Contoh Soal 2 Distribusi Binomial

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

$\begin{array}{ll}\\ 7.&\textrm{Pada pengundian terhadap mata uang identik},\\ &\textrm{sebanyak 10 kali, peluang distribusi binomial} \\ &\textrm{untuk mendapatkan 7 muka gambar adalah}\: ....\\ &\begin{array}{lllllll}\\ \textrm{a}.\quad \displaystyle 0,2653&&\textrm{d}.\quad \displaystyle 0,7522\\ \textrm{b}.\quad \displaystyle \color{red}0,1172&\textrm{c}.\quad \displaystyle 0,2653&\textrm{e}.\quad 0,2422 \end{array}\\\\ &\textrm{Jawab}:\\ &\textrm{Uraian berikut sekaligus tambahan}\\ &\textrm{penjelasan pada uraian jawaban}\\ &\color{blue}\textrm{soal no. 6 di atas}\\ &\begin{aligned}f(x)&=P(x;n;p)=\begin{pmatrix} n\\ x \end{pmatrix}p^{x}q^{n-x}\\ &\textrm{Ingat sebuah koin ada 2 muka}\\ &\textrm{yaitu muka gambar (G) dan angka (A)}\\ &\color{red}\textrm{misalkan}\\ &A=\textrm{kejadian muncul muka gambar}\\ &\textrm{maka peluangnya adalah}\: \: \displaystyle \frac{1}{2}\\ &\textrm{Selanjutnya di sini disimbolkan dengan}\: \: \: \color{blue}p=\displaystyle \frac{1}{2}\\ &\color{red}\textrm{Demikian juga misalkan}\\ &B=\textrm{kejadian muncul muka angka}\\ &\textrm{maka peluang juga}\: \displaystyle \frac{1}{2}\\ &\textrm{Di sini dituliskan dengan}\: \: \: \color{blue}q=\displaystyle \frac{1}{2}\\ f(7)&=\begin{pmatrix} 10\\ 7 \end{pmatrix}\left ( \displaystyle \frac{1}{2} \right )^{7}\left ( \displaystyle \frac{1}{2} \right )^{10-7}\\ &=\begin{pmatrix} 10\\ 7 \end{pmatrix}\left ( \displaystyle \frac{1}{2} \right )^{7}\left ( \displaystyle \frac{1}{2} \right )^{3}\\ &=\displaystyle \frac{10!}{7!\times (10-7)!}\left ( \displaystyle \frac{1}{2} \right )^{7+3}\\ &=\displaystyle \frac{10.9.8.\not{7!}}{\not{7!}.3.2.1}\left ( \displaystyle \frac{1}{1024} \right ) \\ &=\color{red}0,1172 \end{aligned} \end{array}$

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

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

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