Distribusi Binomial (Matematika Peminatan kelas XII SMA/MA)

 $\text{A. Pendahuluan Distribusi Binomial}$

$\begin{array}{rl}& \{\begin{array}{c}(1)\text{Review}\{\begin{array}{l}\text{Peluang}\{\begin{array}{l}\text{Populasi}\\ \text{Sampel}\{\begin{array}{l}\text{Acak}\\ \text{Bukan Acak}.\end{array}\end{array}\\ \text{Kombiasi}& \end{array}\\ (2)\text{Variabel Acak}\{\begin{array}{ll}\text{Diskrit}& .\\ \text{Kontinue}& \end{array}\\ (3)\text{Distribusi}\{\begin{array}{ll}\text{Distribusi Peluang Variabel Acak}& \\ \text{Fungsi Distribusi Kumulatif}& \\ \text{Variabel Acak Binomial}& \\ \text{Distribusi Binomial}\end{array}\end{array}\end{array}$

$\text{Penjelasan}$

$\begin{array}{|cll|}\hline \text{No}& \text{Istilah}& \text{Penjelasan}\\ 1& \text{Statistika}& \text{Ilmu tentang pengumpulan, pengolahan},\\ & & \text{penganalisaan serta penarikan kesimpulan}\\ & & \text{data. Selanjutnya akan dibagi dua yaitu}\\ & & \text{deskriptif dan inferensia}\\ 2& \text{Statistik}& \text{Kumpulan data/ukuran sampel}\\ 3& \text{Parameter}& \text{Ukuran populasi}\\ 4& \text{Populasi}& \text{Keseluruhan/semua anggota objek/data}\\ 5& \text{Sampel}& \text{Subjek/Objek yang mewakili populasi}\\ 6& \text{Sesus}& \text{Penelitian seluruh data (populasi)}\\ 7& \text{Tekik}& \text{Cara pengambilan data terbatas pada}\\ & \text{Sampling}& \text{sebagian saja dari populasi yang diteliti}\\ \hline\end{array}$.

$\text{B. Kombinasi, Peluang, dan Variabel Acak}$.

Untuk memulai bahasan ini kita sertakan pengertian yang berkaitan dengan kombinasi yaitu adalah permutasi. Perhatikanlah tabel berikut

$\begin{array}{|lll|}\hline \text{Istilah}& \text{Permutasi}& \text{Kombinasi}\\ \text{Definisi}& \begin{array}{rl}& \text{Permutasi r unsur dari n unsur}\\ & \text{adalah banyaknya kemungkinan}\\ & \text{urutan r unsur yang dipilih}\\ & \text{dari n unsur yang tersedia}.\\ & \text{Tiap unsur berbeda dan}r\le n\\ & \end{array}& \begin{array}{rl}& \text{Kombinasi r unsur dan n unsur}\\ & \text{adalah banyaknya kemungkinan}\\ & \text{tidak terurut dalam pemilihan}\\ & \text{r unsur yang diambil dari n}\\ & \text{unsur yang tersedia. Tiap unsur}\\ & \text{berbeda dan}r\le n\end{array}\\ \text{Tipe}& \begin{array}{rl}& \text{Bentuk khusus kaidah}\\ & \text{perkalian}\end{array}& \begin{array}{rl}& \text{Bentuk khusus dari bentuk}\\ & \text{permutasi}\end{array}\\ \text{Notasi}& {}_n{P}_r,P_n^r,\text{atau}P(n,k)& {}_n{C}_r,C_r^n,(\genfrac{}{}{0ex}{}{n}{r}),\text{atau}C(n,r)\\ \text{Rumus}& P(n,r)={\displaystyle \frac{n!}{(n-r)!}}& (\genfrac{}{}{0ex}{}{n}{r})=C(n,r)={\displaystyle \frac{n!}{r!(n-r)!}}\\ \hline\end{array}$.

$\begin{array}{rl}& \text{Sebagai catatan bahwa}\\ & n!=1\times 2\times 3\times \cdots \times (n-1)\times n\end{array}$

Selanjutnya yang akan kita bahas berkaitan bab ini adalah kombinasi beserta contohnya. Perhatikan pula tabel berikut

$\begin{array}{|cc|}\hline \text{Kombinasi}& \text{Kombinasi dalam}\\ \text{dengan pengulangan}& \text{Binom Newton}\\ \begin{array}{rl}& C(n+r-1,r)\\ & =C(n+r-1,n-1)\\ & (\genfrac{}{}{0ex}{}{n+r-1}{r})\\ & =(\genfrac{}{}{0ex}{}{n+r-1}{n-1})\end{array}& \begin{array}{rl}& (x+y{)}^n\\ & =\sum _{k=o}^n(\genfrac{}{}{0ex}{}{n}{r})x^{n-k}{y}^k\\ \\ & \text{Koefisien untuk}\\ & x^{n-k}{y}^k,\text{yaitu}\\ & \text{suku ke}-(k+1)\\ & \text{adalah}(\genfrac{}{}{0ex}{}{n}{r})\end{array}\\ \hline\end{array}$.

serta

$\text{CONTOH SOAL}$

$\begin{array}{l}\\ 1.& \text{Tentukanlah nilai}\\ & \begin{array}{l}\\ \text{a}.3!& \text{e}.{\displaystyle \frac{6!}{4!}}& \text{i}.{\displaystyle \frac{2!}{0!}+\frac{3!}{1!}+\frac{4!}{2!}}\\ \text{b}.5!& \text{f}.{\displaystyle \frac{10!}{6!}}& \text{j}.{\displaystyle \frac{2!}{0!}\times \frac{3!}{1!}+\frac{4!}{2!}}\\ \text{c}.0!+1!+2!+3!& \text{g}.{\displaystyle \frac{7!}{3!\times 4!}}& \text{k}.{\displaystyle \frac{3\times 4!}{3!(5!-5!)}}\\ \text{d}.(2!)!+(3!)!& \text{h}.{\displaystyle \frac{13!}{12!+12!}}& \text{l}.{\displaystyle \frac{3!+5!+7!}{4!+6!}}\end{array}\\ \\ & \text{Jawab}:\\ \\ & \begin{array}{c}\\ \text{a}.3!=3.2.1=6\\ \text{b}.5!=5.4.3.2.1=120\\ \begin{array}{rl}\text{c}.0!+1!+2!+3!& =1+1+2+6\\ & =10\end{array}\\ \begin{array}{rl}\text{d}.(2!)!+(3!)!& =2!+6!\\ & =2+720\\ & =722\end{array}\\ \text{e}.{\displaystyle \frac{6!}{4!}=\frac{720}{24}=30\text{atau}{\displaystyle \frac{6!}{4!}={\displaystyle \frac{6.5.\text{⧸}4.\text{⧸}3.\text{⧸}2.\text{⧸}1}{\text{⧸}4.\text{⧸}3.\text{⧸}2.\text{⧸}1}=6.5=30}}}\\ \text{f}.{\displaystyle \frac{10!}{6!}=\frac{10.9.8.7.6.5.4.3.2.1}{6.5.4.3.2.1}=....(\text{silahkan diselesaikan sendiri})}\\ \text{g}.{\displaystyle \frac{7!}{3!\times 4!}=\frac{7.6.5.4.3.2.1}{(3.2.1)\times (4.3.2.1)}=....(\text{silahkan juga diselesaikan sendiri})}\\ ⋮\\ (\text{silahkan selanjutnya diselesaikan sendiri})\end{array}\end{array}$

$\begin{array}{l}\\ 2.& \text{Sederhanakanlah}\\ & \begin{array}{l}\\ \text{a}.{\displaystyle \frac{n!}{(n-1)!}}& \text{e}.{\displaystyle \frac{1}{n!}+\frac{n}{(n+1)!}-\frac{1}{(n-1)!}}\\ \text{b}.{\displaystyle \frac{(n+2)!}{(n+1)!}}& \text{f}.{\displaystyle \frac{(4n)!}{(4n+1)!}+\frac{(4n)!}{(4n-1)!}}\\ \text{c}.{\displaystyle \frac{(2n)!}{(2n+1)!}}& \text{g}.{\displaystyle \frac{1}{n}-\frac{n!}{(n-1).(n-2)!}}\\ \text{d}.{\displaystyle \frac{(n+2)!}{(n^2+3n+2)}}& \text{h}.1.1!+2.2!+3.3!+4.4!+5.5!+...+n.n!\end{array}\\ \\ & \text{Jawab}:\\ \\ & \begin{array}{c}\\ \text{a}.{\displaystyle \frac{n!}{(n-1)!}=\frac{n.(n-1)!}{(n-1)!}=n}\\ \text{b}.{\displaystyle \frac{(n+2)!}{(n+1)!}=\frac{(n+2).(n+1)!}{(n+1)!}=n+2}\\ \text{c}.{\displaystyle \frac{(2n)!}{(2n+1)!}=\frac{(2n)!}{(2n+1).(2n)!}=\frac{1}{2n+1}}\\ \text{d}.{\displaystyle \frac{(n+2)!}{n^2+3n+2}=\frac{(n+2)!}{(n+2).(n+1)}=\frac{(n+2).(n+1).n!}{(n+2).(n+1)}=n!}\\ ⋮\\ (\text{silahkan selanjutnya diselesaikan sendiri sebagai latihan})\\ ⋮\\ \begin{array}{rl}\text{h}.& 1.1!+2.2!+3.3!+4.4!+5.5!+...+n.n!\\ & =(2-1).1!+(3-1).2!+(4-1).3!+(5-1).4!+...+(n+1-1).n!\\ & =2.1!+3.2!+4.3!+5.4!+...+(n+1).n!-1!-2!-3!-4!-...-n!\\ & =2!+3!+4!+5!+...+(n+1)!-(1!+2!+3!+4!+...+n!)\\ & =(n+1)!-1\end{array}\end{array}\end{array}$

$\begin{array}{l}\\ 3.& \text{Sederhanakanlah bentuk penjumlahan berikut}\\ & {\displaystyle \frac{3}{1!+2!+3!}+\frac{4}{2!+3!+4!}+\frac{5}{3!+4!+5!}+\cdots +{\displaystyle \frac{100}{98!+99!+100!}}}\\ \\ & \text{Jawab}:\\ \\ & \begin{array}{rl}& \text{Perhatikan}\text{bahwa}\\ & {\displaystyle \frac{3}{1!+2!+3!}=\frac{3}{1+2+6}=\frac{3}{9}=\frac{1}{3}\times \frac{2}{2}=\frac{2}{1\times 2\times 3}=\frac{2}{3!}=\frac{3-1}{3!}=\frac{3}{3!}-\frac{1}{3!}=\frac{3}{2!\times 3}-\frac{1}{3!}=\frac{1}{2!}-\frac{1}{3!}}\\ & \text{sehingga}\\ & \frac{3}{1!+2!+3!}=\frac{1}{2!}-\frac{1}{3!}\\ & {\displaystyle \frac{4}{2!+3!+4!}=\cdots =\frac{1}{3!}-\frac{1}{4!}}\\ & {\displaystyle \frac{5}{3!+4!+5!}=\cdots =\frac{1}{4!}-\frac{1}{5!}}\\ & ⋮\\ & {\displaystyle \frac{100}{98!+99!+100!}=\cdots =\frac{1}{99!}-\frac{1}{100!}}\\ & ---------------------------\\ & =\frac{1}{2!}-\frac{1}{100!}\end{array}\end{array}$.