Contoh Soal 3 Peluang Kejadian

 $\begin{array}{ll}11.& \text{Jika kejadian}A\text{dan}B\text{adalah dua kejadian}\\ & \text{dengan}P(A)={\displaystyle \frac{8}{15},P(B)={\displaystyle \frac{7}{12},\text{dan}}}\\ & P(A|B)={\displaystyle \frac{4}{7},\text{maka nilai}P(B|A)=....}\\ & \begin{array}{l}\\ \text{a}.& {\displaystyle \frac{8}{45}}& & & \text{d}.& {\displaystyle \frac{5}{8}}\\ \\ \text{b}.& {\displaystyle \frac{1}{3}}& \text{c}.& {\displaystyle \frac{3}{8}}& \text{e}.& {\displaystyle \frac{7}{15}}\end{array}\\ \\ & \text{Jawab}:\\ & \begin{array}{rl}& \text{Kejadian di atas adalah contoh}\\ & \text{kejadian}\mathbf{\text{tidak saling bebas (bersyarat)}}.\\ & \text{Diketahui bahwa}\\ & P(A)={\displaystyle \frac{8}{15},P(B)={\displaystyle \frac{7}{12},P(A|B)={\displaystyle \frac{4}{7}}}}\\ & \text{Ditanyakan nilai}P(B|A)=...?\\ & \text{maka}\\ & \begin{array}{rl}P(A\cap B)& =P(A\cap B)\\ P(A)\times P(B|A)& =P(B)\times P(A|B)\\ P(B|A)& ={\displaystyle \frac{P(B)\times P(A|B)}{P(A)}}\\ & ={\displaystyle \frac{\left({\displaystyle \frac{7}{12}}\right)\times \left({\displaystyle \frac{4}{7}}\right)}{{\displaystyle \frac{8}{15}}}}\\ & ={\displaystyle \frac{{\displaystyle \frac{1}{3}}}{{\displaystyle \frac{8}{15}}}}\\ & ={\displaystyle \frac{1}{3}\times \frac{15}{8}}\\ & ={\displaystyle \frac{5}{8}}\end{array}\end{array}\end{array}$

Contoh Soal 2 Peluang Kejadian

 $\begin{array}{ll}6.& \text{Sebuah kantong berisi 7 kelereng merah,}\\ & \text{5 kelereng hijau, dan 4 kelereng biru}\\ & \text{Diambil sebuah kelereng secara acak.}\\ & \text{Peluang yang terambil merah atau hijau}\\ & \text{adalah}....\\ & \begin{array}{l}\\ \text{a}.& {\displaystyle \frac{5}{16}}& & & \text{d}.& {\displaystyle \frac{3}{4}}\\ \\ \text{b}.& {\displaystyle \frac{7}{16}}& \text{c}.& {\displaystyle \frac{1}{2}}& \text{e}.& {\displaystyle \frac{2}{3}}\end{array}\\ \\ & \text{Jawab}:\\ & \begin{array}{rl}& \text{Kejadian di atas adalah contoh}\\ & \text{kejadian}\mathbf{\text{saling lepas}}.\text{Misalkan}\\ & A=\text{kejadian terambil 1 kelereng merah}\\ & n(A)=C(7,1)=\left(\begin{array}{c}7\\ 1\end{array}\right)=7\\ & B=\text{kejadian terambil 1 kelereng hijau}\\ & n(B)=C(5,1)=\left(\begin{array}{c}5\\ 1\end{array}\right)=5\\ & S=\text{semua dianggap identik}\\ & n(S)=C(16,1)=\left(\begin{array}{c}16\\ 1\end{array}\right)=16\\ & \text{maka}\\ & P(A\cup B)=P(A)+P(B)\\ & ={\displaystyle \frac{n(A)}{n(S)}+\frac{n(B)}{n(S)}}\\ & ={\displaystyle \frac{7}{16}+\frac{5}{16}=\frac{12}{16}={\displaystyle \frac{3}{4}}}\end{array}\end{array}$

$\begin{array}{ll}7.& \text{Dari 100 orang yang mengikuti kegiatan}\\ & \text{jalan santai terdapat 60 orang memakai}\\ & \text{topi dan 45 orang yang berkacamata.}\\ & \text{Peluang bahwa seorang yang dipilih dari}\\ & \text{kelompok orang itu memakai topi dan}\\ & \text{kacamata adalah}....\\ & \begin{array}{l}\\ \text{a}.& {\displaystyle \frac{1}{20}}& & & \text{d}.& {\displaystyle \frac{11}{20}}\\ \\ \text{b}.& {\displaystyle \frac{2}{5}}& \text{c}.& {\displaystyle \frac{9}{20}}& \text{e}.& {\displaystyle \frac{3}{5}}\end{array}\\ \\ & \text{Jawab}:\\ & \text{Perhatikanlah ilustrasi}\mathbf{\text{Diagram Venn}}\\ & \text{berikut ini}\\ & \begin{array}{|ccl|}\hline \begin{array}{|l|}\hline \text{S}=100\\ \hline\end{array}& & \\ & \text{A}\text{B}& \\ & \begin{array}{|lcr|}\hline 60-n& n& 45-n\\ \hline\end{array}& \\ & & \\ \hline\end{array}\\ & \begin{array}{rl}& \text{Kejadian di atas adalah contoh}\\ & \text{kejadian}\mathbf{\text{tidak saling lepas}}.\\ & A=\text{kejadian terpilih seorang bertopi}\\ & n(A)=C(60,1)=\left(\begin{array}{c}60\\ 1\end{array}\right)=60\\ & B=\text{kejadian terpilih seorang berkacamata}\\ & n(B)=C(45,1)=\left(\begin{array}{c}45\\ 1\end{array}\right)=45\\ & A\cap B=\text{terpilih seorang bertopi dan}\\ & \text{berkacamata}\\ & n(A\cap B)=x\\ & S=\text{semua dianggap identik}\\ & n(S)=C(100,1)=\left(\begin{array}{c}100\\ 1\end{array}\right)=100\\ & \text{maka}\\ & P(A\cup B)=P(A)+P(B)-P(A\cap B)\\ & {\displaystyle \frac{n(A\cup B)}{n(S)}={\displaystyle \frac{n(A)}{n(S)}+\frac{n(B)}{n(S)}-{\displaystyle \frac{n(A\cap B)}{n(S)}}}}\\ & {\displaystyle \frac{100}{100}=\frac{60}{100}+\frac{45}{100}-\frac{x}{100}}\\ & {\displaystyle \frac{x}{100}={\displaystyle \frac{105}{100}-\frac{100}{100}}}\\ & ={\displaystyle \frac{5}{100}=\frac{1}{20}}\end{array}\end{array}$

$\begin{array}{ll}8.& \text{Diketahui dua buah kotak A dan B}\\ & \text{berisi 5 bola putih dan 3 bola merah.}\\ & \text{Kotak B berisi 4 bola putih dan 2 bola}\\ & \text{merah. Jika diambil secara acak 1 kotak,}\\ & \text{kemudian diambil secara acak 1 bola dari}\\ & \text{kotak tersebut, maka peluang terambilnya}\\ & \text{bola putih adalah}....\\ & \begin{array}{l}\\ \text{a}.& {\displaystyle \frac{5}{16}}& & & \text{d}.& {\displaystyle \frac{1}{2}}\\ \\ \text{b}.& {\displaystyle \frac{1}{3}}& \text{c}.& {\displaystyle \frac{7}{16}}& \text{e}.& {\displaystyle \frac{31}{48}}\end{array}\\ \\ & \text{Jawab}:\\ & \begin{array}{rl}& \text{Kejadian di atas adalah kejadian}\\ & \mathbf{\text{saling lepas}}\text{dari dua kejadian Q}\\ & \text{dan R. Misalkan}:\\ & \text{Pada kotak A}\\ & Q=\text{Terambil 1 bola putih di kotak A}\\ & n(Q)=C(5,1)=\left(\begin{array}{c}5\\ 1\end{array}\right)=5\\ & S_Q=\text{Terambil 1 bola saja di kotak A}\\ & n(S_Q)=C(8,1)=\left(\begin{array}{c}8\\ 1\end{array}\right)=8\\ & \text{Pada kotak B}\\ & R=\text{Terambil 1 bola putih di kotak B}\\ & n(R)=C(4,1)=\left(\begin{array}{c}4\\ 1\end{array}\right)=4\\ & S_R=\text{Terambil 1 bola saja di kotak B}\\ & n(S_R)=C(6,1)=\left(\begin{array}{c}6\\ 1\end{array}\right)=6\\ & \text{Karena kejadian pengambilan sebuah}\\ & \text{bola putih di atas adalah dari pilihan}\\ & \text{dua buah kotak yang ada, maka peluang}\\ & \text{pengambilannya adalah 1 dari 2 kotak}\\ & \text{peluang kejadian ini adalah}={\displaystyle \frac{1}{2}.}\\ & \text{Sehingga peluang kasus di atas adalah}:\\ & {\displaystyle \frac{1}{2}P(Q\cup R)={\displaystyle \frac{1}{2}(P(Q)+P(R))}}\\ & ={\displaystyle \frac{1}{2}\left({\displaystyle \frac{n(Q)}{n(S_Q)}+\frac{n(R)}{n(S_R)}}\right)}\\ & ={\displaystyle \frac{1}{2}\left({\displaystyle \frac{5}{8}+\frac{4}{6}}\right)}\\ & ={\displaystyle \frac{1}{2}\left({\displaystyle \frac{31}{24}}\right)={\displaystyle \frac{31}{48}}}\end{array}\end{array}$

$\begin{array}{ll}9.& \text{Kotak I berisi 3 bola merah dan 2 bola }\\ & \text{putih. Kotak II berisi 3 bola hijau dan 5}\\ & \text{biru. Dari tiap-tiap kotak diambil 2 bola}\\ & \text{sekaligus secara acak. Peluang terambil 2}\\ & \text{bola merah pada kotak I dan 2 bola biru}\\ & \text{dari kotak II adalah}....\\ & \begin{array}{l}\\ \text{a}.& {\displaystyle \frac{1}{10}}& & & \text{d}.& {\displaystyle \frac{3}{8}}\\ \\ \text{b}.& {\displaystyle \frac{3}{28}}& \text{c}.& {\displaystyle \frac{4}{15}}& \text{e}.& {\displaystyle \frac{57}{140}}\end{array}\\ \\ & \text{Jawab}:\\ & \begin{array}{rl}& \text{Kejadian di atas adalah kejadian}\\ & \mathbf{\text{saling bebas}}\text{dari dua kejadian A}\\ & \text{dan B. Misalkan}:\\ & \text{Pada kotak I}\\ & A=\text{Terambil 2 bola merah di kotak I}\\ & n(A)=C(3,2)=\left(\begin{array}{c}3\\ 2\end{array}\right)=3\\ & S=\text{Terambil 2 bola saja di kotak I}\\ & n(S)=C(5,2)=\left(\begin{array}{c}5\\ 2\end{array}\right)=10\\ & \text{Pada kotak II}\\ & B=\text{Terambil 2 bola biru di kotak II}\\ & n(B)=C(5,2)=\left(\begin{array}{c}5\\ 2\end{array}\right)=10\\ & S=\text{Terambil 2 bola saja di kota II}\\ & n(S)=C(8,2)=\left(\begin{array}{c}8\\ 2\end{array}\right)=28\\ & \text{Sehingga peluang kasus di atas adalah}:\\ & {\displaystyle P(A\cap B)=P(A)\times P(B)}\\ & ={\displaystyle \frac{n(A)}{n(S)}\times \frac{n(B)}{n(S)}}\\ & ={\displaystyle \frac{3}{10}\times \frac{10}{28}}\\ & ={\displaystyle \frac{3}{28}}\end{array}\end{array}$

$\begin{array}{ll}10.& \text{Jika kejadian}A\text{dan}B\text{dapat terjadi secara}\\ & \text{bersamaan. Jika}P(A)=0,6,P(B)=0.75,\\ & \text{dan}P(A\cap B)=0,43,\text{maka}P(A\cup B)=....\\ & \begin{array}{l}\\ \text{a}.& {\displaystyle 0,98}& & & \text{d}.& {\displaystyle 0,92}\\ \\ \text{b}.& {\displaystyle 0,96}& \text{c}.& {\displaystyle 0,94}& \text{e}.& {\displaystyle 0,91}\end{array}\\ \\ & \text{Jawab}:\\ & \begin{array}{rl}& \text{Kejadian di atas adalah contoh}\\ & \text{kejadian}\mathbf{\text{tidak saling lepas}}.\\ & \text{Diketahui bahwa}\\ & P(A)=0,6,P(B)=0,75,P(A\cap B)=0,43\\ & \text{Ditanyakan nilai}P(A\cup B)=...?\\ & \text{maka}\\ & P(A\cup B)=P(A)+P(B)-P(A\cap B)\\ & =0,6+0,75-0,43\\ & =0,92\end{array}\end{array}$

PKKM MA Futuhiyah Jeketro 2022

Contoh Soal 1 Peluang Kejadian

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

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

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

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

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

Contoh Soal 4 Kaidah Pencacahan

 $\begin{array}{ll}16.& \text{Diketahui himpunan yang terdiri dari 5}\\ & \text{huruf vokal dan 10 huruf konsonan yang}\\ & \text{semuanya berlainan. Dari himpunan itu}\\ & \text{disusun suatu kata yang terdiri dari 2}\\ & \text{huruf vokal dan 3 konsonan. Banyak kata}\\ & \text{yang dapat disusun sebanyak}....\\ & \begin{array}{l}\\ \text{a}.& {\displaystyle 144.000}& & & \text{d}.& {\displaystyle 72.000}\\ \\ \text{b}.& {\displaystyle 126.000}& \text{c}.& {\displaystyle 96.000}& \text{e}.& {\displaystyle 36.000}\end{array}\\ \\ & \text{Jawab}:\\ & \begin{array}{rl}& \text{Diketahui bahwa ingin menyusun}\\ & \mathbf{\text{5 huruf dengan susunan berbeda}}\\ & \mathit{\text{yang tersusun dari}}\\ & \text{2 dari 5 vokal berbeda disusun, dan}\\ & \text{3 dari 10 konsonan berbeda juga disusun}\\ & \text{maka banyak susunan kata terbentuk}:\\ & \text{Seperti menyusun 5 objek (kombinasi)}\\ & \text{2 benda dari 5 benda, atau 3 }\\ & \text{benda yang terbentuk dari 5}\\ & \text{objek yg tidak identik(permutasi)}\\ & \mathbf{\text{Cara pertama}}\\ & =C((2+3),2)\times P(5,2)\times P(10,5)\\ & ={\displaystyle \frac{5!}{2!\times 3!}\times \frac{5!}{(5-2)!}\times \frac{10!}{(10-3)!}}\\ & ={\displaystyle \frac{5!}{2!\times 3!}\times \frac{5!}{3!}\times \frac{10!}{7!}}\\ & =10\times 60\times 720\\ & =144.000\\ & \mathbf{\text{Cara kedua}}\\ & =C((2+3),3)\times P(5,2)\times P(10,5)\\ & ={\displaystyle \frac{5!}{3!\times 2!}\times \frac{5!}{(5-2)!}\times \frac{10!}{(10-3)!}}\\ & ={\displaystyle \frac{5!}{3!\times 2!}\times \frac{5!}{3!}\times \frac{10!}{7!}}\\ & =10\times 60\times 720\\ & =144.000\end{array}\end{array}$

Contoh Soal 3 Kaidah Pencacahan

 $\begin{array}{ll}11.& \text{Berikut ini nilainya tidak sama dengan}\\ & C(7,5)\text{adalah}....\\ \\ & (i){\displaystyle \frac{7!}{5!(7-5)!}}\\ \\ & (ii)C(6,1)\\ \\ & (iii){\displaystyle \frac{P(7,5)}{5!}}\\ \\ & (iv)\left(\begin{array}{c}6\\ 1\end{array}\right)\\ & \begin{array}{l}\\ \text{a}.& {\displaystyle (i),(ii),\mathrm{\&}(iii)}& & & \text{d}.& {\displaystyle \text{hanya}(i)}\\ \\ \text{b}.& {\displaystyle (i)\mathrm{\&}(iii)}& \text{c}.& (ii)\mathrm{\&}(iv)& \text{e}.& {\displaystyle \text{hanya}(iv)}\end{array}\\ \\ & \text{Jawab}:\\ & C(7,5)={\displaystyle \frac{P(7,5)}{5!}=\frac{7!}{5!(7-5)!}}\end{array}$

$\begin{array}{ll}12.& \text{Nilai}n\text{yang memenuhi persamaan}\\ & \left(\begin{array}{c}100\\ 45\end{array}\right)=\left(\begin{array}{c}100\\ 5n\end{array}\right)\text{adalah}....\\ & \begin{array}{l}\\ \text{a}.& {\displaystyle 15}& & & \text{d}.& {\displaystyle 12}\\ \\ \text{b}.& {\displaystyle 14}& \text{c}.& {\displaystyle 13}& \text{e}.& {\displaystyle 11}\end{array}\\ \\ & \text{Jawab}:\\ & \begin{array}{rl}& \text{Diketahui bahwa}\\ & \left(\begin{array}{c}100\\ 45\end{array}\right)=\left(\begin{array}{c}100\\ 5n\end{array}\right),\text{maka}\\ & 45+5n=100\\ & 5n=100-45=55\\ & n={\displaystyle \frac{55}{5}=11}\end{array}\end{array}$

$\begin{array}{ll}13.& \text{Koefisien suku ke-4 dari}(2x-3{)}^4\\ & \begin{array}{l}\\ \text{a}.& {\displaystyle -216}& & & \text{d}.& {\displaystyle 81}\\ \\ \text{b}.& {\displaystyle -96}& \text{c}.& {\displaystyle 16}& \text{e}.& {\displaystyle 216}\end{array}\\ \\ & \text{Jawab}:\\ & \begin{array}{rl}& \text{Diketahui bahwa}\\ & (2x-3{)}^4={\displaystyle \sum _{i=0}^4\left(\begin{array}{c}4\\ i\end{array}\right)(2x{)}^{4-i}(-3{)}^i}\\ & \text{Suku ke-4-nya adalah}:r=4.\\ & \text{Suku ke-r}=\left(\begin{array}{c}n\\ r-1\end{array}\right)a^{n-r+1}{b}^{r-1}\\ & \text{Sehingga suku ke-4 adalah}:\\ & =\left(\begin{array}{c}4\\ 4-1\end{array}\right)(2x{)}^{4-4+1}(-3{)}^{4-1}\\ & =\left(\begin{array}{c}4\\ 3\end{array}\right)(2x{)}^1(-3{)}^3\\ & ={\displaystyle \frac{4!}{3!\times 1!}2x(-27)}\\ & =-4.2.27x\\ & =-216\end{array}\end{array}$

$\begin{array}{ll}14.& \text{Bentuk sederhana dari}{\displaystyle \sum _{r=1}^{n}r{\displaystyle \left(\begin{array}{c}n\\ r\end{array}\right)}}\\ & \text{dengan}\left(\begin{array}{c}n\\ r\end{array}\right)={\displaystyle \frac{n!}{r!(n-r)!}\text{adalah}....}\\ & \begin{array}{l}\\ \text{a}.& {\displaystyle 2^{n+1}}& & & \text{d}.& {\displaystyle 3^n}\\ \\ \text{b}.& {\displaystyle n{2}^{n-1}}& \text{c}.& {\displaystyle n{2}^n}& \text{e}.& {\displaystyle 3^{n+1}}\end{array}\\ \\ & \text{Jawab}:\\ & \begin{array}{rl}{\displaystyle \sum _{r=1}^{n}r{\displaystyle \left(\begin{array}{c}n\\ r\end{array}\right)}}& ={\displaystyle \sum _{r=1}^{n}r{\displaystyle \frac{n!}{r!(n-r)!}}}\\ & ={\displaystyle \sum _{r=1}^{n}r{\displaystyle \frac{n(n-1)!}{r(r-1)!(n-r)!}}}\\ & =n{\displaystyle \sum _{r=1}^n{\displaystyle \frac{(n-1)!}{(r-1)!(n-r)!}}}\\ & =n{\displaystyle \sum _{r=1}^n{\displaystyle \frac{(n-1)!}{(r-1)!((n-1)-(r-1))!}}}\\ & ={\displaystyle \sum _{r=1}^n\left(\begin{array}{c}n-1\\ r-1\end{array}\right)}\\ & =n{.2}^{r-1}\end{array}\end{array}$

$\begin{array}{ll}15.& \text{Banyaknya diagonal segi 6 adalah}....\\ & \begin{array}{l}\\ \text{a}.& {\displaystyle 15}& & & \text{d}.& {\displaystyle 9}\\ \\ \text{b}.& {\displaystyle 14}& \text{c}.& {\displaystyle 10}& \text{e}.& {\displaystyle 6}\end{array}\\ \\ & \text{Jawab}:\\ & \begin{array}{rl}& \text{Banyak diagonal segi}-n\text{adalah}:\\ & C(n,2)-n.\text{Jika seperti soal dengan}\\ & n=6,\text{maka}\\ & C(6,2)={\displaystyle \frac{6!}{2!\times 4!}=\frac{6\times 5\times \text{⧸}4!}{2\times 1\times \text{⧸}4!}=15}\\ & \text{Sehingga}\\ & C(6,2)-6=15-6=9\end{array}\end{array}$

Contoh Soal 2 Kaidah Pencacahan

 $\begin{array}{ll}6.& \text{Banyaknya cara milih 4 orang dari 10 orang }\\ & \text{anggota jika salah seorang di antaranya}\\ & \text{selalu terpilih adalah}....\\ & \begin{array}{l}\\ \text{a}.& {\displaystyle 72}& & & \text{d}.& {\displaystyle 504}\\ \\ \text{b}.& {\displaystyle 84}& \text{c}.& {\displaystyle 252}& \text{e}.& {\displaystyle 3024}\end{array}\\ \\ & \text{Jawab}:\\ & \begin{array}{rl}& \text{Cara memilih}=\text{Kombinasi}=C(10-1,4-1)\\ & \text{karena 1 orang di antaranya selalu ada/terpilih}\\ & =C(9,3)\\ & =(\genfrac{}{}{0ex}{}{9}{3})\\ & ={\displaystyle \frac{9!}{3!\times (9-3)!}}\\ & ={\displaystyle \frac{9\times 8\times 7\times \text{⧸}6!}{3\times 2\times \times \text{⧸}6!}}\\ & ={\displaystyle \frac{9.8.7}{3.2}}\\ & =84\end{array}\end{array}$

$\begin{array}{ll}7.& \text{Banyaknya cara menyusun huruf-huruf dari}\\ & \text{kata "SEMARANG" adalah}....\\ & \begin{array}{l}\\ \text{a}.& {\displaystyle 1680}& & & \text{d}.& {\displaystyle 20320}\\ \\ \text{b}.& {\displaystyle 6720}& \text{c}.& {\displaystyle 20160}& \text{e}.& {\displaystyle 40320}\end{array}\\ \\ & \text{Jawab}:\\ & \begin{array}{rl}& \text{Penyelesaian di atas dapat diselesaikan}\\ & \text{baik dengan permutasi maupun kombinasi}\\ & \text{Susunan huruf berbeda yang diambil dari}\\ & \text{kata "SEMARANG" adalah}:\\ & \{\begin{array}{ll}\text{S}& =1\\ \text{E}& =1\\ \text{M}& =1\\ \text{A}& =2\\ \text{R}& =1\\ \text{N}& =1\\ \text{G}& =1\end{array}\\ & \text{Jumlah huruf ada 8 buah}\\ & \text{Dengan cara permutasi}\\ & \begin{array}{rl}P(n;n_1,n_2,n_2,...,n_r)& ={\displaystyle \frac{n!}{n_1!.n_2!.n_3!...n_r!}}\\ P(8;1,1,1,2,1,1,1)& ={\displaystyle \frac{8!}{1!.1!.1!.2!.1!.1!.1!}}\\ & ={\displaystyle \frac{40.320}{2}}\\ & =20.160\end{array}\\ & \text{Dengan cara kombinasi}\\ & \begin{array}{rl}C(n;...)& ={\displaystyle \frac{n!}{n_1!.n_2!.n_3!...n_r!}}\\ C(8;...)& ={\displaystyle (\genfrac{}{}{0ex}{}{8}{1}).(\genfrac{}{}{0ex}{}{7}{1}).(\genfrac{}{}{0ex}{}{6}{1}).(\genfrac{}{}{0ex}{}{5}{2}).(\genfrac{}{}{0ex}{}{3}{1}).(\genfrac{}{}{0ex}{}{2}{1})}\\ & ={\displaystyle 8.7.6.{\displaystyle \frac{5.4}{2}.3.2}}\\ & ={\displaystyle \frac{40.320}{2}}\\ & =20.160\end{array}\end{array}\end{array}$

$\begin{array}{ll}8.& \text{Jumlah susunan dari sebelas huruf}\\ & \mathbf{\text{MISSISSIPPI}}\\ & \text{Banyak susunan berbeda dari semua}\\ & \text{huruf di atas jika keempat huruf}\mathbf{\text{I}}\\ & \text{selalu tampil berdampingan}\\ & \begin{array}{l}\\ \text{a}.& {\displaystyle {\displaystyle \frac{9!}{2!4!}}}& & & \text{d}.& {\displaystyle \frac{6!}{2!4!}}\\ \\ \text{b}.& {\displaystyle \frac{8!}{2!4!}}& \text{c}.& {\displaystyle \frac{7!}{2!4!}}& \text{e}.& {\displaystyle \frac{5!}{2!4!}}\end{array}\\ \\ & \text{National University of Singapore}\\ & \text{Sample Test Entrance Examination}\\ \\ & \text{Jawab}:\\ & \begin{array}{rl}& \text{Pandang semua huruf}I\text{dianggap 1}\\ & \text{maka perhitungannnya}\\ & P(8;1,1,4,2)={\displaystyle \frac{8!}{2!4!}}\end{array}\end{array}$

$\begin{array}{ll}9.& \text{Nilai dari}P(4,2)\times P(5,3)=....\\ & \begin{array}{l}\\ \text{a}.& {\displaystyle 12}& & & \text{d}.& {\displaystyle 480}\\ \\ \text{b}.& {\displaystyle 48}& \text{c}.& {\displaystyle 60}& \text{e}.& {\displaystyle 720}\end{array}\\ \\ & \text{Jawab}:\\ & \begin{array}{rl}& P(4,2)\times P(5,3)\\ & ={\displaystyle \frac{4!}{(4-2)!}\times \frac{5!}{(5-3)!}}\\ & ={\displaystyle \frac{4!}{2!}\times \frac{5!}{2!}}\\ & ={\displaystyle \frac{4.3.\text{⧸}2!}{\text{⧸}2!}\times \frac{5.4.3.\text{⧸}2!}{\text{⧸}2!}}\\ & =720\end{array}\end{array}$

$\begin{array}{ll}10.& \text{Nilai}n\text{jika}P(n+1,3)=P(n,4)\\ & \text{adalah}....\\ & \begin{array}{l}\\ \text{a}.& {\displaystyle 3}& & & \text{d}.& {\displaystyle 6}\\ \\ \text{b}.& {\displaystyle 4}& \text{c}.& {\displaystyle 5}& \text{e}.& {\displaystyle 7}\end{array}\\ \\ & \text{Jawab}:\\ & \begin{array}{rl}P(n+1,3)& =P(n,4)\\ {\displaystyle \frac{(n+1)!}{((n+1)-3)!}}& ={\displaystyle \frac{n!}{(n-4)!}}\\ {\displaystyle \frac{(n+1)!}{(n-2)!}}& =\frac{n!}{(n-4)!}\\ {\displaystyle \frac{(n+1).\text{⧸}n!}{(n-2).(n-3).\text{⧸}(n-4)!}}& ={\displaystyle \frac{\text{⧸}n!}{\text{⧸}(n-4)!}}\\ {\displaystyle \frac{n+1}{n^2-5n+6}}& =1\\ n^2-5n+6& =n+1\\ n^2-6n+5& =0\\ (n-1)(n-5)& =0\\ n=1\text{atau}n=5& \end{array}\end{array}$

Contoh Soal 1 Kaidah Pencacahan

 $\begin{array}{ll}1.& \text{Nilai dari}{\displaystyle \frac{1}{14!}-\frac{10}{15!}+\frac{4}{16!}}\\ & \begin{array}{l}\\ \text{a}.& {\displaystyle \frac{114}{16!}}& & & \text{d}.& {\displaystyle \frac{9}{16!}}\\ \\ \text{b}.& {\displaystyle \frac{108}{16!}}& \text{c}.& {\displaystyle \frac{84}{16!}}& \text{e}.& {\displaystyle \frac{4}{16!}}\end{array}\\ \\ & \text{Jawab}:\\ & \begin{array}{rl}{\displaystyle \frac{1}{14!}}& -{\displaystyle \frac{10}{15!}+\frac{4}{16!}}\\ & ={\displaystyle \frac{15\times 16}{14!\times 15\times 16}-\frac{10\times 16}{15!\times 16}+\frac{4}{16!}}\\ & ={\displaystyle \frac{240}{16!}-\frac{160}{16!}+\frac{4}{16!}}\\ & ={\displaystyle \frac{84}{16!}}\end{array}\end{array}$.

$\begin{array}{ll}2.& {\displaystyle \frac{(n+1)!}{(n-1)!}=....}\\ & \begin{array}{l}\\ \text{a}.& {\displaystyle n}& & & \text{d}.& {\displaystyle n^2-n}\\ \\ \text{b}.& {\displaystyle n-1}& \text{c}.& {\displaystyle n+1}& \text{e}.& {\displaystyle n^2+n}\end{array}\\ \\ & \text{Jawab}:\\ & \begin{array}{rl}{\displaystyle \frac{(n+1)!}{(n-1)!}}& ={\displaystyle \frac{(n+1)n(n-1)!}{(n-1)!}}\\ & =(n+1)n\\ & =n^2+1\end{array}\end{array}$.

$\begin{array}{ll}3.& \text{Permutasi 4 unsur dari 11 unsur}\\ & \text{adalah}....\\ & \begin{array}{l}\\ \text{a}.& {\displaystyle 7980}& & & \text{d}.& {\displaystyle 7290}\\ \\ \text{b}.& {\displaystyle 7920}& \text{c}.& {\displaystyle 7820}& \text{e}.& {\displaystyle 7280}\end{array}\\ \\ & \text{Jawab}:\\ & \begin{array}{rl}P(n,r)& ={\displaystyle \frac{n!}{(n-r)!}}\\ P(11,4)& ={\displaystyle \frac{11!}{(11-4)!}}\\ & ={\displaystyle \frac{11!}{7!}=\frac{11\times 10\times 9\times 8\times \text{⧸}7!}{\text{⧸}7!}}\\ & =7920\end{array}\end{array}$

$\begin{array}{ll}4.& \text{Empat siswa dan dua siswi akan duduk}\\ & \text{berdampingan. Apabila siswi selalu duduk}\\ & \text{paling pinggir, banyak cara mereka duduk}\\ & \text{adalah}....\\ & \begin{array}{l}\\ \text{a}.& {\displaystyle 24}& & & \text{d}.& {\displaystyle 64}\\ \text{b}.& {\displaystyle 48}& \text{c}.& {\displaystyle 56}& \text{e}.& {\displaystyle 72}\end{array}\\ \\ & \text{Jawab}:\\ & \begin{array}{rl}& \text{Total ada 6 anak; 4 siswa, 2 siswi}\\ & \text{Karena ini posisi orang, maka dan semuanya}\\ & \text{tidak identik, maka dapat diurutkan}\\ & \text{Sehingga rumus yang dipergunakan adalah}\\ & \text{permutasi, yaitu}:\\ & \text{Perhatikan posisi mereka}\\ & \mathbf{\text{Posisi pertama}}\\ & \begin{array}{|cccccc|}\hline (1)& (2)& (3)& (4)& (5)& (6)\\ \text{A}& ◻& ◻& ◻& ◻& \text{B}\\ \hline\end{array}\\ & =P(1,1)\times P(4,4)\times P(1,1)=24\\ & \mathbf{\text{Posisi kedua}}\\ & \begin{array}{|cccccc|}\hline (1)& (2)& (3)& (4)& (5)& (6)\\ \text{B}& ◻& ◻& ◻& ◻& \text{A}\\ \hline\end{array}\\ & =P(1,1)\times P(4,4)\times P(1,1)=24\\ & \text{Total}=24+24=48\end{array}\end{array}$

$\begin{array}{ll}5.& \text{Jika}P(7,r)=210,\text{maka nilai}r\\ & \begin{array}{l}\\ \text{a}.& {\displaystyle 2}& & & \text{d}.& {\displaystyle 5}\\ \\ \text{b}.& {\displaystyle 3}& \text{c}.& {\displaystyle 4}& \text{e}.& {\displaystyle 6}\end{array}\\ \\ & \text{Jawab}:\\ & \begin{array}{rl}P(7,r)& ={\displaystyle \frac{7!}{(7-r)!}}\\ 210& ={\displaystyle \frac{7!}{(7-r!)}}\\ (7-r)!& ={\displaystyle \frac{7!}{210}=\frac{7\times 6\times 5\times 4\times 3\times 2\times 1}{7\times 5\times 3\times 2\times 1}}\\ (7-r)!& =6.4=24\\ (7-r)!& =4!\\ 7-r& =4\\ r& =7-4\\ r& =3\end{array}\end{array}$

Pengujian Hipotesis

D. Uji Hipotesis

D. 1 Pengertian

Dalam suatu penyelidikan berkaitan suatu permasalahan untuk penarikan suatu kesimpulan diperlukan adanya dugaan atau dalam bahasa matematika dinamakan istilah hipotesis. Hipotesis berasal dari bahasa Yunani Hupo yang berarti sementara dan Thesis yang berarti pernyataan atau dugaan. Sehingga secara bahasa memiliki arti dugaan sementara. Selanjutnya, karena hipotesis ini masih berupa jawaban sementara, maka hipotesis ini harus diuji kebenaranya dan prosesnya dinamakan uji hipotesis. Uji hipotesis yang dibahasa di sini adalah pengujian berkaitan dengan rata-rata  $\mu$ pada sebuah sampel. Jika hasil yang didapatkan dalam penelitian nantinya, jauh berbeda dengan yang diharapkan berdasarkan hipotesis, maka hipotesis ditolak, demikian sebaliknya jika sesuai, maka hipotesis diterima.

D. 2 Jenis-Jenis Hipotesis

Ada dua jenis hipotetsis yaitu:

• Hipotesis nol ($H_0$) yang terkandung makna tidak memiliki perbedaan
• Hipotesis alternatif  ($H_1$) dengan pengertian terdapat tidak sama atau ada perbedaan.

D. 3 Langkah-Langkah Pengujian Hipotesis.

Berikut prosedur pengujian hipotesis

$.\begin{array}{rl}& \text{Merumuskan}{H}_0\text{dan}{H}_1\\ & \Downarrow \\ & \text{Menentukan daerah kritis}\\ & (\text{taraf signifikansi/kepercayaan})\\ & \Downarrow \\ & \text{Menentukan nilai statistik uji}\\ & \Downarrow \\ & \text{Menentukan keputusan uji}\\ & \Downarrow \\ & \text{Penarikan kesimpulan}\end{array}$.

D.4 Bentuk Pengujian Hipotesis

Ada 3 macam, yaitu, uji dua pihak, uji pihak kanan, dan uji pihak kiri.

$\text{CONTOH SOAL}$.

Peluang Kejadian Majmuk (Lanjutan Materi 3 Kelas XII Matematika Wajib)

 $\begin{array}{rl}\text{E. 4}.& \text{Peluang Kejadian Saling Lepas}\end{array}$

Kasus ini terjadi jika dua kejadian tidak mungkin secara bersamaan. Jika kejadian A dan kejadian B saling lepas seperti ini maka   $(A\cap B)=\varnothing$. Untuk penentukan besar peluangnya adalah sebagai berikut

$\begin{array}{rl}P(A\cup B)& =P(A)+P(B)\\ \\ \mathbf{\text{Keterang}}& \mathbf{\text{an}}:\\ P(A\cup B)& =\text{Peluang kejadian A atau B}\\ P(A)& =\text{Peluang kejadian A}\\ P(B)& =\text{Peluang kejadian B}\end{array}$

$\begin{array}{rl}\text{E. 5}.& \text{Peluang Kejadian Tidak Saling Lepas}\end{array}$

Kasus ini terjadi jika dua kejadian dapat terjadi secara bersamaan. 

$\begin{array}{rl}P(A\cup B)& =P(A)+P(B)-P(A\cap B)\\ \\ \mathbf{\text{Keterang}}& \mathbf{\text{an}}:\\ P(A\cup B)& =\text{Peluang kejadian A atau B}\\ P(A\cap B)& =\text{Peluang kejadian A dan B}\\ P(A)& =\text{Peluang kejadian A}\\ P(B)& =\text{Peluang kejadian B}\end{array}$

$\text{CONTOH SOAL}$

$\begin{array}{l}\\ 1.& \text{Sebuah dadu dilempar, Peluang }\\ & \text{munculnya mata dadu 5 atau 6 adalah}\\ \\ & \mathbf{\text{Jawab}}:\\ & \begin{array}{rl}A& =\text{Kejadian muncul mata dadu 5}=\left\{5\right\}\\ & n(A)=1\\ B& =\text{Kejadian muncul mata dadu 6}=\left\{6\right\}\\ & n(B)=1\\ S& =\text{Semua mata dadu}=\{1,2,3,4,5,6\}\\ P& (A\cup B)=P(A)+P(B)\\ P& (A\cup B)={\displaystyle \frac{n(A)}{n(S)}+\frac{n(B)}{n(S)}={\displaystyle \frac{1}{6}+\frac{1}{6}=\frac{2}{6}=\frac{1}{3}}}\end{array}\\ & \begin{array}{rl}& \mathbf{\text{Atau}}\\ & \text{Soal di atas dapat dikerjakan dengan rumus}\\ & P(A\cup B)=P(A)+P(B)-P(A\cap B)\\ & \text{Karena antara kejadian muncul mata}\\ & \text{dadu 5 dan mat dadu 6 tidak ada irisannya}\\ & \text{maka irisannya haruslah bernilai}0\text{atau}\\ & \text{nilai}n(A\cap B)=0,\text{maka}\\ & P(A\cup B)=P(A)+P(B)-P(A\cap B)\\ & P(A\cup B)={\displaystyle \frac{n(A)}{n(S)}+\frac{n(B)}{n(S)}-\frac{n(A\cap B)}{n(S)}}\\ & ={\displaystyle \frac{1}{6}+\frac{1}{6}-\frac{0}{6}=\frac{2}{6}={\displaystyle \frac{1}{3}}}\end{array}\end{array}$

$\begin{array}{l}\\ 2.& \text{Dua buah dadu dilempar sekali. Peluang}\\ & \text{munculnya mata dadu berjumlah 3 atau 10}\\ \\ & \mathbf{\text{Jawab}}:\\ & \begin{array}{rl}& \text{Perhatikan tabel berikut}\\ & \begin{array}{|ccccccc|}\hline ◻& 1& 2& 3& 4& 5& 6\\ 1& (1,1)& (1,2)& (1,3)& (1,4)& (1,5)& (1,6)\\ 2& (2,1)& (2,2)& (2,3)& (2,4)& (2,5)& (2,6)\\ 3& (3,1)& (3,2)& (3,3)& (3,4)& (3,5)& (3,6)\\ 4& (4,1)& (4,2)& (4,3)& (4,4)& (4,5)& (4,6)\\ 5& (5,1)& (5,2)& (5,3)& (5,4)& (5,5)& (5,6)\\ 6& (6,1)& (6,2)& (6,3)& (6,4)& (6,5)& (6,6)\\ \hline\end{array}\\ & \text{Misal}\\ & A=\text{Kejadian jumlah mata dadu 3}\\ & \Rightarrow n(A)=2\\ & B=\text{Kejadian jumlah mata dadu 10}\\ & \Rightarrow n(B)=3\\ & S=\text{Semua mata dadu}\Rightarrow n(S)=36\\ & \text{Gunakan rumus}\\ & P(A\cup B)=P(A)+P(B)-P(A\cap B)\\ & P(A\cup B)={\displaystyle \frac{n(A)}{n(S)}+\frac{n(A)}{n(S)}-\frac{n(A\cap B)}{n(S)}}\\ & ={\displaystyle \frac{2}{36}+\frac{3}{36}-\frac{0}{36}}\\ & ={\displaystyle \frac{5}{36}}\end{array}\end{array}$

$\begin{array}{l}\\ 3.& \text{Seratus kartu yang diberi diberi nomor urut}\\ & \text{1 sampai 100 diambil sebuah saja}.\\ & \text{Tentukanlah peluang}\\ & \text{a}.\text{muncul kelipatan 4}\\ & \text{b}.\text{muncul kelipatan 6}\\ & \text{c}.\text{muncul kelipatan 4 atau 6}\\ \\ & \mathbf{\text{Jawab}}:\\ & \begin{array}{rl}& S=\{1,2,3,4,\cdots ,100\}\Rightarrow n(S)=100\\ & A=\text{Kejadian muncul kelipatan 4}\\ & \iff A=\{4\times 1,4\times 2,\cdots ,4\times 25\}\Rightarrow n(A)=25\\ & B=\text{Kejadian muncul kelipatan 6}\\ & \iff B=\{6\times 1,6\times 2,\cdots ,6\times 16\}\Rightarrow n(B)=16\\ & \text{Tentunya kesamaan antara kejadian A dan B}\\ & \text{dan ini dilambangkan dengan}(A\cap B),\text{maka}\\ & (A\cap B)=\{12\times 1,12\times 2,\cdots ,12\times 8\}\\ & \Rightarrow n(A\cap B)=8\\ & \text{Selanjutnya adalah}:\\ & P(A\cup B)=P(A)+P(B)-P(A\cap B)\\ & ={\displaystyle \frac{n(A)}{n(S)}+\frac{n(B)}{n(S)}-\frac{n(A\cap B)}{n(S)}}\\ & ={\displaystyle \frac{25}{100}+\frac{16}{100}-\frac{8}{100}}\\ & ={\displaystyle \frac{33}{100}}\end{array}\end{array}$

DAFTAR PUSTAKA

1. Kanginan, M., Terzalgi, Y. 2014. Matematika untuk SMA/MA Kelas XI (Wajib). Bandung: SEWU.
2. Kartini, Suprapto, Subandi, Setiyadi, U. 2005. Matematika untuk SMA dan MA Program Studi Ilmu Alam Kelas XI. Klaten: INTAN PARIWARA.
3. Sharma, dkk. 2017. Jelajah Matematika 3 SMA Kelas XII Program Wajib. Jakarta: YUDHISTIRA.

Peluang Kejadian Majmuk (Lanjutan Materi 2 Kelas XII Matematika Wajib)

 $\begin{array}{rl}\text{E}.& \text{Peluang Kejadian Majmuk}\end{array}$

$\begin{array}{rl}\text{E. 1}.& \text{Peluang Komplemen Suatu Kejadian}\end{array}$

Komplemen suatu kejadian A misalnya adalah kejadian tidak terjadinya A dan dinotasikan dengan  $A^{\prime }\text{atau}{A}^c$

Selanjutnya peluang kejadian bukan A dituliskan dengan  $P(A^{\prime })\mathrm{atau}P(A^c)$ dan 

$P(A^{\prime })=1-P(A)\mathrm{atau}P(A^{\prime })+P(A)=1$

$\text{CONTOH SOAL}$

$\begin{array}{l}\\ 1.& \text{Dua puluh kartu diberi angka}1,2,3,\cdots ,20\\ & \text{Setelah semuanya bernomor kemudian kartu}\\ & \text{tersebut dikocok. Jika sebuah kartu diambil}\\ & \text{secara acak, maka peluang bahwa kartu yang}\\ & \text{termabil bukan angka prima}\\ 2.& \text{Jika sebuah keluarga merencanakan kelahiran}\\ & \text{dengan 4 anak anak. Peluang paling sedikit}\\ & \text{memiliki satu anak laki-laki}\\ \\ & \text{Jawab}:\\ 1.& A=\text{Kejadian nomor prima}\\ & A=\{2,3,5,7,11,13,17,19\}\Rightarrow n(A)=8\\ & \text{Peluang terambilnya sebuah kartu prima}:\\ & P(A)={\displaystyle \frac{C(8,1)}{C(20,1)}=\frac{8}{20}}\\ & \text{Sehingga peluang termabilnya 1 kartu}\\ & \text{bukan prima adalah}:\\ & P(A^{\prime })=1-P(A)=1-{\displaystyle \frac{8}{20}=\frac{12}{20}=\frac{3}{5}}\\ 2.& S=\text{Kejadian total kelahiran 4 anak}\\ & n(S)=C(2,1)\times C(2,1)\times C(2,1)\times C(2,1)\\ & \iff n(S)=2\times 2\times 2\times 2=16\text{susunan anak}\\ & =\{LLLL,LLLP,LLPL,LLPP,\cdots ,PPPP\}\\ \\ & \text{Jika}B=\text{Kejadian Kelahiran tanpa anak laki-laki}\\ & \text{atau kejadian semuanya perempuan=PPPP=1}\\ & \text{hanya akan terjadi 1 dari 16},\\ & \text{maka}n(B)=1.\text{Sehingga}\\ & P(B)={\displaystyle \frac{n(B)}{n(S)}=\frac{1}{16}}\\ & \text{Peluang kejadian kelahiran tanpa anak}\\ & \text{perempuan adalah}:\\ & P(B^{\prime })=1-P(B)=1-{\displaystyle \frac{1}{16}=\frac{15}{16}}\\ \\ & \text{Atau dengan cara langsung pun bisa sebenarnya}\\ & B^{\prime }=\text{Kejadian lahir minimal satu laki-laki}\\ & \text{maka}n(B^{\prime })=15\\ & P(B^{\prime })={\displaystyle \frac{n(B^{\prime })}{n(S)}=\frac{15}{16}}\end{array}$

$\begin{array}{rl}\text{E. 2}.& \text{Peluang Dua Kejadian Saling Bebas}\end{array}$

Dua kejadian dianggap saling bebas jika munculnya kejadian yang pertama tidak mempengaruhi peluang munculnya kejadian yang kedua.

$\begin{array}{rl}P(A\cap B)& =P(A)\times P(B)\\ \\ \mathbf{\text{Keteran}}& \mathbf{\text{gan}}:\\ P(A\cap B)& =\text{Peluang kejadian A dan B}\\ P(A)& =\text{Peluang kejadian A}\\ P(B)& =\text{Peluang kejadian B}\end{array}$

$\begin{array}{rl}\text{E. 3}.& \text{Peluang Dua Kejadian Tidak Saling Bebas}\end{array}$

Dua kejadian disebut dua kejadian tidak saling bebas jika munculnya kejadian pertama akan mempengaruhi peluang munculnya kejadian yang kedua, demikian sebaliknya. Selanjutnya kasus ini dinamakan peluang dua kejadian bersyarat

$\begin{array}{r}\{\begin{array}{ll}\bullet P(A|B)={\displaystyle \frac{P(A\cap B)}{P(B)}}& ,\text{dengan}P(B)\ne 0\\ \text{Peluang kejadian B}& \text{yang pertama terjadi}\\ \\ \bullet P(B|A)={\displaystyle \frac{P(A\cap B)}{P(A)}}& ,\text{dengan}P(A)\ne 0\\ \text{Peluang kejadian A}& \text{yang pertama terjadi}\end{array}\end{array}$

$\text{CONTOH SOAL}$

$\begin{array}{l}\\ 3.& \text{Sebuah kotak berisi 5 bola merah dan 6 bola}\\ & \text{biru. Dari kotak tersebut diambil bola satu}\\ & \text{persatu}\\ & \text{a}.\text{Tentukanlah peluang Jika bola pertama}\\ & \text{terambil merah lalu dikembalikan lalu}\\ & \text{terambil biru}\\ & \text{b}.\text{Tentukan peluang jika tanpa dikembalikan}\\ & \text{pada kasus 3.a di atas}\\ \\ & \text{Jawab}:\\ & \begin{array}{rl}& \text{Misalkan}\\ & X=\text{kejadian terambil merah pada proses 1}\\ & Y=\text{kejadian terambil biru pada proses 2}\\ & \text{a}.P(X\cap Y)=P(X)\times P(Y)\\ & ={\displaystyle \frac{C(5,1)}{C(11,1)}\times \frac{C(6,1)}{C(11,1)}={\displaystyle \frac{5\times 6}{11\times 11}=\frac{30}{121}}}\\ & \text{b}.P(X\cap Y)=P(X)\times P(Y)=P(X)\times P(Y|X)\\ & ={\displaystyle \frac{C(5,1)}{C(11,1)}\times \frac{C(6,1)}{C(10,1)}={\displaystyle \frac{5\times 6}{11\times 10}=\frac{3}{11}}}\\ & \text{ingat saat tanpa pengembalian, maka}\\ & \text{jumlah bola total berkurang 1}\end{array}\end{array}$

Materi Lanjutan Distribusi Normal (Matematika Peminatan Kelas XII)

C. Transformasi Suatu Variabel Random Berdistribusi Normal

Dalam menentukan luas suatu variabel berdistribusi normal ke dalam variabel random berdistribusi normal baku dengan jalan mentransformasikannya

Adapun luasnya sama yaitu: 

$\begin{array}{rl}& \begin{array}{rl}\text{P}({\text{x}}_1<X<{\text{x}}_2)& ={\displaystyle {\int }_{{\text{x}}_1}^{{\text{x}}_2}{\displaystyle \frac{1}{\sigma \sqrt{2\pi }}{e}^{-\frac{1}{2}{\left({\displaystyle \frac{\text{x}-\mu }{\sigma }}\right)}^2}d\text{x}}}\\ & ={\displaystyle {\int }_{{\text{z}}_1}^{{\text{z}}_2}{\displaystyle \frac{1}{\sqrt{2\pi }}{e}^{-\frac{1}{2}{\text{Z}}^2}d\text{x}}}\\ & =\text{P}({\text{z}}_1<\text{Z}<{\text{z}}_2)\end{array}\\ \\ & \text{Luas di atas adalah hasil tranformasi}\\ & \text{variabel acak X}\sim \text{N}(\mu ,\sigma )\text{ke}\text{Z}\sim \text{N}(0,1)\\ & \text{dengan}\text{z}={\displaystyle \frac{\text{x}-\mu }{\sigma }}\end{array}$.

$\text{CONTOH SOAL}$.

$\begin{array}{l}\\ 1.& \text{Diketahui variabel acak Z berdistribusi}\\ & \text{normal}\text{N}(0,1)\text{dan X berdistribusi}\\ & \text{normal N}(18,5).\text{Tentukanlah besar}\\ & \text{peluang berikut}\\ & \text{a}.\text{P}(\text{Z}>0,68)\\ & \text{b}.\text{P}(\text{X}<20)\\ & \text{c}.\text{P}(0,36<\text{Z}<1,42)\\ & \text{d}.\text{P}(17<\text{X}<18,5)\\ \\ & \mathbf{\text{Jawab}}:\\ & \begin{array}{rl}\text{a}.\text{P}(\text{Z}>0,68)& =0,5-\text{P}(0<\text{Z}<0,68)\\ & =0,5-0,2517=0,2483\end{array}\\ & \begin{array}{rl}\text{b}.& \text{Transformasi}\text{x}=20,\text{dengan}\{\begin{array}{ll}\mu & =18\\ \sigma & =5\end{array}\\ & \text{z}={\displaystyle \frac{\text{x}-\mu }{\sigma }={\displaystyle \frac{20-18}{5}=0,4,\text{maka}}}\\ & \begin{array}{rl}\text{P}(\text{X}<20)& =\text{P}(\text{Z}<0,4)\\ & =0,5+\text{P}(0<\text{Z}<0,4)\\ & =0,5+0,1554=0,6554\end{array}\end{array}\\ & \begin{array}{rl}\text{c}.\text{P}(0,36<\text{Z}<1,42)& =\text{P}(0<\text{Z}<1,42)-\text{P}(0<\text{Z}<0,36)\\ & =0,4222-0,1406=0,2816\end{array}\\ & \begin{array}{rl}\text{d}.& \text{Transformasi}{\text{x}}_1=17,\text{dan}{\text{x}}_2=18,5\\ & \{\begin{array}{ll}\mu & =18\\ \sigma & =5\end{array}\\ & {\text{z}}_1={\displaystyle \frac{\text{x}-\mu }{\sigma }={\displaystyle \frac{17-18}{5}=-0,2,\text{dan}}}\\ & {\text{z}}_2={\displaystyle \frac{\text{x}-\mu }{\sigma }={\displaystyle \frac{18,5-18}{5}=0,1,\text{maka}}}\\ & \begin{array}{rl}\text{P}(17<\text{X}<18,5)& =\text{P}(-0,2<\text{Z}<0,1)\\ & =\text{P}(0<\text{Z}<0,2)+\text{P}(0<\text{Z}<0,1)\\ & =0,0793+0,0398=0,1191\end{array}\end{array}\end{array}$.

$\begin{array}{l}\\ 2.& \text{Diketahui variabel acak X berdistribusi}\\ & \text{normal memiliki rata-rata 16 dan simpangan}\\ & \text{baku}1,4.\text{Hitunglah besar peluang dari}\\ & \text{a}.\text{P}(\text{X}\le 18,8)\\ & \text{d}.\text{P}(12,1\le \text{X}\le 16,3)\\ \\ & \mathbf{\text{Jawab}}:\\ & \begin{array}{rl}\text{a}.& \text{Transformasi}\text{x}=18,8,\text{dengan}\{\begin{array}{ll}\mu & =16\\ \sigma & =1,4\end{array}\\ & \text{z}={\displaystyle \frac{\text{x}-\mu }{\sigma }={\displaystyle \frac{18,8-16}{1,4}=2,\text{maka}}}\\ & \begin{array}{rl}\text{P}(\text{X}<18,8)& =\text{P}(\text{Z}<2)\\ & =0,5+\text{P}(0<\text{Z}<2)\\ & =0,5+0,4772=0,9772\end{array}\end{array}\\ & \begin{array}{rl}\text{b}.& \text{Transformasi}{\text{x}}_1=12,1,\text{dan}{\text{x}}_2=16,3\\ & \{\begin{array}{ll}\mu & =16\\ \sigma & =1,4\end{array}\\ & {\text{z}}_1={\displaystyle \frac{\text{x}-\mu }{\sigma }={\displaystyle \frac{12,1-16}{1,4}=-2,79,\text{dan}}}\\ & {\text{z}}_2={\displaystyle \frac{\text{x}-\mu }{\sigma }={\displaystyle \frac{16,3-16}{1,4}=0,21,\text{maka}}}\\ & \begin{array}{rl}\text{P}(12,1<\text{X}<16,3)& =\text{P}(-2,79<\text{Z}<0,21)\\ & =\text{P}(0<\text{Z}<2,79)+\text{P}(0<\text{Z}<0,21)\\ & =0,4974+0,0832=0,5806\end{array}\end{array}\end{array}$ 

$\begin{array}{l}\\ 3.& \text{Di sebuah MA dengan 1000 siswa diperoleh data}\\ & \text{rata-rata berat bada siswanya}54\text{Kg}\text{dan simpangan}\\ & \text{baku}8\text{Kg}.\text{Jika data tersebut berdistribusi normal}\\ & \text{tentukan bsnysk siswa yang memiliki berat badan}\\ & \text{a}.\text{lebih dari 70 Kg}\\ & \text{b}.\text{antara 40 Kg sampai 50 Kg}\\ \\ & \mathbf{\text{Jawab}}:\\ & \begin{array}{rl}\text{a}.& \text{Transformasi}\text{x}=70,\text{dengan}\{\begin{array}{ll}\mu & =54\\ \sigma & =8\end{array}\\ & \text{z}={\displaystyle \frac{\text{x}-\mu }{\sigma }={\displaystyle \frac{70-54}{8}=\frac{16}{8}=2}}\\ & \text{Dari tabel diperoleh}\\ & P(0<\text{Z}<2)=0,4772,\text{selanjutnya}\\ & \begin{array}{rl}\text{P}(\text{X}>70)& =\text{P}(\text{Z}>2)\\ & =0,5-\text{P}(0<\text{Z}<2)\\ & =0,5-0,4772=0,0228\end{array}\\ & \text{Selanjutnya frekuensi harapan dalam hal ini}\\ & \begin{array}{rl}{f}_h(\text{X}>70)& =1000\times P(\text{X}>70)\\ & =1000\times 0,0228\\ & =22,8\\ & \approx 23\end{array}\\ & \text{Jadi, ada sebanyak 23 anak dengan berat badan}\\ & >\text{70 Kg}\end{array}\\ & \begin{array}{rl}\text{b}.& \text{Transformasi}{\text{x}}_1=40,\text{dan}{\text{x}}_2=50\\ & \{\begin{array}{ll}\mu & =54\\ \sigma & =8\end{array}\\ & {\text{z}}_1={\displaystyle \frac{\text{x}-\mu }{\sigma }={\displaystyle \frac{40-54}{8}=-1,75,\text{dan}}}\\ & {\text{z}}_2={\displaystyle \frac{\text{x}-\mu }{\sigma }={\displaystyle \frac{50-54}{8}=-0,5,\text{maka}}}\\ & \begin{array}{rl}\text{P}(40<\text{X}<50)& =\text{P}(-1,75<\text{Z}<-0,5)\\ & =\text{P}(0,5<\text{Z}<1,75)\\ & =\text{P}(0<\text{Z}<1,75)-\text{P}(0<\text{Z}<0,5)\\ & =0,4599-0,1915=0,2684\end{array}\\ & \text{Selanjutnya frekuensi harapan dalam hal ini}\\ & \begin{array}{rl}{f}_h(40<\text{X}<50)& =1000\times P(40<\text{X}<50)\\ & =1000\times 0,2684\\ & =268,4\\ & \approx 268\end{array}\\ & \text{Jadi, ada sebanyak 268 anak dengan berat badan}\\ & \text{antara 40 samapi 50 Kg}\end{array}\end{array}$.

$\begin{array}{l}\\ 4.& \text{Diketahui Nilai-nilai ujian penerimaan pegawai baru}\\ & \text{diperoleh mean 78 dan deviasi standarnya 6. Jika}\\ & \text{hanya}12,5\mathrm{\%}\text{calon yang akan diterima, maka nilai}\\ & \text{terendah yang lolos jika distribusinya normal}\\ \\ & \mathbf{\text{Jawab}}:\\ & \begin{array}{rl}& \text{Transformasi}\text{X}=\text{x}\\ & \text{dengan}\{\begin{array}{ll}\mu & =78\\ \sigma & =6,\text{dan}\\ \text{P}(\text{X}>x)& =12,5\mathrm{\%}=0,125\end{array}\\ & \begin{array}{rl}\text{P}(\text{X}>\text{x})& =\text{P}(\text{Z}>0)-\text{P}(0<\text{Z}<\text{z})\\ 0,125& =0,5-\text{P}(0<\text{Z}<\text{z})\\ \text{P}(0<\text{Z}<\text{z})& =0,5-0,125=0,375\end{array}\\ & \text{Dari tabel diperoleh}\text{P}(0<\text{Z}<\text{z})=0,375,\\ & \text{selanjutnya didapatkan nilai}z=1,15\\ & \text{Sehingga nilai terendah x yang diterima}\\ & \text{z}={\displaystyle \frac{\text{x}-\mu }{\sigma }}\\ & \iff \text{x}=\text{z}\sigma +\mu =(1,15).6+78=84,9\approx 85\end{array}\end{array}$.

DAFTAR PUSTAKA

1. Tasari, Aksin, N., Miyanto, Muklis. 2016. Matematika untuk SMA/MA Kelas XII Peminatan Matematika dan Ilmu-Ilmu Alam. Klaten: INTAN PARIWARA.

Contoh Soal 15 (Segitiga dan Ketaksamaan)

$\begin{array}{l}\\ 71.& (\mathbf{\text{IMO 2001}})\\ & \text{Diberikan}a,b\text{dan}c\text{bilangan real positif, tunjukkan}\\ & {\displaystyle \frac{a}{\sqrt{a^2+8bc}}+\frac{b}{\sqrt{b^2+8ca}}+\frac{c}{\sqrt{c^2+8ab}}\ge 1}\\ \\ & \mathbf{\text{Bukti}}:\\ & \begin{array}{rl}& \text{Dengan Ketaksamaan Holder pilih}\\ & {\lambda }_1={\lambda }_2={\lambda }_3={\displaystyle \frac{1}{3}\left({\displaystyle \sum _{i=1}^3{\lambda }_i=1}\right)\text{akan diperoleh}}\\ & {\left({\displaystyle \sum _{\text{siklik}}^{.}a(a^2+8bc)}\right)}^{\frac{1}{3}}{\left({\displaystyle \sum _{\text{siklik}}^{.}{\displaystyle \frac{a}{\sqrt{a^2+8bc}}}}\right)}^{\frac{1}{3}}{\left({\displaystyle \sum _{\text{siklik}}^{.}{\displaystyle \frac{a}{\sqrt{a^2+8bc}}}}\right)}^{\frac{1}{3}}\ge ({\displaystyle \sum _{\text{siklik}}^{.}a)}\\ & \text{dapat juga dituliskan lebih sederhana}\\ & {\left({\displaystyle \sum _{\text{siklik}}^{.}a(a^2+8bc)}\right)}^{\frac{1}{3}}{\left({\displaystyle \sum _{\text{siklik}}^{.}{\displaystyle \frac{a}{\sqrt{a^2+8bc}}}}\right)}^{\frac{2}{3}}\ge \left({\displaystyle \sum _{\text{siklik}}^{.}a}\right)\\ & \iff \left({\displaystyle \sum _{\text{siklik}}^{.}a(a^2+8bc)}\right){\left({\displaystyle \sum _{\text{siklik}}^{.}{\displaystyle \frac{a}{\sqrt{a^2+8bc}}}}\right)}^2\ge {\left({\displaystyle \sum _{\text{siklik}}^{.}a}\right)}^3\\ & \iff {\left({\displaystyle \sum _{\text{siklik}}^{.}{\displaystyle \frac{a}{\sqrt{a^2+8bc}}}}\right)}^2\ge {\displaystyle \frac{{\left({\displaystyle \sum _{\text{siklik}}^{.}a}\right)}^3}{\left({\displaystyle \sum _{\text{siklik}}^{.}a(a^2+8bc)}\right)}}\\ & \text{Perhatikan bahwa}\\ & \bullet {\left({\displaystyle \sum _{\text{siklik}}^{.}a}\right)}^3=(a+b+c{)}^3\ge a^3+b^3+c^3+24abc\\ & \bullet \left({\displaystyle \sum _{\text{siklik}}^{.}a(a^2+8bc)}\right)=a^3+b^3+c^3+24abc\\ & \text{maka}\\ & {\left({\displaystyle \sum _{\text{siklik}}^{.}{\displaystyle \frac{a}{\sqrt{a^2+8bc}}}}\right)}^2\ge {\displaystyle \frac{a^3+b^3+c^3+24abc}{a^3+b^3+c^3+24abc}=1}\\ & {\displaystyle \sum _{\text{siklik}}^{.}{\displaystyle \frac{a}{\sqrt{a^2+8bc}}\ge 1◼}}\end{array}\end{array}$.

$.\begin{array}{rl}& \mathbf{\text{Sebagai catatan}}\\ & (a+b+c{)}^3=a^3+b^3+c^3+3a^{2}b+3a{b}^2\\ & +3b^{2}c+3b{c}^2+3a^{2}c+3a{c}^2+6abc\\ & (a+b+c{)}^3-(a^3+b^3+c^3)=3a^{2}b+3a{b}^2\\ & +3b^{2}c+3b{c}^2+3a^{2}c+3a{c}^2+6abc\\ & \text{Dengan AM-GM akan diperoleh bentuk}\\ & (a+b+c{)}^3-(a^3+b^3+c^3)\\ & \ge 3(3\sqrt[3]{(abc{)}^3}+3\sqrt[3]{(abc{)}^3})+6abc\\ & \ge 3(3abc+3abc)+6abc=24abc\\ & \text{Sehingga}\\ & (a+b+c{)}^3\ge a^3+b^3+c^3+24abc\end{array}$.

$\begin{array}{l}\\ 72.& \text{Jika}a,b,\text{dan}c\text{bilangan real positif}\\ & \text{dengan}a+b+c=1\text{tunjukkan bahwa}\\ & {\displaystyle \frac{1}{a(b+c{)}^2}+\frac{1}{b(a+c{)}^2}+\frac{1}{c(a+b{)}^2}\ge {\displaystyle \frac{81}{4}}}\\ \\ & \mathbf{\text{Bukti}}:\\ & \begin{array}{rl}& \text{Dengan Ketaksamaan Holder pilih}\\ & {\lambda }_1={\lambda }_2={\lambda }_3={\lambda }_4={\displaystyle \frac{1}{4}\left({\displaystyle \sum _{i=1}^4{\lambda }_i=1}\right)\text{akan diperoleh}}\\ & (a+b+c)(b+c+a+c+a+b)(b+c+a+c+a+b)\left({\displaystyle \frac{1}{a(b+c{)}^2}+\frac{1}{b(a+c{)}^2}+\frac{1}{c(a+b{)}^2}}\right)\\ & \ge (1^{{.}^{\frac{1}{4}}}+1^{{.}^{\frac{1}{4}}}+1^{{.}^{\frac{1}{4}}}{)}^4=3^4=81\\ & \iff (1)(2)(2)\left({\displaystyle \frac{1}{a(b+c{)}^2}+\frac{1}{b(a+c{)}^2}+\frac{1}{c(a+b{)}^2}}\right)\ge 81\\ & \iff {\displaystyle \frac{1}{a(b+c{)}^2}+\frac{1}{b(a+c{)}^2}+\frac{1}{c(a+b{)}^2}\ge {\displaystyle \frac{81}{4}◼}}\end{array}\end{array}$.

$\begin{array}{l}\\ 73.& \text{Buktikan bahwa setiap bilangan real}\\ & \text{positif}a,b\text{dan}c\text{berlaku}\\ & a^2+b^2+c^2\ge ab+ac+bc\\ \\ & \mathbf{\text{Bukti}}\\ & \begin{array}{rl}& \text{Alternatif 1}\\ & \text{Perhatikan bahwa}(a-b{)}^2\ge 0\\ & (a-c{)}^2\ge 0,\text{dan}(b-c{)}^2\ge 0\\ & \text{adalah benar, maka}\\ & (a-b{)}^2=a^2-2ab+b^2\ge 0\\ & \iff a^2+b^2\ge 2ab.....(1)\\ & \text{Dengan cara yang kurang lebih sama}\\ & \text{akan didapatkan}\\ & \bullet a^2+c^2\ge 2ac.....(2)\\ & \bullet b^2+c^2\ge 2bc.....(1)\\ & \text{Jika ketaksamaan}(1),(2),\mathrm{\&}(3)\text{dijumlahkan}\\ & \text{akan didapatkan bentuk}\\ & 2a^2+2b^2+2c^2\ge 2ab+2ac+2bc\\ & \iff a^2+b^2+c^2\ge ab+ac+bc◼\\ & \text{Alternatif 2}\\ & \text{Dengan ketaksamaan}\mathbf{\text{Cauchy-Schwarz}}\\ & (a^2+b^2+c^2)(b^2+c^2+a^2)\ge (ab+bc+ca{)}^2\\ & \iff a^2+b^2+c^2\ge ab+ac+bc◼\\ & \text{Alternatif 3}\\ & \text{Untuk barisan}(a,b,c),\text{asumsikan}a\ge b\ge c\\ & \text{maka dengan}\mathbf{\text{Ketaksamaan Renata}}\text{diperoleh}\\ & a.a+b.b+c.c\ge ab+bc+ca\\ & a^2+b^2+c^2\ge ab+ac+bc◼\\ & \text{Alternatif 4}\\ & \text{Dengan}\mathbf{\text{Ketaksamaan Schur}}\text{saat}r=0,\\ & \text{yaitu}\\ & a^r(a-b)(a-c)+b^r(b-a)(b-c)+c^r(c-a)(c-b)\ge 0\\ & \iff a^0(a-b)(a-c)+b^0(b-a)(b-c)+c^0(c-a)(c-b)\ge 0\\ & \iff (a-b)(a-c)+(b-a)(b-c)+(c-a)(c-b)\ge 0\\ & \iff a^2+b^2+c^2-ab-ac-bc\ge 0\\ & \iff a^2+b^2+c^2\ge ab+ac+bc◼\end{array}\end{array}$.

$\begin{array}{l}\\ 74.& \mathbf{\text{(IMO 1964)}}\\ & \text{Jika}a,b,c\text{adalah panjang sisi-sisi segitiga}\\ & \text{tunjukkan bahwa}\\ & a^2(b+c-a)+b^2(a+c-b)+c^2(a+b-c)\le 3abc\\ \\ & \mathbf{\text{Bukti}}:\\ & \text{Dengan}\mathbf{\text{Ketaksamaan Schur}}\text{saat}r=1.\\ & a^3+b^3+c^3+3abc\ge ab(a+b)+bc(b+c)+ca(c+a)\\ & \iff 3abc\ge ab(a+b)-a^3+bc(b+c)-b^3+ca(c+a)-c^3\\ & \iff 3abc\ge a^{2}b+a{b}^2-c^3+b^{2}c+b{c}^2-b^3+c^{2}a+c{a}^2-c^3\\ & \iff 3abc\ge a^{2}b+c{a}^2-a^3+a{b}^2+b^{2}c-b^3+c^{2}a+b{c}^2-c^3\\ & \iff 3abc\ge a^2(b+c-a)+b^2(a+c-b)+c^2(a+b-c)\\ & \text{atau}\\ & \iff a^2(b+c-a)+b^2(a+c-b)+c^2(a+b-c)\le 3abc◼\end{array}$ .

$\begin{array}{l}\\ 75.& \mathbf{\text{(IMO 2000)}}\\ & \text{Jika}a,b,c\text{bilangan real positif dengan}\\ & abc=1,\text{tunjukkan bahwa}\\ & (a+1-{\displaystyle \frac{1}{b}})(b+1-{\displaystyle \frac{1}{c}})(c+1-{\displaystyle \frac{1}{a}})\le 1\\ \\ & \mathbf{\text{Bukti}}:\\ & \text{Misalkan}a={\displaystyle \frac{x}{y},b=\frac{y}{z},\text{dan}c={\displaystyle \frac{z}{x},\text{maka}}}\\ & \left({\displaystyle \frac{x}{y}+1-\frac{z}{y}}\right)\left({\displaystyle \frac{y}{z}+1-\frac{x}{z}}\right)\left({\displaystyle \frac{z}{x}+1-\frac{y}{x}}\right)\le 1\\ & \iff \left({\displaystyle \frac{x+y-z}{y}}\right)\left({\displaystyle \frac{y+z-x}{z}}\right)\left({\displaystyle \frac{z+x-y}{x}}\right)\le 1\\ & \iff (x+y-z)(y+z-x)(z+x-y)\le xyz,\text{atau}\\ & \iff xyz\ge (x+y-z)(y+z-x)(z+x-y)\\ & \text{Bentuk terakhir memenuhi bentuk kedua dari}\\ & \mathbf{\text{Ketaksamaan Schur}}\text{saat}r=1.\\ & \text{Jadi},\\ & (a+1-{\displaystyle \frac{1}{b}})(b+1-{\displaystyle \frac{1}{c}})(c+1-{\displaystyle \frac{1}{a}})\le 1◼\end{array}$.