Contoh Soal dan Pembahasan Distribusi Binomial (Bagian 3)

 $\begin{array}{ll}\\ 11.&\textrm{Suatu tes dengan pilihan jawaban }\\ &\textrm{benar-salah berjumlah 8 soal}\\ &\textrm{Supaya lulus tes, peserta diharuskan }\\ &\textrm{menjawab benar minimal 50}\%\\ &\textrm{Peluang seseorang dianggap lulus tes }\\ &\textrm{adalah}\: ....\\ &\textrm{a}.\quad \displaystyle 0,2188\qquad\qquad\quad\qquad \quad\textrm{d}.\quad 0,6367\\ &\textrm{b}.\quad \displaystyle \color{red}0,2734\quad \: \color{black}\textrm{c}.\quad 0,3633\quad\quad \textrm{e}.\quad 0,7266\\\\ &\textrm{Jawab}:\\ &\begin{aligned}&p=\textbf{Peluang benar}=\displaystyle \frac{1}{2},\qquad \textrm{dan}\: \: \\ &q=\textbf{Peluang Salah}=1-\displaystyle \frac{1}{2}=\frac{1}{2}\\ &f(x)=P(X=x)=\begin{pmatrix} n\\ x \end{pmatrix}p^{x}.q^{n-x}\\ &\textrm{maka}\\ &P\left ( X=50\%(8)=4 \right )=\begin{pmatrix} 8\\ 4 \end{pmatrix}\times \left ( \displaystyle \frac{1}{2} \right )^{4}\times \left ( \frac{1}{2} \right )^{8-4}\\ &\qquad =\displaystyle \frac{8!}{4!\times 4!}\left ( \displaystyle \frac{1}{2} \right )^{4+4}\\ &\qquad =70\times \displaystyle \frac{1}{256}\\ &\qquad =\color{red}0,2734 \end{aligned} \end{array}$

$\begin{array}{ll}\\ 12.&\textrm{Sebuah kotak berisi 20 bola dengan }\\ &\textrm{rincian 12 boal berwarna kuning dan }\\ &\textrm{sisanya berwarna hijau. Dari kotak} \\ &\textrm{diambil 6 bola secara acak. Peluang}\\ &\textrm{terambil 4 bola hijau adalah}....\\ &\textrm{a}.\quad \displaystyle 0,1238\quad\quad\qquad\qquad \qquad\textrm{d}.\quad 0,8132\\ &\textrm{b}.\quad \color{red}\displaystyle 0,1382\: \quad \color{black}\textrm{c}.\quad 0,3110\quad\quad \textrm{e}.\quad 0,9590\\\\ &\textrm{Jawab}:\\ &\begin{aligned}&p=\textbf{Peluang bola kuning}\\ &\: \: =\displaystyle \frac{C_{1}^{12}}{C_{1}^{20}}=\displaystyle \frac{12}{20}=\frac{3}{5},\\ &q=\textbf{Peluang bola hijau}=1-\displaystyle \frac{3}{5}=\frac{2}{5}\\ &f(x)=P\left ( X=x \right )=\begin{pmatrix} n\\ x \end{pmatrix}p^{x}.q^{n-x}\\ &\textrm{maka}\\ &f(4)=\begin{pmatrix} 6\\ 4 \end{pmatrix}\times \left ( \displaystyle \frac{2}{5} \right )^{4}\times \left ( \frac{3}{5} \right )^{6-4}\\ &\qquad =\displaystyle \frac{6!}{2!\times 4!}\left ( \displaystyle \frac{16}{625} \right )\times \left ( \displaystyle \frac{9}{25} \right )\\ &\qquad =15\times \displaystyle \frac{144}{15625}=\frac{2160}{15625}\\ &\qquad =\color{red}0,1382 \end{aligned} \end{array}$

$\begin{array}{ll}\\ 13.&\textrm{Dua dadu dilambungkan 5 kali}\\ &\textrm{Peluang muncul pasangan mata dadu}\\ &\textrm{berjumlah 4 sampai dengan 7 }\\ &\textrm{sebanyak 4 kali adalah}\: ....\\ &\textrm{a}.\quad \displaystyle 0,1503\: \: \: \: \qquad\qquad\quad\quad \quad\textrm{d}.\quad 0,1583\\ &\textrm{b}.\quad \displaystyle 0,1553\quad \textrm{c}.\quad \color{red}0,1563\quad\quad \color{black}\textrm{e}.\quad 0,1593\\\\ &\textrm{Jawab}:\\ &\begin{aligned}&p=\textbf{Peluang mata dadu berjumlah 4 sampai 7}\\ &\: \: =\displaystyle \frac{18}{36}=\frac{1}{2},\qquad \textrm{dan}\: \: \\ &q=\textbf{Peluang bola hijau}=1-\displaystyle \frac{1}{2}=\frac{1}{2}\\ &f(x)=P\left ( X=x \right )=\begin{pmatrix} n\\ x \end{pmatrix}p^{x}.q^{n-x}\\ &f(4)=P\left ( X=4 \right )=\begin{pmatrix} 5\\ 4 \end{pmatrix}\times \left ( \displaystyle \frac{1}{2} \right )^{4}\times \left ( \frac{1}{2} \right )^{5-4}\\ &\qquad =\displaystyle \frac{5!}{1!\times 4!}\left ( \displaystyle \frac{1}{16} \right )\times \left (\frac{1}{2} \right )\\ &\qquad =5\times \displaystyle \frac{1}{32}=\frac{5}{32}\\ &\qquad =\color{red}0,1563 \end{aligned} \end{array}$

$\begin{array}{ll}\\ 14.&\textrm{Peluang seseorang sembih dari }\\ &\textrm{penyakit jantung adalah 0,6}\\ &\textrm{Jika 7 orang penderita ini menjalani }\\ &\textrm{operasi, maka peluang 3 sampai}\\ &\textrm{6 orang sembuh adalah}... .\\ &\textrm{a}.\quad \displaystyle 0,0629\qquad\qquad\quad\qquad \quad\textrm{d}.\quad \color{red}0,6822\\ &\textrm{b}.\quad \displaystyle 0,2613\quad \textrm{c}.\quad 0,2898\quad\quad \: \textrm{e}.\quad 0,9720\\\\ &\textrm{Jawab}:\\ &\begin{aligned}&p=\textbf{Peluang sembuh}=0,6,\qquad \textrm{maka}\: \: \\ &q=\textbf{Peluang tidak sembuh}=1-0,6=0,4\\ &f(x)=P\left ( X=x \right )=\begin{pmatrix} n\\ x \end{pmatrix}p^{x}.q^{n-x}\\ &\textrm{maka}\\ &P\left ( 3\leq X\leq 6 \right )=P\left ( X\leq 6 \right )-P\left ( X\leq 3 \right )\\ &=C_{4}^{7}(0,6)^{4}(0,4)^{3}+C_{5}^{7}(0,6)^{5}(0,4)^{2}+C_{6}^{7}(0,6)^{6}(0,4)^{1}\\ &=35\times 0,0082944+21\times 0,0124416+7\times 0,0186624\\ &=0,290304+0,2612736+0,1306368\\ &=\color{red}0,6822144 \end{aligned} \end{array}$

$\begin{array}{ll}\\ 15.&\textrm{Peluang seseorang mendapatkan reaksi }\\ &\textrm{buruk setelah disuntik adalah 0,0005}\\ &\textrm{Dari 4000 orang yang disuntik, maka }\\ &\textrm{peluang seseorang mendapatkan reaksi}\\ & \textrm{ada 2 orang adalah}.....\\ &\textrm{a}.\quad \displaystyle \frac{1}{2}e^{-2}\\ &\textrm{b}.\quad e^{-2}\\ &\textrm{c}.\quad \color{red}2e^{-2}\\ &\textrm{d}.\quad \displaystyle \frac{1}{2}e^{2}\\ &\textrm{e}.\quad 2e^{2}\\\\ &\textrm{Jawab}:\\ &\begin{aligned}&\textrm{Di atas adalah contoh kasus }\\ &\textrm{permasalahan}\: \: \textbf{Distribusi Poisson}\\ &P\left ( X=x \right )=f(x)=\left\{\begin{matrix} \displaystyle \frac{e^{-\lambda }.\lambda ^{x}}{x!}\: \: ,\: \: x=0,1,2,3,\cdots \\\ 0,\quad \textrm{untuk}\: \: x\: \: \textrm{yang lain} \end{matrix}\right.\\ &P\left ( X=2 \right )=\displaystyle \frac{e^{-np}.(np)^{2}}{2!}\\ &\qquad =\displaystyle \frac{e^{-(4000.0,0005)}.(4000.0,0005)^{2}}{2!}\\ &\qquad =\displaystyle \frac{e^{-2}.2^{2}}{2}\\ &\qquad =\color{red}2e^{-2} \end{aligned} \end{array}$