Contoh Soal 1 Peluang Kejadian Majmuk

$\begin{array}{ll}1.& \text{Banyak anggota ruang sampel dari}\\ & \text{pelemparan sebuah dadu dan dua }\\ & \text{keping mata uang secara bersamaan}\\ & \text{adalah}....\\ & \begin{array}{l}\\ \text{a}.& {\displaystyle 4}& & & \text{d}.& {\displaystyle 24}\\ \\ \text{b}.& {\displaystyle 6}& \text{c}.& {\displaystyle 12}& \text{e}.& {\displaystyle 36}\end{array}\\ \\ & \text{Jawab}:\\ & \begin{array}{rl}1& \text{mata dadu}=P(6,1)=6\\ 2& \text{keping mata uang}=P(2,1)\times P(2,1)=4\\ \text{R}& \text{uang sampelnya adalah}:6\times 4=24\end{array}\end{array}$

$\begin{array}{ll}2.& \text{Setumpuk kartu remi diambil sebuah}\\ & \text{kartu secara acak. Peluang agar kartu}\\ & \text{yang terambil bukan kartu king}\\ & \text{adalah}....\\ & \begin{array}{l}\\ \text{a}.& {\displaystyle 0}& & & \text{d}.& {\displaystyle {\displaystyle \frac{12}{13}}}\\ \\ \text{b}.& {\displaystyle {\displaystyle \frac{1}{13}}}& \text{c}.& {\displaystyle \frac{1}{2}}& \text{e}.& {\displaystyle 1}\end{array}\\ \\ & \text{Jawab}:\\ & \begin{array}{rl}& \text{Misalkan}A=\text{Kejadian muncul kartu king}\\ & n(A)=\text{banyak kartu king ada}=4\\ & n(S)=\text{total kartu}=4\times 13\\ & A^{\prime }=\text{kejadian muncul bukan kartu king}\\ & \text{maka peluangnya bukan kartu king}:\\ & P(A^{\prime })=1-P(A)=1-{\displaystyle \frac{4}{4\times 13}=\frac{12}{13}}\end{array}\end{array}$

$\begin{array}{ll}3.& \text{Sebuah dadu dilempar sekali. Peluang}\\ & \text{muncul mata dadu 3 atau lebih adalah}....\\ & \begin{array}{l}\\ \text{a}.& {\displaystyle {\displaystyle \frac{1}{6}}}& & & \text{d}.& {\displaystyle {\displaystyle \frac{3}{5}}}\\ \\ \text{b}.& {\displaystyle {\displaystyle \frac{1}{3}}}& \text{c}.& {\displaystyle \frac{1}{2}}& \text{e}.& {\displaystyle \frac{2}{3}}\end{array}\\ \\ & \text{Jawab}:\\ & \begin{array}{rl}& \text{Misal}A=\text{muncul mata dadu 3 atau lebih}\\ & A=\{3,4,5,6\}\\ & S=\{1,2,3,4,5,6\}\\ & \text{maka}n(A)=4\text{dengan}(S)=6\\ & P(A)={\displaystyle \frac{n(A)}{n(S)}=\frac{4}{6}=\frac{2}{3}}\end{array}\end{array}$

$\begin{array}{ll}4.& \text{Sebuah dadu dan sebuah mata uang logam}\\ & \text{dilempar bersama-sama. Peluang muncul}\\ & \text{gambar pada mata uang dan mata 1 pada}\\ & \text{dadu adalah}....\\ & \begin{array}{llllll}\text{a}.& {\displaystyle \frac{1}{12}}& & & \text{d}.& {\displaystyle \frac{1}{3}}\\ \\ \text{b}.& {\displaystyle \frac{1}{6}}& \text{c}.& {\displaystyle \frac{1}{4}}& \text{e}.& {\displaystyle \frac{1}{2}}\end{array}\\ \\ & \text{Jawab}:\\ & \begin{array}{rl}& \mathbf{\text{Cara pertama}}\\ & \text{Perhatikan tabel berikut}\\ & \begin{array}{|ccccccc|}\hline ◻& 1& 2& 3& 4& 5& 6\\ A& (A,1)& (A,2)& (A,3)& (A,4)& (A,5)& (A,6)\\ G& (G,1)& (G,2)& (G,3)& (G,4)& (G,5)& (G,6)\\ \hline\end{array}\\ & \text{dari tabel di atas didapatkan bahwa}:\\ & A=\text{kejadian muncul mata 1 pada dadu}\\ & n(A)=2\\ & B=\text{kejadian muncul gambar pada uang}\\ & n(B)=6\\ & A\cap B=\text{kejadian muncul 1 pada dadu}\\ & \text{gambar pada koin}\\ & n(A\cap B)=1\\ & \text{dengan}n(S)=12,\\ & \text{maka peluang muncul mata 1 dan gambar}\\ & P(A\cap B)={\displaystyle \frac{n(A\cap B)}{n(S)}=\frac{1}{12}}\\ & \mathbf{\text{Cara kedua}}\\ & \text{Karena ini dua kejadian}\mathit{\text{saling bebas}}\\ & \text{maka}\\ & P(A\cap B)=P(A)\times P(B)\\ & ={\displaystyle \frac{n(A)}{n(S)}\times \frac{n(B)}{n(S)}}\\ & ={\displaystyle \frac{2}{12}\times \frac{6}{12}=\frac{12}{144}=\frac{1}{12}}\end{array}\end{array}$

$\begin{array}{ll}5.& \text{Peluang Dika lulus ujian adalah}0,75\text{dan}\\ & \text{peluang Tutik lulus ujian adalah}0,80.\\ & \text{Peluang keduanya lulus ujian adalah}....\\ & \begin{array}{l}\\ \text{a}.& {\displaystyle 0,4}& & & \text{d}.& {\displaystyle 0,7}\\ \\ \text{b}.& {\displaystyle 0,5}& \text{c}.& {\displaystyle 0,6}& \text{e}.& {\displaystyle 0,8}\end{array}\\ \\ & \text{Jawab}:\\ & \begin{array}{rl}& \text{Dua kejadian ini adalah}\mathit{\text{saling bebas}}\\ & \text{Misal}A=\text{Kejadian Dika lulus}\\ & n(A)=\cdots \text{tidak diketahui, tetapi}\\ & P(A)=0,75={\displaystyle \frac{3}{4},\mathbf{\text{diketahui}}}\\ & \text{dan}B=\text{Tutik lulus}\\ & n(B)=\cdots \text{juga tidak diketahui}\\ & P(B)=0,8={\displaystyle \frac{4}{5}}\\ & P(A\cap B)=P(A)\times P(B)\\ & ={\displaystyle \frac{n(A)}{n(S)}\times \frac{n(B)}{n(S)}}\\ & ={\displaystyle \frac{3}{4}\times \frac{4}{5}=\frac{3}{5}=0,6}\end{array}\end{array}$