard deviations ke andar hota hai.

Verify: $20$ correct hai $(20 - 10) /2.739 \approx 3.65$ standard deviations mean se upar — pure guessing se bahut unlikely, isliye score of 20 real knowledge signal karta hai. Units: "correct answers", ek count. Mean $10 < n = 40$ . ✓

Mean aur $\pm 2 σ$ band Figure 3 mein distribution ke upar draw kiya gaya hai.

Example 5 — Cell E (degenerate $p$ )

Worked example Jab randomness gayab ho jaaye

Ek biased coin mein $p = 1$ hai (hamesha heads). Ise $n = 7$ baar flip kiya jaata hai. $X =$ number of heads ki distribution nikalo. Phir $p = 0$ ke liye repeat karo.

Forecast: agar yeh hamesha heads land karta hai, to kitni uncertainty bachi hai?

PMF mein $p = 1$ plug karo. $P (X = k) = (k 7) (1)^{k} (0)^{7 - k}$ . Yeh step kyun? Hum formula ko uski boundary par test karte hain dekhne ke liye ki woh sane rehti hai ya nahi.
Zero har term ko kill karta hai $k = 7$ ko chhodkar. $k < 7$ ke liye factor $(0)^{7 - k} = 0$ hai; sirf $k = 7$ mein $(0)^{0} = 1$ milta hai (upar wala convention). To $P (X = 7) = 1$ , baaki sab probabilities $0$ . Yeh step kyun? $0$ ka factor kisi bhi term ko wipe out kar deta hai, isliye saari probability usi ek outcome par pile ho jaati hai jisme failure factor nahi hai.
Summary stats check karo. $E [X] = n p = 7$ , $Var = n p (1 - p) = 7 \cdot 1 \cdot 0 = 0$ . Yeh step kyun? Zero variance ka matlab koi spread nahi — outcome certain hai, $7$ par spike se match karta hai.
Mirror case $p = 0$ . Ab $k > 0$ wale har term mein $(0)^{k} = 0$ hai; sirf $k = 0$ bachta hai, $P (X = 0) = 1$ , aur $Var = n \cdot 0 \cdot 1 = 0$ . Yeh step kyun? Success aur failure swap karke same zero-kills-terms logic se, saari mass "no heads" par land karti hai.

Verify: dono variances $0$ hain — parent note se "max at $p = 0.5$ , zero at the ends" rule exactly hold karta hai. Distribution ek single point par collapse ho jaati hai, koi randomness nahi bachti, forecast se match karta hai. ✓

Example 6 — Cell F (extreme $k$ , dono tails)

Worked example Coin experiment ke dono tails

Ek fair coin $n = 5$ baar flip karo ( $p = 0.5$ ). $P (X = 0)$ aur $P (X = 5)$ nikalo, aur confirm karo ki dono equal hain.

Forecast: fair coin ke liye, kya "all tails" aur "all heads" equally likely hone chahiye?

Bottom tail $k = 0$ . $(0 5) = 1$ , to $P (X = 0) = (0.5)^{0} (0.5)^{5} = (0.5)^{5} = \frac{1}{32} \approx 0.03125$ . Yeh step kyun? $k = 0$ "all failures" corner hai — ek arrangement, paanch failure factors.
Top tail $k = 5$ . $(5 5) = 1$ , to $P (X = 5) = (0.5)^{5} = \frac{1}{32} \approx 0.03125$ . Yeh step kyun? $k = n$ "all successes" corner hai — phir ek arrangement, paanch success factors.
Yahan equal kyun? Kyunki $p = 1 - p = 0.5$ : successes aur failures swap karne par har term ki value unchanged rehti hai. Yeh step kyun? Symmetry sirf $p = 0.5$ par aati hai; biased coins mein tails differ karti hain.

Verify: dono $\frac{1}{32}$ equal hain. Agar $p = 0.3$ hota: $P (X = 0) = 0. 7^{5} \approx 0.1681$ lekin $P (X = 5) = 0. 3^{5} \approx 0.00243$ — ab equal nahi, confirm karta hai ki symmetry $p = 0.5$ ke liye special hai. ✓

Example 7 — Cell G (NOT binomial)

Worked example Without-replacement ka trap

Ek box mein $10$ balls hain: $4$ red, $6$ blue. Tum $3$ balls without replacement draw karte ho. $P (exactly 2 red)$ nikalo. Kya binomial use kar sakte hain?

Forecast: kya $p$ (red ki chance) har draw mein same rehti hai?

BINS test karo. Ek red ball remove karne ke baad sirf $3$ reds bachti hain $9$ mein se — isliye $p$ change ho jaata hai $\frac{4}{10}$ se $\frac{3}{9}$ par. Yeh step kyun? "Same $p$ " aur "Independent" dono conditions fail hoti hain; binomial yahan illegal hai.
Hypergeometric distribution par switch karo. Favourable over total count karo: $P (2 red) = \frac{( 2 4 ) ( 1 6 )}{( 3 10 )} = \frac{6 \cdot 6}{120} = \frac{36}{120} = 0.3.$ Yeh step kyun? $4$ reds mein se $2$ aur $6$ blues mein se $1$ choose karo, $10$ mein se $3$ choose karne ke sab tareekon par.
Galat binomial answer se compare karo. Naively $p = 0.4$ ke saath: $(2 3) (0.4)^{2} (0.6) = 0.288$ — close hai lekin galat hai. Yeh step kyun? Dono numbers side by side rakhne se pata chalta hai ki dependent draw par binomial force karne se kitni badi error hoti hai.

Verify: sahi answer $0.3$ hai; binomial shortcut $0.288$ deta hai, $0.012$ ki error. Chhota $n$ population size ke relative to kyon dono kareeb hain. ✓

Example 8 — Cell H (Poisson limit)

Worked example Lambe document mein rare typos

Ek typist har word mein $p = 0.002$ probability se error karta hai, $n = 1500$ words mein (independent). Poisson distribution approximation use karke $P (exactly 2 errors)$ nikalo.

Forecast: bada $n$ , chhota $p$ — average mein kitni errors, aur exactly 2 kitni likely?

Cell H pehchano. $n = 1500$ bahut bada hai, $p = 0.002$ bahut chhota hai, lekin $λ = n p = 3$ moderate hai. Yeh step kyun? $n = 1500$ ke saath binomial terms sum karna painful hai; Poisson limit ise ek clean formula se replace karta hai.
$λ = n p = 1500 \cdot 0.002 = 3$ set karo. Yeh errors ki expected number hai. Yeh step kyun? Poisson approximation poori tarah apne mean $λ$ se governed hoti hai, jo binomial mean $n p$ se match hona chahiye.
Poisson PMF apply karo $P (X = k) = e^{- λ} \frac{λ ^{k}}{k !}$ : $P (X = 2) = e^{- 3} \frac{3 ^{2}}{2 !} = e^{- 3} \cdot \frac{9}{2} \approx 0.2240.$ Yeh step kyun? Jaise $n \to \infty, p \to 0$ with $n p = λ$ fixed, binomial Poisson ki taraf converge karta hai, isliye hum simpler formula se answer padh sakte hain.

Verify: exact binomial deta hai $(2 1500) (0.002)^{2} (0.998)^{1498} \approx 0.2242$ — teen decimals tak agree karta hai, approximation confirm hoti hai. ✓

Example 9 — Cell I (Normal approximation)

Worked example Continuity correction ke saath polling

$n = 100$ voters ke poll mein, har voter ek measure ko $p = 0.5$ probability se support karta hai. Normal approximation to binomial use karke $P (X \leq 45)$ nikalo.

Forecast: $45$ mean $50$ se neeche hai — kya yeh $50%$ se well under hona chahiye?

Rule of thumb check karo. Continuity-corrected Normal tabhi trustworthy hai jab $n p \geq 10$ aur $n (1 - p) \geq 10$ dono hon. Yahan $n p = 50$ aur $n (1 - p) = 50$ , dono comfortably $10$ se upar. Yeh step kyun? Yeh conditions guarantee karti hain ki distribution itni tall aur symmetric hai ki ek smooth bell uske bars fit kar sake; check skip karna woh tarika hai jisse log approximation ko skewed, small- $n$ cases par misuse karte hain.
Cell I pehchano. $n = 100$ bada hai aur $p = 0.5$ central hai, isliye distribution ek bell curve ke kareeb hai — haath se $46$ PMF terms sum karna absurd hai, aur Normal curve ise ek lookup mein de deta hai. Yeh step kyun? Yahan Normal limit choose karna hi ek 46-term sum ko ek single standardised lookup mein badal deta hai.
Mean aur variance match karo. $μ = n p = 50$ , $σ = n p (1 - p) = 25 = 5$ . Yeh step kyun? Approximating bell ko binomial ka centre aur spread share karna chahiye, warna woh galat jagah par baith jaayegi.
Continuity correction. Discrete $X \leq 45$ continuous $X \leq 45.5$ ban jaata hai. Yeh step kyun? Har integer ke upar width 1 ka ek poora bar hota hai; $45.5$ tak half-bar include karo taaki area match kare.
$Z$ use karke standardise karo. Maano $Z$ standard normal random variable hai, $Z \sim N (0, 1)$ — ek bell curve centred at $0$ with standard deviation $1$ . Hum apne count ko mean subtract karke aur $σ$ se divide karke $Z$ -value mein convert karte hain: $z = \frac{45.5 - 50}{5} = - 0.9.$ Yeh step kyun? Standardise karne se hum kisi bhi Normal probability ko ek universal $Z$ -table se padh sakte hain, har $μ, σ$ ke liye nayi curve ki jagah.
Probability padho. Standard-normal table se, $P (Z \leq - 0.9) \approx 0.1841$ . Yeh step kyun? Ek single $z$ -value milne ke baad, uske left ka tail area hi woh answer hai jo hum chahte the.

Verify: exact binomial sum $P (X \leq 45) \approx 0.1841$ bhi deta hai — continuity-corrected Normal ekdum sahi hai. Aur $0.184 < 0.5$ , forecast se match karta hai ki $45$ mean se neeche hai. ✓

Half-bar shift aur bell overlay Figure 4 mein dikhaya gaya hai.

Active recall

Recall Kaunsa cell "at least one" hai, aur complement kyun?

Cell B. "At least one" kai $k$ values span karta hai, lekin uska complement ek single easy term $P (X = 0) = (1 - p)^{n}$ hai, isliye $P (X \geq 1) = 1 - (1 - p)^{n}$ .

Recall Example 7 ko binomial kyun nahi banana, aur kya replace karta hai?

Without replacement draw karne se har draw mein $p$ change ho jaata hai (aur independence bhi toot jaati hai), isliye "Same $p$ " condition fail hoti hai. Iske bajay Hypergeometric distribution use karo.

Recall Poisson vs Normal kab use karte hain?

Poisson tab jab $n$ bahut bada ho aur $p$ tiny ho with $λ = n p$ moderate (rare events). Normal tab jab $n$ bada ho aur $p$ extreme na ho taaki $n p \geq 10$ aur $n (1 - p) \geq 10$ ho (bell-shaped) — continuity correction yaad rakho aur yeh ki $Z \sim N (0, 1)$ .

Connections

Binomial distribution — PMF, mean, variance — parent formulas jo upar har example use karta hai.
Bernoulli distribution — in examples mein har single trial.
Binomial Theorem — kyun PMF ka sum 1 hota hai.
Linearity of expectation — Example 4 mein mean.
Variance and covariance — Example 4 mein variances kyun add hote hain.
Poisson distribution — Example 8 mein rare-event limit.
Normal approximation to binomial — Example 9 mein large- $n$ limit.
Hypergeometric distribution — Example 7 mein sahi model.