Contoh Soal 1 Peluang Kejadian Majmuk

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

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

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

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

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