Binomial distribution — PMF, mean, variance: Question bank

2.7.13 · D5 · Maths › Statistics & Probability — Intermediate › Binomial distribution — PMF, mean, variance

Ye page Binomial distribution — PMF, mean, variance ki child hai. Ye Bernoulli distribution, Linearity of expectation, Variance and covariance, Binomial Theorem, aur boundary neighbours Poisson distribution, Normal approximation to binomial, aur Hypergeometric distribution pe lean karta hai.

True or false — justify

Recall Binomial ka mean hamesha ek integer hota hai kyunki ye successes count karta hai.

False. Count $X$ hamesha ek integer hota hai, lekin uska average $n p$ zaroori nahi ho. Example: $n = 5, p = 0.5$ deta hai mean $2.5$ — ek aisa average jo koi kabhi actually observe nahi karta. ::: False — $n p$ ek average hai, isliye $2.5$ jaisi values bilkul theek hain, chahe har individual outcome successes ka ek whole number hi kyun na ho.

Recall Agar

$X \sim Bin (n, p)$ toh $n - X \sim Bin (n, 1 - p)$ . ::: True — "number of successes" aur "number of failures" roles swap kar lete hain; failures usi $n$ independent trials mein $1 - p$ probability se hoti hain, isliye $n - X$ binomial hai with success prob $1 - p$ .

Recall

$n$ ko double karne se mean aur standard deviation dono double ho jaate hain. ::: False — mean $n p$ double hota hai, lekin SD $n p (1 - p)$ hai, isliye $n$ double karne se SD sirf $2$ se multiply hoti hai. Spread, mean se slower badhti hai, isliye bade samples proportionally "tighter" ho jaate hain.

Recall Fixed

$n$ ke liye, variance tab sabse zyada hoti hai jab mean sabse zyada hoti hai. ::: False — mean $n p$ , $p \to 1$ hone par $n$ tak badhta rehta hai, lekin variance $n p (1 - p)$ , $p = 0.5$ par peak karti hai aur $p = 1$ par $0$ ho jaati hai. Maximum spread aur maximum mean alag-alag $p$ par hote hain.

Recall Ek binomial random variable negative values le sakta hai agar

$p$ chhota ho. ::: False — $X$ successes count karta hai, isliye ye $p$ ki parwah kiye bina ${0, 1, \dots, n}$ par rehta hai. Chhota $p$ sirf probability ko $0$ ke paas pile karta hai, usse neeche nahi.

Recall Do independent binomials

$Bin (n_{1}, p) + Bin (n_{2}, p)$ ko jodhne se $Bin (n_{1} + n_{2}, p)$ milta hai. ::: True — lekin sirf isliye kyunki unka same $p$ hai. Tab ye sirf $n_{1} + n_{2}$ independent trials ka ek bada pool hai with common success prob $p$ . Alag-alag $p$ ke saath sum bilkul bhi binomial nahi hota.

Recall Binomial ke liye

$P (X = k)$ aur $P (X = n - k)$ hamesha equal hote hain. ::: Aam taur par False — ye sirf tab equal hote hain jab $p = 0.5$ ho, jahan distribution symmetric hoti hai. $p \neq = 0.5$ ke liye factor $p^{k} (1 - p)^{n - k}$ unbalanced hota hai, isliye dono probabilities differ karti hain.

Recall

$X$ ki sabse likely single value hamesha $n p$ ke sabse paas wali hoti hai. ::: True (ek caveat ke saath) — mode $⌊(n + 1) p ⌋$ hota hai, jo $n p$ par ya uske bilkul baad hota hai. Jab $(n + 1) p$ ek whole number ho toh do adjacent equally-likely modes hote hain.

Spot the error

Recall "Kyunki 4 quizzes mein se har ek

$p = 0.7$ se pass hoti hai, toh charon pass karne ka chance $4 \times 0.7 = 2.8$ hai." ::: Probabilities kabhi $1$ se zyada nahi hoti, isliye $2.8$ bakwaas hai. Tum independent probabilities ko multiply karte ho, add nahi: $0. 7^{4} \approx 0.24$ . Add karna "all of" (AND → product) ko expected count se confuse karne ki galti hai.

Recall "

$P (X = 3) = p^{3} (1 - p)^{n - 3}$ kyunki yahi teen successes phir failures ki probability hai." ::: Ye ek specific ordering ki probability hai. Kai orderings exactly 3 successes deti hain, isliye tumhe unhe count karne ke liye $(3 n)$ se multiply karna hoga. $(k n)$ drop karna sabse common PMF mistake hai.

Recall "Var

$(X) = n p (1 - p)$ , isliye standard deviation bhi $n p (1 - p)$ hai." ::: SD variance ka square root hota hai: $σ = n p (1 - p)$ . Variance ko hi SD report karna dono ko confuse karta hai — aur spread ki wrong units deta hai.

Recall "Main 5-card poker hand mein aces ki number ke liye

$Bin (52, \frac{4}{52})$ use karunga." ::: Without replacement draw karne ka matlab hai ki har card ke baad $p$ change hoti hai, isliye trials dependent hain — ye hypergeometric hai, binomial nahi. Saath hi, trials ki number $52$ nahi, $5$ draws hai.

Recall "Mean

$n p (1 - p)$ hai kyunki mean aur variance dono $n p$ se bane hain." ::: Mean $n p$ hai jisme koi $(1 - p)$ factor nahi — ye linearity of expectation se aata hai, jahan har trial average par $p$ add karta hai. Sirf variance mein extra $q = 1 - p$ hota hai.

Recall "'At least one success' ke liye main

$P (X = 1) + P (X = 2) + \dots + P (X = n)$ sum karta hoon; koi shortcut nahi hai." ::: Hai: $P (X \geq 1) = 1 - P (X = 0) = 1 - (1 - p)^{n}$ . Complement ek lambi sum ko ek subtraction se replace kar deta hai. Ye miss karna correctness ka nahi, work-efficiency ka trap hai.

Recall "Kyunki expectation ke liye independence chahiye,

$E [X] = n p$ sirf tab hold karta hai jab trials independent hon." ::: Linearity of expectation dependent variables ke liye bhi hold karta hai, isliye $E [X] = n p$ ko koi independence nahi chahiye. Independence variance formula $n p (1 - p)$ ko chahiye (taaki covariances vanish ho jaayein).

Why questions

Recall Variance ko independence kyun chahiye lekin mean ko nahi?

::: $E [\sum X_{i}] = \sum E [X_{i}]$ hamesha (linearity). Lekin $Var (\sum X_{i}) = \sum Var (X_{i}) + \sum_{i \neq = j} Cov (X_{i}, X_{j})$ ; sirf independence har covariance ko $0$ force karti hai taaki variances simply add ho jayein.

Recall Variance

$p = 0.5$ par maximise kyun hoti hai? ::: $p (1 - p)$ , $p$ mein ek downward parabola hai jo $p = 0.5$ par peak karti hai. Intuitively, $p = 0.5$ maximum uncertainty hai — tum ek single trial predict nahi kar sakte — jabki $p = 0$ ya $1$ certain hain, isliye unka spread zero hota hai.

Recall

$(k n)$ kyun appear hota hai, aur exactly woh formula kyun? ::: Ye count karta hai ki $n$ ordered trials mein $k$ successes ke kitne arrangements exist karte hain; har arrangement ki identical probability $p^{k} (1 - p)^{n - k}$ hoti hai, isliye hum count se multiply karte hain. $(k n) = \frac{n !}{k ! ( n - k )!}$ un ways ki sankhya hai jisse choose kiya jaaye ki kaun se slots succeed karein.

Recall PMF exactly

$1$ tak sum kyun karta hai? ::: $(k n) p^{k} (1 - p)^{n - k}$ ko saare $k$ par sum karna Binomial Theorem ka $(p + (1 - p))^{n} = 1^{n} = 1$ expansion hai. Binomial expansion ki algebra literally wahi bookkeeping hai jo probabilities mein hoti hai.

Recall Single Bernoulli trial ki variance

$p (1 - p)$ kyun hai aur kuch square root wala kyun nahi? ::: Ek indicator $X_{i}$ ke liye, $X_{i}^{2} = X_{i}$ , isliye $E [X_{i}^{2}] = p$ aur $Var = E [X_{i}^{2}] - (E [X_{i}])^{2} = p - p^{2} = p (1 - p)$ . Shortcut $X_{i}^{2} = X_{i}$ hi ise itna clean banata hai.

Recall Large

$n$ ke liye binomial ko Normal se approximate kyun kiya ja sakta hai? ::: $X$ kai independent identical Bernoullis ka sum hai, isliye Central Limit Theorem uski shape ko bell curve ki taraf le jaata hai — Normal approximation to binomial matching mean $n p$ aur variance $n p (1 - p)$ ke saath.

Recall "Rare event, many trials" regime mein binomial Poisson kyun ban jaata hai?

::: Jab $n \to \infty$ aur $p \to 0$ ke saath $n p = λ$ fixed rakha jaaye, toh $(k n) p^{k} (1 - p)^{n - k}$ formula limit karke $\frac{λ ^{k} e ^{- λ}}{k !}$ ban jaata hai — ye Poisson distribution hai. Ye kai chances par rare successes ke counts describe karta hai.

Edge cases

Recall

$Bin (1, p)$ reduce hoke kya banta hai? ::: Ek single Bernoulli trial: $X$ , $1$ prob $p$ se aur $0$ prob $1 - p$ se. Mean $p$ , variance $p (1 - p)$ — har binomial ka atomic building block.

Recall Jab

$p = 0$ ho toh distribution kya hoti hai? ::: $X$ certainty ke saath $0$ hota hai — koi trial succeed nahi kar sakta. Mean $= n \cdot 0 = 0$ aur variance $= n \cdot 0 \cdot 1 = 0$ ; "randomness" bilkul khatam ho gayi.

Recall Jab

$p = 1$ ho toh distribution kya hoti hai? ::: $X = n$ certainty ke saath — har trial succeed karta hai. Mean $= n$ , variance $= n \cdot 1 \cdot 0 = 0$ . Phir ek degenerate, spike distribution hai jisme koi spread nahi.

Recall Aam taur par

$P (X = 0)$ aur $P (X = n)$ kya hain? ::: $P (X = 0) = (1 - p)^{n}$ (saari failures) aur $P (X = n) = p^{n}$ (saari successes), kyunki $(0 n) = (n n) = 1$ har ek ke liye ek hi ordering chodta hai.

Recall

$Bin (0, p)$ — zero trials — ka kya hoga? ::: Sirf ek possible value hai $X = 0$ , probability $1$ ke saath. Mean aur variance dono $0$ hain; count karne ke liye kuch hai hi nahi. Ek valid lekin trivial degenerate case hai.

Recall Kya do alag

$(n, p)$ pairs same mean de sakte hain lekin spread mein bilkul alag? ::: Haan — $Bin (100, 0.05)$ aur $Bin (5, 1.0)$ dono ka mean $5$ hai, lekin pehle ki variance $4.75$ hai jabki doosre ki $0$ . Equal means shape ke baare mein kuch nahi kehte.

Connections

Bernoulli distribution — $n = 1$ edge case jis par upar ke har trap reduce hote hain.
Linearity of expectation — wajah ki mean independence ignore karta hai.
Variance and covariance — wajah ki variance nahi karta.
Binomial Theorem — kyun probabilities ek tak sum karti hain.
Poisson distribution — rare-event limiting edge case.
Normal approximation to binomial — large- $n$ limiting shape.
Hypergeometric distribution — "without replacement" trap.