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Probabilitas

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Diberikan $n$ bilangan bulat positif dan $0<t<1$. Jika dipilih $a_1,a_2,...,a_n$ sehingga $a_1+a_2+...+a_n=1$, berapakah probabilitas $max(a_1,a_2,...,a_n)\ge t$?

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Ah salah, sorry (canceled)

Spoiler

*) Jika $nt \le 1$ (termasuk saat n=1) $=> P=1$.

Trivial pk pigeonhole

 

*) Kasus $nt>1$.

Pandang kasus komplemenny, yakni Q= peluang $max(a_1, \cdots , a_n)<t$.

Statement ini ekuiv dg semua bil $a_1$ sp $a_n$ krg dr t, yg memenuhi $\sum{a}=1$.

Perhatikan bhw $1=a_1+(a_2+...+a_n)>t+(n-1)\min{(a_2,a_3,...a_n)}$

$\min{(a_2,a_3,...a_n)}> \frac{1-t}{n-1}$ (krn n>1)

Dg asumsi yg sama $\min{(a_1,a_2,...a_{n-1})} > \frac{1-t}{n-1}$.

Nah, dr 2 ketaksamaan ini, pasti $min(a)=a_i$ terletak di min 1 dr 2 set $(a_1,...a_{n-1})f dan$ (a_2,...,a_n) $.

Jd $\min(a) > \frac{1-t}{n-1}$ dan $a_x > \frac{1-t}{n-1}$ utk stp indeks x.

 

Perhatikan bhw $a \in (\frac{1-t}{n-1}, t)$ adl syarat minimum. Ini karena, ketika dipilih sembarang $a_1$ sp $a_{n-1}$, nilai $a_n$ mengikuti dr jumlah total =1, dan bs dicek bhw nilai $a_n$ selalu berada di interval ini juga.

Pdhl peluang memilih a d interval ini dr kemungkinan interval$ [0,1] $adl $\frac{t-\frac{1-t}{n-1}}{1-0}=\frac{(nt-1)}{(n-1)}$.

Spt yg dijelaskn td, nilai $a_n $bs mengikuti dr sembarang pemilihan$ a_1$ sp$ a_n-1$.

Jd Q, peluang $max<t $adl $(\frac{(nt-1)}{(n-1)})^{n-1}$.

 

Lalu $ P=1-Q= 1-(\frac{(nt-1)}{(n-1)})^{n-1}$.

CMIIW

 

Edited by alb

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