Contoh Soal 3 Distribusi Binomial

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

$\begin{array}{l}\\ 12.& \text{Sebuah kotak berisi 20 bola dengan }\\ & \text{rincian 12 boal berwarna kuning dan }\\ & \text{sisanya berwarna hijau. Dari kotak}\\ & \text{diambil 6 bola secara acak. Peluang}\\ & \text{terambil 4 bola hijau adalah}....\\ & \text{a}.{\displaystyle 0,1238\text{d}.0,8132}\\ & \text{b}.{\displaystyle 0,1382\text{c}.0,3110\text{e}.0,9590}\\ \\ & \text{Jawab}:\\ & \begin{array}{rl}& p=\mathbf{\text{Peluang bola kuning}}\\ & ={\displaystyle \frac{C_1^{12}}{C_1^{20}}={\displaystyle \frac{12}{20}=\frac{3}{5},}}\\ & q=\mathbf{\text{Peluang bola hijau}}=1-{\displaystyle \frac{3}{5}=\frac{2}{5}}\\ & f(x)=P(X=x)=\left(\begin{array}{c}n\\ x\end{array}\right)p^x.q^{n-x}\\ & \text{maka}\\ & f(4)=\left(\begin{array}{c}6\\ 4\end{array}\right)\times {\left({\displaystyle \frac{2}{5}}\right)}^4\times {\left(\frac{3}{5}\right)}^{6-4}\\ & ={\displaystyle \frac{6!}{2!\times 4!}\left({\displaystyle \frac{16}{625}}\right)\times \left({\displaystyle \frac{9}{25}}\right)}\\ & =15\times {\displaystyle \frac{144}{15625}=\frac{2160}{15625}}\\ & =0,1382\end{array}\end{array}$

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

$\begin{array}{l}\\ 14.& \text{Peluang seseorang sembih dari }\\ & \text{penyakit jantung adalah 0,6}\\ & \text{Jika 7 orang penderita ini menjalani }\\ & \text{operasi, maka peluang 3 sampai}\\ & \text{6 orang sembuh adalah}....\\ & \text{a}.{\displaystyle 0,0629\text{d}.0,6822}\\ & \text{b}.{\displaystyle 0,2613\text{c}.0,2898\text{e}.0,9720}\\ \\ & \text{Jawab}:\\ & \begin{array}{rl}& p=\mathbf{\text{Peluang sembuh}}=0,6,\text{maka}\\ & q=\mathbf{\text{Peluang tidak sembuh}}=1-0,6=0,4\\ & f(x)=P(X=x)=\left(\begin{array}{c}n\\ x\end{array}\right)p^x.q^{n-x}\\ & \text{maka}\\ & P(3\le X\le 6)=P(X\le 6)-P(X\le 3)\\ & =C_4^7(0,6{)}^4(0,4{)}^3+C_5^7(0,6{)}^5(0,4{)}^2+C_6^7(0,6{)}^6(0,4{)}^1\\ & =35\times 0,0082944+21\times 0,0124416+7\times 0,0186624\\ & =0,290304+0,2612736+0,1306368\\ & =0,6822144\end{array}\end{array}$

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