Exercises — Binomial distribution — PMF, mean, variance

BINS conditions check karo: Binary, Independent, fixed Number, Same $p$.

• (a) Binomial. Har flip head/tail hai (Binary), flips ek doosre ko influence nahi karte (Independent), exactly $12$ flips hain (Number fixed), aur $p=0.5$ har baar hai (Same $p$). Toh $X\sim \text{Bin}(12,0.5)$.
• (b) NOT binomial. Card nikalne se deck badal jaati hai, toh $p$(red) har draw mein shift hoti hai aur draws dependent hain. Yeh without-replacement wala cousin hai → Hypergeometric distribution.
• (c) NOT binomial. Trials ki sankhya pehle se fixed nahi hai — aap tab rukते ho jab six aata hai. Yeh ek geometric waiting-time question hai, fixed-$n$ count nahi.
• (d) Binomial (approximately). Binary Yes/No hai, essentially constant $p$ aur independence hai kyunki population $30$ ke relative bahut badi hai, aur $n=30$ fixed hai. Toh $X\sim \text{Bin}(30,p)$.

$n=10$ (shots ki fixed sankhya), $p=0.8$ (har shot par same, independent, made/missed binary hai). Toh $X\sim \text{Bin}(10,0.8)$, aur $X$ koi bhi integer $k=0,1,2,\dots ,10$ ho sakta hai.

KYA: exactly $k=2$ successes in $n=6$ trials with $p=0.4$, $q=0.6$. KAISE: PMF mein plug karo. $P(X=2)=\left(\genfrac{}{}{0ex}{}{6}{2}\right)(0.4{)}^2(0.6{)}^4.$ $\left(\genfrac{}{}{0ex}{}{6}{2}\right)=\frac{6!}{2!4!}=15$ (2 successes rakhne ke tarike). Phir $(0.4{)}^2=0.16$ aur $(0.6{)}^4=0.1296$. $P(X=2)=15\times 0.16\times 0.1296=\mathrm{0.31104.}$

$E[X]=np=200\times 0.05=10$ defects on average. $\text{Var}(X)=np(1-p)=200\times 0.05\times 0.95=9.5$. $\sigma =\sqrt{9.5}\approx 3.082$. Isko samjho: average $10$ defects expect karo, usually $\pm 3$ ke andar. $25$ defects ki batch $(25-10)/3.08\approx 4.9$ standard deviations high hai — yeh sach mein alarm ka signal hai.

"$X\le 1$" ka matlab hai $X=0$ ya $X=1$; yeh alag alag outcomes hain, toh inki probabilities add karo. $P(X=0)=\left(\genfrac{}{}{0ex}{}{4}{0}\right)(0.5{)}^0(0.5{)}^4=\frac{1}{16},P(X=1)=\left(\genfrac{}{}{0ex}{}{4}{1}\right)(0.5{)}^1(0.5{)}^3=4\cdot \frac{1}{16}=\frac{4}{16}.$ $P(X\le 1)=\frac{1}{16}+\frac{4}{16}=\frac{5}{16}=\mathrm{0.3125.}$

$P(X\ge 1)=P(1)+P(2)+\cdots +P(50)$ compute karna $50$ terms ka kaam hai. Iske bajaye complement use karo: $P(X\ge 1)=1-P(X=0)=1-\left(\genfrac{}{}{0ex}{}{50}{0}\right)(0.02{)}^0(0.98{)}^{50}=1-(0.98{)}^{50}.$ $(0.98{)}^{50}\approx 0.36417$, toh $P(X\ge 1)\approx 1-0.36417=0.63583$. Complement kyun jeetता hai: "$X=0$" ek single term $(1-p{)}^n$ hai — ek line se pachaas ki jagah kaam ho jaata hai.

Range $\{2,3,4\}$ ko add karna ya remove karna — dono tarike yahan theek hain. Directly: $P(X=2)=\left(\genfrac{}{}{0ex}{}{5}{2}\right)(0.6{)}^2(0.4{)}^3=10(0.36)(0.064)=0.2304,$ $P(X=3)=\left(\genfrac{}{}{0ex}{}{5}{3}\right)(0.6{)}^3(0.4{)}^2=10(0.216)(0.16)=0.3456,$ $P(X=4)=\left(\genfrac{}{}{0ex}{}{5}{4}\right)(0.6{)}^4(0.4{)}^1=5(0.1296)(0.4)=\mathrm{0.2592.}$ Sum: $0.2304+0.3456+0.2592=0.8352$. Complement se check: bahar ka part hai $\{0,1,5\}$: $P(0)=0.4^5=0.01024$, $P(1)=5(0.6)(0.4^4)=0.0768$, $P(5)=0.6^5=0.07776$; total $0.16480$, aur $1-0.16480=0.8352$ ✓.

Conditional probability: $P(A\mid B)=\frac{P(A\cap B)}{P(B)}$. Yahan $A=\{X=3\}$, $B=\{X\ge 1\}$, aur $A\subset B$ toh $A\cap B=A$. $P(X\ge 1)=1-P(X=0)=1-(0.5{)}^3=1-\frac{1}{8}=\frac{7}{8},P(X=3)=(0.5{)}^3=\frac{1}{8}.$ $P(X=3\mid X\ge 1)=\frac{1/8}{7/8}=\frac{1}{7}\approx \mathrm{0.142857.}$

Pure system ki failure = saare servers down. Har server $q=0.1$ probability se down hota hai. $P(\text{all down})=0.1^n\le 0.0001=10^{-4}.$ Base-10 log lo: $n{\log }_{10}(0.1)\le {\log }_{10}(10^{-4})$, yaani $-n\le -4$, toh $n\ge 4$. Sabse chhota integer $n=4$ hai. Check karo: $0.1^4=0.0001$, jisse $P(\text{at least one up})=1-0.0001=0.9999$ ✓.

Maano $X\sim \text{Bin}(8,0.25)$ = jeetne ki sankhya. Net profit $=5X-1(8-X)=6X-8$. Linearity of expectation se, $E[\text{profit}]=6E[X]-8=6(np)-8=6(8\cdot 0.25)-8=6(2)-8=4$. Expected net profit =\4$.\ast \ast Matlab:\ast \ast averageparaapgainkarteho,jopooritarah$E[X]=np=2$ wins se driven hai.

$\lambda =np=500\times 0.004=2$. Exact: $P(X=2)=\left(\genfrac{}{}{0ex}{}{500}{2}\right)(0.004{)}^2(0.996{)}^{498}$. $\left(\genfrac{}{}{0ex}{}{500}{2}\right)=124750$, $(0.004{)}^2=1.6\times 10^{-5}$, $(0.996{)}^{498}\approx 0.13629$. $P(X=2)\approx 124750\times 1.6\times 10^{-5}\times 0.13629\approx 0.27201$. Poisson: $P=\frac{{\lambda }^2{e}^{-\lambda }}{2!}=\frac{4e^{-2}}{2}=2e^{-2}\approx 0.27067$. Dono lagbhag $3$ decimals tak agree karte hain — expected hai, kyunki $n$ large hai aur $p$ tiny hai aur $np$ moderate hai. Yahi woh regime hai jahan binomial $\to$ Poisson.

Dono PMF values likho aur divide karo — factorials telescope ho jaate hain: $\frac{P(k)}{P(k-1)}=\frac{\left(\genfrac{}{}{0ex}{}{n}{k}\right)p^k{q}^{n-k}}{\left(\genfrac{}{}{0ex}{}{n}{k-1}\right)p^{k-1}{q}^{n-k+1}}=\frac{\left(\genfrac{}{}{0ex}{}{n}{k}\right)}{\left(\genfrac{}{}{0ex}{}{n}{k-1}\right)}\cdot \frac{p}{q}.$ Ab $\frac{\left(\genfrac{}{}{0ex}{}{n}{k}\right)}{\left(\genfrac{}{}{0ex}{}{n}{k-1}\right)}=\frac{n!/(k!(n-k)!)}{n!/((k-1)!(n-k+1)!)}=\frac{(k-1)!(n-k+1)!}{k!(n-k)!}=\frac{n-k+1}{k}$. Toh ratio hai $\frac{n-k+1}{k}\cdot \frac{p}{q}$. ✓ Mode deduce karo: PMF tab badhti hai ($P(k)>P(k-1)$) jab tak yeh ratio $>1$ hai: $(n-k+1)p>k(1-p)⟺np+p-kp>k-kp⟺k<(n+1)p.$ Toh probabilities $k=\lfloor (n+1)p\rfloor$ tak badhti hain aur uske baad ghatti hain — mode $\lfloor (n+1)p\rfloor$ hai (aur agar $(n+1)p$ integer hai, toh woh aur $(n+1)p-1$ dono tie karte hain). Staircase figure dekho.

$n>0$ fix karo. $f(p)=p(1-p)=p-p^2$ maximise karo. Complete the square: $p-p^2=-\left(p^2-p\right)=-{\left(p-\frac{1}{2}\right)}^2+\frac{1}{4}.$ Kyunki ${\left(p-\frac{1}{2}\right)}^2\ge 0$ hai aur subtract ho raha hai, $f(p)\le \frac{1}{4}$ hai, aur equality tab hoti hai jab $p=\frac{1}{2}$. Isliye $\text{Var}=np(1-p)\le \frac{n}{4}$, $p=\frac{1}{2}$ par maximise hoti hai, aur ${\text{Var}}_{\max }=\frac{n}{4}$ hai; aur $\text{Var}=0$ jab $p=0$ ya $p=1$ (koi randomness nahi). Parabola figure dekho.

Conceptual proof (sabse clean). $X$ $m$ independent trials mein successes count karta hai jahan har trial mein success chance $p$ hai; $Y$ aur $n$ aise trials mein successes count karta hai. Saare $m+n$ trials pool karo: woh independent hain, binary hain, number fixed hai, aur sab same $p$ share karte hain — yeh exactly BINS setup hai $\text{Bin}(m+n,p)$ ke liye. Total count $X+Y$ wahi binomial hai. ∎ Consistency checks (inke liye $p$ equal chahiye, aur variance ke liye independence): $E[X+Y]=mp+np=(m+n)p✓,\text{Var}(X+Y)=mp(1-p)+np(1-p)=(m+n)p(1-p)✓.$ Caveat: agar dono $p$ alag hote, toh pooling Same $p$ violate karta aur $X+Y$ binomial nahi hota — sum theorem identical $p$ par depend karta hai.