Lanjutan Materi Distribusi Peluang Diskrit (Matematika Peminatan Kelas XII)

$\color{blue}\begin{aligned}\textrm{C. 1}.\quad&\textrm{Distribusi Peluang Diskrit} \end{aligned}$

$\begin{aligned} &\textrm{Misalkan}\: \: X\: \: \textrm{adalah variabel acak diskrit}\\ &\textrm{dari nilai}\: :\: \: x_{1},\: x_{2},\: x_{3},\: x_{4},\: \cdots \: ,\: x_{k},\: \textrm{dan}\\ &P\: \textrm{adalah seluruh nilai peluang untuk}\: :\\ &p_{1},\: p_{2},\: p_{3},\: p_{4},\: \cdots \: ,p_{k}, \textrm{maka nilai untuk}\\ &\color{blue}p_{1}+ p_{2}+ p_{3}+ p_{4}+ \cdots +p_{k}=1\\ &\textbf{dan}\\ &\textrm{Fungsi}\: \: f(x) =P(X=x)\: \: \textrm{yang mempunyai}\\ &\textrm{nilai}\: \: p_{1},\: p_{2},\: p_{3},\: p_{4},\: \cdots \: ,p_{k},\: \textrm{pada variabel}\\ &X=x_{1},\: x_{2},\: x_{3},\: x_{4},\: \cdots \: ,\: x_{k},\: \textrm{disebut fungsi}\\ &\textrm{kepekatan peluang dari variabel acak}\: \: X.\\ &\textrm{Selanjutnya jika kita gambar grafik}\: \: f(x)\\ &\textrm{terhadap}\: \: x,\: \textrm{maka kita akan grafik yang}\\ &\textrm{dinamakan dengan}\: \: \color{red}\textbf{grafik peluang} \end{aligned}$

Suatu fungsi  $f(x)=P(X=x)$  disebut fungsi peluang (probabilitas) dari  $X$, jika memenuhi syarat-syarat:

$\color{blue}\begin{matrix} (\textrm{i})\quad f(x)\geq 0\: \: \: \textrm{untuk semua}\: \: x\qquad\qquad\qquad\qquad\qquad\qquad\: \:  \\\\ (\textrm{ii})\quad \sum_{i=1}^{n}f\left ( x_{i} \right )=\color{red}f(x_{1})+f(x_{2})+f(x_{3})+...+f(x_{n})=\color{black}1 \end{matrix}$

$\LARGE\colorbox{yellow}{CONTOH SOAL}$

$\begin{array}{ll}\\ 1.&\textrm{Pada percobaan melempar 3 koin identik}\\ &\textrm{sekaligus bersama-sama. Variabel acak}\\ &\textrm{dalam hal ini pada kejadian muncul sisi}\\ &\textrm{gambar, tentukan}\\ &\textrm{a}.\: \: \textrm{distribusi peluangnya}\\ &\textrm{b}.\: \: \textrm{tabel fungsi peluangnya}\\ &\textrm{c}.\: \: \textrm{grafik fungsi peluangnya}\\\\ &\textbf{Jawab}:\\ &\begin{aligned}&\textrm{Diketahui dari soal}\: \: \color{red}\textrm{variabel acak}\\ &\textrm{pada kejadian di atas adalah munculnya}\\ &\textrm{sisi gambar pada pelemparan 3 koin}\\ &\textrm{maka} \end{aligned}\\ &\begin{aligned}\textrm{a}.\quad&\textrm{Distribusi peluangnya}\\ &\begin{array}{|l|c|c|c|c|c|c|c|c|}\hline \textrm{Sampel}&AAA&AA\color{red}G&A\color{red}G\color{black}A&A\color{red}GG&\color{red}G\color{black}AA&\color{red}G\color{black}A\color{red}G&\color{red}GG\color{black}A&\color{red}GGG\\\hline \textrm{Muncul}\: \color{red}(G)&0&1&1&2&1&2&2&3\\\hline \end{array}\\ \textrm{b}.\quad&\textrm{Tabel fungsi peluangnya}\\ &x=\textrm{muncul kejadian sisi gambar}\: \: \color{red}(G)\\ &\begin{array}{|c|c|c|c|c|c|}\hline x&0&1&2&3&\color{red}\textrm{Jumlah}\\\hline f(x)&\displaystyle \frac{1}{8}&\displaystyle \frac{3}{8}&\displaystyle \frac{3}{8}&\displaystyle \frac{1}{8}&\color{red}1\\\hline \end{array}\\ \textrm{c}.\quad&\textrm{Grafik fungsi peluangnya adalah}\\ & \end{aligned} \end{array}$

 

$\begin{array}{ll}\\ 2.&\textrm{Pada sebuah kotak terdapat 2 kelereng}\\ &\textrm{biru dan 4 kelereng merah. Tiga kereng}\\ &\textrm{diambil secara acak. Tentukanlah distribusi}\\ &\textrm{peluang}\: \: \color{red}x\: \: \color{black}\textrm{jika}\: \: \color{red}x\: \: \color{black}\textrm{menyatakan banyaknya}\\ &\textrm{terambilnya bola biru}\\\\ &\textbf{Jawab}:\\ &\begin{array}{|l|c|}\hline \qquad\qquad\textrm{Nama}&\textrm{Perhitungan}\\\hline \textrm{Banyak}&\\ \textrm{titik sampel}&\begin{aligned}C_{3}^{6}&=\displaystyle \frac{6!}{3!(6-3)!}=20 \end{aligned}\\\hline \textrm{Banyak cara}&\\ \textrm{mendapatkan bola biru}&C_{x}^{2}\\\hline \textrm{Banyak cara}&\\ \textrm{mendapatkan bola merah}&C_{3-x}^{4}\\\hline \end{array} \end{array}$

$.\quad \: \begin{array}{|l|l|}\hline \color{red}\textrm{Distribusi peluang}&\qquad\quad\color{red}\textrm{Perhitungan}\\\hline P(X=x)=f(x)&f(x)=\displaystyle \frac{\displaystyle C_{x}^{2}.C_{3-x}^{4}}{\displaystyle C_{3}^{6}},\\ \textrm{untuk}&\begin{aligned}x&=0,1,2 \end{aligned}\\\hline x=0\Rightarrow P(x=0)&f(x)=\displaystyle \frac{\displaystyle C_{0}^{2}.C_{3-0}^{4}}{\displaystyle C_{3}^{6}}\\ &.\: \: \, \quad=\displaystyle \frac{\displaystyle C_{0}^{2}.C_{3}^{4}}{\displaystyle C_{3}^{6}}=\displaystyle \frac{\displaystyle \frac{2!}{0!2!}\times \frac{4!}{3!1!}}{\displaystyle \frac{6!}{3!3!}}\\ &.\: \: \, \quad=\displaystyle \frac{2!4!3!3!}{2!3!6!}=0,2\\\hline x=1\Rightarrow P(x=1)&f(x)=\displaystyle \frac{\displaystyle C_{1}^{2}.C_{3-1}^{4}}{\displaystyle C_{3}^{6}}\\ &.\: \: \, \quad=\displaystyle \frac{\displaystyle C_{1}^{2}.C_{2}^{4}}{\displaystyle C_{3}^{6}}=\displaystyle \frac{\displaystyle \frac{2!}{1!1!}\times \frac{4!}{2!2!}}{\displaystyle \frac{6!}{3!3!}}\\ &.\: \: \, \quad=\displaystyle \frac{2!4!3!3!}{2!2!6!}=0,6\\\hline x=2\Rightarrow P(x=2)&f(x)=\displaystyle \frac{\displaystyle C_{2}^{2}.C_{3-2}^{4}}{\displaystyle C_{3}^{6}}\\ &.\: \: \, \quad=\displaystyle \frac{\displaystyle C_{2}^{2}.C_{1}^{4}}{\displaystyle C_{3}^{6}}=\displaystyle \frac{\displaystyle \frac{2!}{2!0!}\times \frac{4!}{1!3!}}{\displaystyle \frac{6!}{3!3!}}\\ &.\: \: \, \quad=\displaystyle \frac{2!4!3!3!}{2!3!6!}=0,2\\\hline \end{array}$

$\begin{array}{ll}\\ 3.&\textrm{Tunjukkan bahwa fungsi}\: \: P(x)=\displaystyle \frac{x+2}{12}\\ &\textrm{untuk}\: \: x=1,2,\: \textrm{dan}\: \: 3\: \: \textrm{merupakan fungsi}\\ &\textrm{peluang}\\\\ &\textbf{Jawab}:\\ &\begin{aligned}\textrm{Perhatik}&\textrm{an bahwa}\\ \bullet \quad P(1)&=\displaystyle \frac{1+2}{12}=\frac{3}{12}=\frac{1}{4}\\ \bullet \quad P(2)&=\displaystyle \frac{2+2}{12}=\frac{4}{12}=\frac{1}{3} \\ \bullet \quad P(3)&=\displaystyle \frac{5+2}{12}=\frac{5}{12} \\ \textrm{Sehing}&\textrm{ga}\: \: \displaystyle \sum_{i=1}^{3}P(i)=\displaystyle \frac{3}{12}+\frac{4}{12}+\frac{5}{12}=\color{red}\frac{12}{12}=1\\ &\begin{cases} (\textrm{i}) & \textrm{Peluangnya berada}\: \: \color{red}0\leq P(i)\leq 1 \\ (\textrm{ii}) & \textrm{dan nilai totolnya}=\displaystyle \color{red}\sum_{i=1}^{3}P(i)=1 \end{cases}\\ \textrm{Jadi},\: &\textrm{fungsi}\: \: P(x)=\displaystyle \frac{x+2}{12}\: \: \textrm{untuk}\: \: x=1,2,\: \textrm{dan}\: \: 3\\ &\textbf{merupakan fungsi peluang} \end{aligned} \end{array}$

$\begin{array}{ll}\\ 4.&\textrm{Diketahui fungsi peluang adalah}\: \: P(x)=\displaystyle \frac{\color{red}m}{x+1}\\ &\textrm{untuk}\: \: x=0,1,2,\: \textrm{dan}\: \: 3\: .\: \textrm{Tentukanlah}\\ &\textrm{a}.\: \: \textrm{nilai}\: \: \color{red}m\\ &\textrm{b}.\: \: \textrm{nilai}\: \: \color{red}P(x\leq 2)\\\\ &\textbf{Jawab}:\\ &\begin{aligned}\textrm{a}.\quad&\displaystyle \sum_{i=0}^{3}P(i)=\color{blue}1\\ &\Leftrightarrow \displaystyle \frac{\color{red}m}{0+1}+\frac{\color{red}m}{1+1}+\frac{\color{red}m}{2+1}+\frac{\color{red}m}{3+1}=1\\ &\Leftrightarrow \color{red}m\color{black}+\displaystyle \frac{\color{red}m}{2}+\frac{\color{red}m}{3}+\frac{\color{red}m}{4}=1\\ &\Leftrightarrow \left (\displaystyle \frac{12+6+4+3}{12} \right )\color{red}m\color{black}=1\\ &\Leftrightarrow \color{red}m\color{black}=\displaystyle \frac{12}{25}\\ \textrm{b}.\quad&P(x\leq 2)=P(x=0)+P(x=1)+P(x=2)\\ &\Leftrightarrow \color{red}m\color{black}+\displaystyle \frac{\color{red}m}{2}+\frac{\color{red}m}{3}=1\\ &\Leftrightarrow \left ( \displaystyle \frac{6+3+2}{6} \right )\color{red}m\color{black}=\displaystyle \frac{11}{6}\color{red}m\\ &\Leftrightarrow \quad =\displaystyle \frac{11}{6}\left ( \displaystyle \frac{12}{25} \right )=\frac{22}{25} \end{aligned} \end{array}$

$\begin{array}{ll}\\ 5.&\textrm{Diketahui fungsi}\: \: f(x)=\begin{cases} \displaystyle \frac{x}{6} &\textrm{untuk}\: \: x=1,2,3 \\\\ 0 &\textrm{untuk}\: \: x\: \: \textrm{yang lain} \end{cases}\\ &\textrm{adalah suatu fungsi peluang/probabilitas}\\ &\textrm{dari pubah/variabel acak}\: \: X.\: \: \textrm{Tentukanlah}\\ &\textrm{a}.\quad \textrm{distribusi peluangnya untuk}\: \: X\\ &\textrm{b}.\quad P(X=2),\: P(X< 3),\: \textrm{dan}\: P(X\geq 2)\\\\ &\textbf{Jawab}:\\ &\begin{aligned}\textrm{a}.\quad&\textrm{Distribusi peluangnya adalah}:\\ &\begin{array}{|c|c|c|c|c|c|c|c|c|}\hline X=x&1&2&3&4&5&\cdots &\textrm{Jumlah}\\\hline P(X=x)&\displaystyle \frac{1}{6}&\displaystyle \frac{2}{6}&\displaystyle \frac{3}{6}&\color{red}0&\color{red}0&\color{red}0&\color{blue}1\\\hline \end{array}\\ \textrm{b}.\quad &\textrm{Karena}\: \: f(x)=\begin{cases} \color{red}\displaystyle \frac{x}{6} &\textrm{untuk}\: \: x=1,2,3 \\\\ 0 &\textrm{untuk}\: \: x\: \: \textrm{yang lain} \end{cases}\\ &\textrm{maka}\\ &\bullet P(X=2)=\color{red}\displaystyle \frac{2}{6}\\ &\bullet P(X<3)=P(X=1)+P(X=2)\\ &\: \quad\qquad \qquad =\displaystyle \frac{1}{6}+\frac{2}{6}\\ &\: \quad\qquad \qquad =\displaystyle \frac{3}{6}=\color{red}\frac{1}{2}\\ &\bullet P(X\geq 2)=P(X=2)+P(X=3)\\ &\: \quad\qquad \qquad =\displaystyle \frac{2}{6}+\frac{3}{6}\\ &\: \quad\qquad \qquad =\color{red}\displaystyle \frac{5}{6} \end{aligned} \end{array}$

$\begin{array}{ll}\\ 6.&\textrm{Diketahui fungsi peluang variabel}\: \: X\\ &f(x)=\begin{cases} \displaystyle \frac{x+2}{14} &\textrm{untuk}\: \: x=0,1,2,\: \: \textrm{dan}\: \: 3 \\\\ \quad 0 &\textrm{untuk}\: \: x\: \: \textrm{yang lain} \end{cases}\\ &\textrm{Tentukanlah}\\ &\textrm{a}.\quad \textrm{bahwa}\: \: X\: \: \textrm{merupakan variabel acak diskrit}\\ &\textrm{b}.\quad P(X=4),\: F(2),\: P(1<X\leq 3),\\ &\qquad \textrm{dan}\: P(X\geq 1)\: \: \textrm{serta}\: \: P(\left |X-2 \right |\leq 1)\\\\ &\textbf{Jawab}:\\ &\begin{aligned}\textrm{a}.\quad&\textrm{Distribusi peluangnya adalah}:\\ &\begin{array}{|c|c|c|c|c|c|c}\hline X=x&0&1&2&3&\textrm{Jumlah}\\\hline P(X=x)&\displaystyle \frac{2}{14}&\displaystyle \frac{3}{14}&\displaystyle \frac{4}{14}&\displaystyle \frac{5}{14}&1\\\hline \end{array} \\ &\textrm{Karena}\: \: \displaystyle \sum_{x=0}^{3}f(x)=1,\: \textrm{serta}\\ &0\leq \displaystyle \frac{2}{14},\: \frac{3}{14},\: \frac{4}{14},\: \frac{5}{14}<1.\: \textrm{Sehingga syarat}\\ &0\leq f(x)<1\: \: \textrm{dan}\: \: \sum f(x)=1\: \: \: \color{red}\textbf{terpenuhi}\\ &\textrm{Jadi, terbukti}\: \: X\: \: \textrm{adalah variabel acak diskrit}\\ \textrm{b}.\quad&\bullet P(X=4)=f(4)=\color{red}0\\ &\bullet F(2)=P(X\leq 2)\\ &\quad\qquad=P(X=0)+P(X=1)+P(X=2)\\ &\quad\qquad=f(0)+f(1)+f(2)\\ &\quad\qquad=\displaystyle \frac{2}{14}+\frac{3}{14}+\frac{4}{14}=\color{red}\frac{9}{14}\\ &\bullet P(1<X\leq 3)=P(X=2)+P(X=3)\\ &\quad\qquad =f(2)+f(3)=\displaystyle \frac{4}{14}+\frac{5}{14}=\color{red}\displaystyle \frac{9}{14}\\ &\bullet P(X\geq 1)=f(1)+f(2)+f(3)\\ &\quad\qquad =\displaystyle \frac{3}{14}+\frac{4}{14}+\frac{5}{14}=\color{red}\displaystyle \frac{12}{14}\\ &\bullet P(\left | X-2 \right |\leq 1)=P(-1\leq X-2\leq 1)\\ &\quad\qquad =P(1\leq X\leq 3)\\ &\quad\qquad =f(1)+f(2)+f(3)\\ &\quad\qquad =\displaystyle \frac{3}{14}+\frac{4}{14}+\frac{5}{14}=\color{red}\displaystyle \frac{12}{14} \end{aligned} \end{array}$

$\begin{array}{ll}\\ 7.&\textrm{Distribusipeluang acak X disajikan dalam tabel berikut}\\ &\begin{array}{|c|c|c|c|}\hline x&2&3&4\\\hline f(x)&\displaystyle \frac{1}{8}&k+\displaystyle \frac{1}{8}&2k\\\hline \end{array}\\ &\textrm{Jika X merupakan variabel acak diskret, tentukanlah}\\ &\textrm{a}.\quad \textrm{nilai \textit{k}}\\ &\textrm{b}.\quad \textrm{nilai}\: \: \textrm{P}(\textrm{X}\geq 3)-\textrm{F}(3)\\\\ &\textrm{Jawab}:\\ &\begin{aligned}\textrm{a}.\quad \sum f(x)&=f(2)+f(3)+f(4)=1\\ \Leftrightarrow \quad &\displaystyle \frac{1}{8}+k+\frac{1}{8}+2k=1\\ \Leftrightarrow \quad &3k=1-\displaystyle \frac{2}{8}=\frac{6}{8}\\ \Leftrightarrow \quad &k=\displaystyle \frac{2}{8}=\color{red}\frac{1}{4}\\ \textrm{b}.\quad \textrm{P}(\textrm{X}\, \geq 3&)-\textrm{F}(3)=\textrm{P}(\textrm{X}\geq 3)-\textrm{P}(\textrm{X}\leq 3)\\ &=f(3)+f(4)-\left ( f(2)+f(3) \right )\\ &=f(4)-f(2)\\ &=2\left ( \displaystyle \frac{1}{4} \right )-\frac{1}{8}\\ &=\displaystyle \frac{4}{8}-\frac{1}{8}=\color{red}\frac{3}{8} \end{aligned} \end{array}$

DAFTAR PUSTAKA

1. Noormandiri. 2017. Matematika Jilid 3 untuk SMA/MA Kelas XII Kelompok Peminatan Matematika dan Ilmu-Ilmu Alam. Jakarta: ERLANGGA
2. Kurnia, N., dkk. 2018. Jelajah Matematika SMA Kelas XII Peminatan MIPA. Bogor: YUDHISTIRA.