Binomial Distribution

Context: FIT1058_MOC, FIT2086_MOC · Bernoulli trial (one success/failure experiment) + binomial (successes across $n$ i.i.d. trials) · pmf uses the Binomial Coefficient · mean/variance via linearity of indicator sums · the two-outcome member of the parametric zoo

Quick Revision

• 🎯 Objective: Bernoulli atom: $X\in \{0,1\}$, $\mathrm{Pr}(X=1)=p$ ➔ sum $n$ i.i.d. atoms ⟹ $\mathrm{Pr}(Z=k)=\left(\genfrac{}{}{0ex}{}{n}{k}\right)p^k(1-p{)}^{n-k}$.
• 📦 Core Components: $E(X)=p$, $\mathrm{Var}(X)=p(1-p)$ ➔ $E(Z)=np$ (linearity) ➔ $\mathrm{Var}(Z)=np(1-p)$ (independence).
• ⚡ Key Constraint: indicator-sum trick beats brute-force pmf algebra; linearity needs no independence, variance does.

📝 Core

1. The Bernoulli Atom (Single Trial)

• Definition ➔ $X=1$ with probability $p$ (success), $X=0$ with $1-p$ — any two-outcome experiment (coin, pass/fail, on/off).
• Compact pmf form ➔ $p(x\mid \theta )={\theta }^x(1-\theta {)}^{1-x}$ on $x\in \{0,1\}$, written $X\sim Be(\theta )$ in FIT2086’s $\theta \in [0,1]$ notation ➔ the exponents act as selectors, collapsing to $\theta$ at $x=1$ and $1-\theta$ at $x=0$.
• Moments ➔ $E(X)=p$; $\mathrm{Var}(X)=E(X^2)-E(X{)}^2=p-p^2=p(1-p)$ (since $X^2=X$); variance max at $p=\frac{1}{2}$ (value $0.25$), zero at $p\in \{0,1\}$.
• Sequences ➔ i.i.d. = same $p$, independent trials; the atom of binomial (count successes) and geometric (wait for first).

2. The Binomial Distribution

• Definition ➔ $Z\sim \mathrm{Bin}(n,p)$ (FIT2086 writes $M\sim Bin(\theta ,n)$) = the number of successes $m(\mathbf{x})={\sum }_{j=1}^n{x}_j$ across $n$ i.i.d. Bernoulli trials; the count $M$ is itself an RV over $\{0,1,\dots ,n\}$.
• pmf derivation ➔ one specific $k$-success sequence has probability ${\prod }_{i=1}^{n}p(x_i\mid \theta )=p^k(1-p{)}^{n-k}$; $\left(\genfrac{}{}{0ex}{}{n}{k}\right)$ placements (Binomial Coefficient) ⟹ $\mathrm{Pr}(Z=k)=\left(\genfrac{}{}{0ex}{}{n}{k}\right)p^k(1-p{)}^{n-k}$.
• Why the coefficient is needed ➔ for $1\le k\le n-1$ several sequences carry $k$ successes; $k=2,n=4$ admits the six orderings $1100,1010,1001,0110,0101,0011$ ⟹ $p(m=2\mid \theta )=\left(\genfrac{}{}{0ex}{}{4}{2}\right){\theta }^2(1-\theta {)}^2$.
• Additivity in $n$ (FIT2086) ➔ $M_1\sim Bin(\theta ,n_1)$ and $M_2\sim Bin(\theta ,n_2)$ with the same $\theta$ ⟹ $M_1+M_2\sim Bin(\theta ,n_1+n_2)$ — immediate from the definition as a sum of Bernoulli variates, so pooling batches simply adds trial counts.
• Reading the pmf term by term (the standard “interpret this factor” exam question):

Factor in $\left(\genfrac{}{}{0ex}{}{n}{m}\right){\theta }^m(1-\theta {)}^{n-m}$	Interpretation
${\theta }^m$	probability of seeing $m$ successes
$(1-\theta {)}^{n-m}$	probability of the event not occurring $n-m$ times, i.e. $n-m$ failures
${\theta }^m(1-\theta {)}^{n-m}$	probability of $m$ successes and $n-m$ failures in one particular ordering
$\left(\genfrac{}{}{0ex}{}{n}{m}\right)$	the number of orderings of $m$ successes among $n$ trials
${\sum }_{k=i}^n\left(\genfrac{}{}{0ex}{}{n}{k}\right){\theta }^k(1-\theta {)}^{n-k}$	probability of $i$ or more successes, irrespective of ordering

• Every specific sequence is equally likely when $\theta =\frac{1}{2}$ ➔ $P(\text{a named sequence})={\theta }^n$ regardless of how many $1$s it contains, so $(0,0,1,0,1,1,0,0,1,1)$ and $(0,0,\dots ,0)$ both have $(\frac{1}{2}{)}^{10}=1/1024$ ➔ the $\left(\genfrac{}{}{0ex}{}{n}{m}\right)$ factor is what makes counts unequal while sequences stay equal.

3. Moments via Indicator Sums

• Decompose ➔ $Z={\sum }_{i=1}^n{X}_i$ with Bernoulli $X_i$.
• $E(Z)=np$ ➔ sum of means (linearity — no independence needed).
• $\mathrm{Var}(Z)=np(1-p)$ ➔ variances add only under independence; $\sigma =\sqrt{np(1-p)}$.
• Approximation ➔ large $n$, small $np$ ⟹ [Poisson Distribution|Poisson].
• Shape ➔ mass skewed toward $np$; $\theta$ and $1-\theta$ produce mirror-image distributions.

📊 Exam Execution Trace & Applied Exercises

1. Manual Execution Trace Layout

$Z\sim \mathrm{Bin}(4,\frac{1}{2})$:

Step / State	Quantity	Value
1	$\mathrm{Pr}(Z=2)=\left(\genfrac{}{}{0ex}{}{4}{2}\right)(\frac{1}{2}{)}^4$	$6\cdot \frac{1}{16}=\frac{3}{8}$
2	$E(Z)=np$	$2$
3	$\mathrm{Var}(Z)=np(1-p)$	$1$ ($\sigma =1$)

2. Applied Exercise — telephone sales (FIT2086)

Problem: each call yields a sale with probability $0.08$, independently; the salesman makes $12$ calls, so $M\sim Bin(0.08,12)$. Find (i) $E[M]$; (ii) $P(M\le 2)$; (iii) $P(M\ge 2)$.

$$\begin{array}{rl}\text{(i)}E[M]& =n\theta =12(0.08)=0.96\\ \text{(ii)}P(M\le 2)& =\sum _{m=0}^2\left(\genfrac{}{}{0ex}{}{12}{m}\right)(0.08{)}^m(0.92{)}^{12-m}\\ & =\underset{0.3677}{\underbrace{(0.92{)}^{12}}}+\underset{0.3837}{\underbrace{12(0.08)(0.92{)}^{11}}}+\underset{0.1835}{\underbrace{66(0.08{)}^2(0.92{)}^{10}}}=0.9348\\ \text{(iii)}P(M\ge 2)& =1-P(M\le 1)=1-(0.3677+0.3837)=0.2487\end{array}$$

Final Extracted Output: $0.96$ expected sales; $P(M\le 2)\approx 0.935$; $P(M\ge 2)\approx 0.249$. Key move: “2 or more” is the complement of $M\le 1$, not of $M\le 2$ — the boundary value $m=2$ sits inside both one-sided events, so $P(M\le 2)+P(M\ge 2)>1$.

3. Applied Exercise — fair-coin tails by hand (Studio 2)

Problem: (i) $P(m\ge 2)$ heads in $n=4$ tosses of a fair coin; (ii) $P(m\ge 5)$ ones in $n=10$ Bernoulli$(\frac{1}{2})$ draws.

$$\begin{array}{rl}\text{(i)}P(m\ge 2\mid \theta =\frac{1}{2},n=4)& =\left(\genfrac{}{}{0ex}{}{4}{2}\right){\left(\frac{1}{2}\right)}^2{\left(\frac{1}{2}\right)}^2+\left(\genfrac{}{}{0ex}{}{4}{3}\right){\left(\frac{1}{2}\right)}^3{\left(\frac{1}{2}\right)}^1+\left(\genfrac{}{}{0ex}{}{4}{4}\right){\left(\frac{1}{2}\right)}^4{\left(\frac{1}{2}\right)}^0\\ & ={\left(\frac{1}{2}\right)}^4\left(6+4+1\right)=\frac{11}{16}\\ \text{(ii)}P(m\ge 5)& =1-P(m\le 4)=1-0.377=0.623\text{1 - pbinom(4, 10, 1/2)}\end{array}$$

Final Extracted Output: $11/16=0.6875$; $\approx 0.623$. Key move: with $\theta =\frac{1}{2}$ every term shares the factor $(\frac{1}{2}{)}^n$, so the sum collapses to $\frac{1}{2^n}\sum \left(\genfrac{}{}{0ex}{}{n}{k}\right)$ — pure counting. For (ii), ”$5$ or more” complements $m\le 4$, so pbinom takes $k-1$.

When It Flips: the indicator sum is the showcase of linearity — instant $E=np$. Binomial fixes $n$ and counts successes; the geometric fixes the first success and varies the trial count.

⚙️ In R

The binom suffix (details ➔ R Simulation and Random Sampling)

dbinom(2, size = 12, prob = 0.08)        # P(M = 2)   = 0.1835
pbinom(2, size = 12, prob = 0.08)        # P(M <= 2)  = 0.9348
pbinom(1, 12, 0.08, lower.tail = FALSE)  # P(M >= 2)  = 0.2487
qbinom(0.95, size = 12, prob = 0.08)     # quantile
rbinom(10, size = 12, prob = 0.08)       # 10 random counts
rbinom(10, size = 1,  prob = 0.08)       # BERNOULLI draws: size = 1

💡 Common Mistake: Bernoulli has no separate R family ➔ generate it as a binomial with size = 1. And pbinom(q) is $P(M\le q)$ inclusive — for $P(M\ge 2)$ pass q = 1 with lower.tail = FALSE.

⚠️ Common Mistakes

• 💡 Linearity needs no independence, variance does ➔ $E=np$ always holds; $\mathrm{Var}=np(1-p)$ requires independent trials.
• 💡 Renaming success swaps $p\leftrightarrow 1-p$ ➔ which outcome is “success” is a modelling choice; keep it fixed through the calculation.
• 💡 Additivity requires a shared $\theta$ ➔ $M_1+M_2\sim Bin(\theta ,n_1+n_2)$ holds only when both batches share the success probability; differing $\theta$ leaves no binomial at all.
• 💡 Calling a ≥3-outcome experiment binomial ➔ a football match (win/draw/loss) or a die roll is not a binomial experiment; it becomes one only if you redefine the outcome as a single yes/no (“did they win?”). Continuous measurements never qualify. Selection drill ➔ Parametric Probability Distributions.
• 💡 Confusing “this sequence” with “this many successes” ➔ a named sequence has probability ${\theta }^m(1-\theta {)}^{n-m}$ with no $\left(\genfrac{}{}{0ex}{}{n}{m}\right)$; the coefficient appears only when the event is a count.

🧠 Active Recall

Derive the binomial pmf $\left(\genfrac{}{}{0ex}{}{n}{k}\right)p^k(1-p{)}^{n-k}$.

• Hint: One sequence × count.

Answer

• Short answer: A specific $k$-success sequence has $p^k(1-p{)}^{n-k}$; there are $\left(\genfrac{}{}{0ex}{}{n}{k}\right)$ such sequences.
• Why: Mutually exclusive ➔ sequences are disjoint events; multiply within, add across placements.

How do linearity and independence give $E=np$ and $\mathrm{Var}=np(1-p)$?

• Hint: Indicator decomposition.

Answer

• Short answer: $Z=\sum X_i$; $E(Z)=\sum p=np$; $\mathrm{Var}(Z)=\sum p(1-p)=np(1-p)$.
• Why: Independence for variance only ➔ expectation is linear unconditionally; covariances must vanish for variances to add.

Why is $\mathrm{Var}(X)=p(1-p)$ for a single Bernoulli trial, and where is it maximised?

• Hint: Exploit $X^2=X$.

Answer

• Short answer: $E(X^2)=E(X)=p$ ⟹ $\mathrm{Var}=p-p^2=p(1-p)$; maximised at $p=\frac{1}{2}$ (value $0.25$).
• Why: Maximum uncertainty ➔ a fair trial is the least predictable; degenerate $p\in \{0,1\}$ gives zero variance.