Contoh Soal 2 Distribusi Binomial

 $\begin{array}{l}\\ 6.& \text{Pengundian terhadap mata uang }\\ & \text{yang homogen sebanyak 10 kali}\\ & \text{Peluang untuk mendapatkan 6 }\\ & \text{muka angka adalah}....\\ & \text{a}.0,1172\\ & \text{b}.0,2051\\ & \text{c}.0,2461\\ & \text{d}.0,2651\\ & \text{e}.0,2852\\ \\ & \text{Jawab}:\\ & \begin{array}{rl}& p=\mathbf{\text{Peluang Angka}}={\displaystyle \frac{1}{2},\text{dan}}\\ & q=\mathbf{\text{Bukan Angka}}\\ & =\mathbf{\text{Peluang Gambar}}=1-{\displaystyle \frac{1}{2}=\frac{1}{2}}\\ & f(x)=P(x;n;p)=P(X=x)=\left(\begin{array}{c}n\\ x\end{array}\right)p^x{q}^{n-x}\\ & \text{maka}\\ & f(x)=P(X=x)=\left(\begin{array}{c}n\\ x\end{array}\right)p^x.q^{n-x}\\ & f(6)=P(X=6)=\left(\begin{array}{c}10\\ 6\end{array}\right)\times {\left({\displaystyle \frac{1}{2}}\right)}^6\times {\left(\frac{1}{2}\right)}^{10-6}\\ & ={\displaystyle \frac{10!}{6!\times 4!}{\left({\displaystyle \frac{1}{2}}\right)}^{6+4}}\\ & =210\times {\displaystyle \frac{1}{1024}}\\ & =0,2051\end{array}\end{array}$

$\begin{array}{l}\\ 7.& \text{Pada pengundian terhadap mata uang identik},\\ & \text{sebanyak 10 kali, peluang distribusi binomial}\\ & \text{untuk mendapatkan 7 muka gambar adalah}....\\ & \begin{array}{l}\\ \text{a}.{\displaystyle 0,2653}& & \text{d}.{\displaystyle 0,7522}\\ \text{b}.{\displaystyle 0,1172}& \text{c}.{\displaystyle 0,2653}& \text{e}.0,2422\end{array}\\ \\ & \text{Jawab}:\\ & \text{Uraian berikut sekaligus tambahan}\\ & \text{penjelasan pada uraian jawaban}\\ & \text{soal no. 6 di atas}\\ & \begin{array}{rl}f(x)& =P(x;n;p)=\left(\begin{array}{c}n\\ x\end{array}\right)p^x{q}^{n-x}\\ & \text{Ingat sebuah koin ada 2 muka}\\ & \text{yaitu muka gambar (G) dan angka (A)}\\ & \text{misalkan}\\ & A=\text{kejadian muncul muka gambar}\\ & \text{maka peluangnya adalah}{\displaystyle \frac{1}{2}}\\ & \text{Selanjutnya di sini disimbolkan dengan}p={\displaystyle \frac{1}{2}}\\ & \text{Demikian juga misalkan}\\ & B=\text{kejadian muncul muka angka}\\ & \text{maka peluang juga}{\displaystyle \frac{1}{2}}\\ & \text{Di sini dituliskan dengan}q={\displaystyle \frac{1}{2}}\\ f(7)& =\left(\begin{array}{c}10\\ 7\end{array}\right){\left({\displaystyle \frac{1}{2}}\right)}^7{\left({\displaystyle \frac{1}{2}}\right)}^{10-7}\\ & =\left(\begin{array}{c}10\\ 7\end{array}\right){\left({\displaystyle \frac{1}{2}}\right)}^7{\left({\displaystyle \frac{1}{2}}\right)}^3\\ & ={\displaystyle \frac{10!}{7!\times (10-7)!}{\left({\displaystyle \frac{1}{2}}\right)}^{7+3}}\\ & ={\displaystyle \frac{10.9.8.\text{⧸}7!}{\text{⧸}7!.3.2.1}\left({\displaystyle \frac{1}{1024}}\right)}\\ & =0,1172\end{array}\end{array}$

$\begin{array}{l}\\ 8.& \text{Sebuah uang logam dilempar sebanyak 8}\\ & \text{kali. Peluang muncul gambar sebanyak}\\ & \text{5 kali adalah}....\\ & \begin{array}{l}\\ \text{a}.& {\displaystyle \frac{3}{32}}& & & \text{d}.& {\displaystyle \frac{7}{32}}\\ \\ \text{b}.& {\displaystyle \frac{4}{32}}& \text{c}.& {\displaystyle \frac{5}{32}}& \text{e}.& {\displaystyle \frac{9}{32}}\end{array}\\ \\ & \text{Jawab}:\\ & \begin{array}{rl}f(x)& =P(x;n;p)=\left(\begin{array}{c}n\\ x\end{array}\right)p^x{q}^{n-x}\\ f(5)& =\left(\begin{array}{c}8\\ 5\end{array}\right){\left({\displaystyle \frac{1}{2}}\right)}^5{\left({\displaystyle \frac{1}{2}}\right)}^{8-5}\\ & =\left(\begin{array}{c}8\\ 5\end{array}\right){\left({\displaystyle \frac{1}{2}}\right)}^5{\left({\displaystyle \frac{1}{2}}\right)}^3\\ & ={\displaystyle \frac{8!}{5!\times (8-5)!}{\left({\displaystyle \frac{1}{2}}\right)}^{5+3}}\\ & ={\displaystyle \frac{8.7.6.5!}{5!.3.2.1}\left({\displaystyle \frac{1}{256}}\right)}\\ & ={\displaystyle \frac{8.7}{256}}\\ & ={\displaystyle \frac{7}{32}}\end{array}\end{array}$

$\begin{array}{l}\\ 9.& \text{Pada pelemparan sebuah koin sebanyak 4 kali}\\ & \text{Peluang didapatkannya dua angka pada}\\ & \text{pelemparan tersebut adalah}....\\ & \begin{array}{l}\\ \text{a}.{\displaystyle 0,123}& & \text{d}.{\displaystyle 0,232}\\ \text{b}.{\displaystyle 0,135}& \text{c}.{\displaystyle 0,154}& \text{e}.0,375\end{array}\\ \\ & \text{Jawab}:\\ & \begin{array}{rl}f(x)& =P(x;n;p)=\left(\begin{array}{c}n\\ x\end{array}\right)p^x{q}^{n-x}\\ f(2)& =\left(\begin{array}{c}4\\ 2\end{array}\right){\left({\displaystyle \frac{1}{2}}\right)}^2{\left({\displaystyle \frac{1}{2}}\right)}^{4-2}\\ & =\left(\begin{array}{c}4\\ 2\end{array}\right){\left({\displaystyle \frac{1}{2}}\right)}^2{\left({\displaystyle \frac{1}{2}}\right)}^2\\ & ={\displaystyle \frac{4!}{2!\times (4-2)!}{\left({\displaystyle \frac{1}{2}}\right)}^{2+2}}\\ & ={\displaystyle \frac{4.3.2!}{2!.2.1}\left({\displaystyle \frac{1}{16}}\right)}\\ & =0,375\end{array}\end{array}$

$\begin{array}{l}\\ 10.& \text{Dari data survei didapatkan bahwa}\\ & \text{satu dari lima orang telah berkunjung}\\ & \text{ke dokter dalam sembarang bulan yang}\\ & \text{ditanyakan. Jika 10 orang dipilih secara}\\ & \text{acak, peluang 3 orang telah berkunjung}\\ & \text{ke dokter bulan lalu adalah}....\\ & \begin{array}{l}\\ \text{a}.& {\displaystyle 0,125}& & & \text{d}.& {\displaystyle 0,201}\\ \\ \text{b}.& {\displaystyle 0,174}& \text{c}.& {\displaystyle 0,182}& \text{e}.& {\displaystyle 0,423}\end{array}\\ \\ & \text{Jawab}:\\ & \begin{array}{rl}f(x)& =P(x;n;p)=\left(\begin{array}{c}n\\ x\end{array}\right)p^x{q}^{n-x}\\ f(3)& =\left(\begin{array}{c}10\\ 3\end{array}\right){\left({\displaystyle \frac{1}{5}}\right)}^3{\left({\displaystyle \frac{4}{5}}\right)}^{10-3}\\ & =\left(\begin{array}{c}10\\ 3\end{array}\right){\left({\displaystyle \frac{1}{5}}\right)}^3{\left({\displaystyle \frac{4}{5}}\right)}^7\\ & ={\displaystyle \frac{10!}{3!\times 7!}\left({\displaystyle \frac{1}{125}}\right)\left({\displaystyle \frac{4^7}{5^7}}\right)}\\ & =\cdots \\ & ={\displaystyle 0,201}\end{array}\end{array}$