12.10: Capítulo 10 Soluciones para ejercicios
- Page ID
- 112982
\( \newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} } \)
\( \newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}} \)
\( \newcommand{\id}{\mathrm{id}}\) \( \newcommand{\Span}{\mathrm{span}}\)
( \newcommand{\kernel}{\mathrm{null}\,}\) \( \newcommand{\range}{\mathrm{range}\,}\)
\( \newcommand{\RealPart}{\mathrm{Re}}\) \( \newcommand{\ImaginaryPart}{\mathrm{Im}}\)
\( \newcommand{\Argument}{\mathrm{Arg}}\) \( \newcommand{\norm}[1]{\| #1 \|}\)
\( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\)
\( \newcommand{\Span}{\mathrm{span}}\)
\( \newcommand{\id}{\mathrm{id}}\)
\( \newcommand{\Span}{\mathrm{span}}\)
\( \newcommand{\kernel}{\mathrm{null}\,}\)
\( \newcommand{\range}{\mathrm{range}\,}\)
\( \newcommand{\RealPart}{\mathrm{Re}}\)
\( \newcommand{\ImaginaryPart}{\mathrm{Im}}\)
\( \newcommand{\Argument}{\mathrm{Arg}}\)
\( \newcommand{\norm}[1]{\| #1 \|}\)
\( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\)
\( \newcommand{\Span}{\mathrm{span}}\) \( \newcommand{\AA}{\unicode[.8,0]{x212B}}\)
\( \newcommand{\vectorA}[1]{\vec{#1}} % arrow\)
\( \newcommand{\vectorAt}[1]{\vec{\text{#1}}} % arrow\)
\( \newcommand{\vectorB}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} } \)
\( \newcommand{\vectorC}[1]{\textbf{#1}} \)
\( \newcommand{\vectorD}[1]{\overrightarrow{#1}} \)
\( \newcommand{\vectorDt}[1]{\overrightarrow{\text{#1}}} \)
\( \newcommand{\vectE}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash{\mathbf {#1}}}} \)
\( \newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} } \)
\( \newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}} \)
\(\newcommand{\avec}{\mathbf a}\) \(\newcommand{\bvec}{\mathbf b}\) \(\newcommand{\cvec}{\mathbf c}\) \(\newcommand{\dvec}{\mathbf d}\) \(\newcommand{\dtil}{\widetilde{\mathbf d}}\) \(\newcommand{\evec}{\mathbf e}\) \(\newcommand{\fvec}{\mathbf f}\) \(\newcommand{\nvec}{\mathbf n}\) \(\newcommand{\pvec}{\mathbf p}\) \(\newcommand{\qvec}{\mathbf q}\) \(\newcommand{\svec}{\mathbf s}\) \(\newcommand{\tvec}{\mathbf t}\) \(\newcommand{\uvec}{\mathbf u}\) \(\newcommand{\vvec}{\mathbf v}\) \(\newcommand{\wvec}{\mathbf w}\) \(\newcommand{\xvec}{\mathbf x}\) \(\newcommand{\yvec}{\mathbf y}\) \(\newcommand{\zvec}{\mathbf z}\) \(\newcommand{\rvec}{\mathbf r}\) \(\newcommand{\mvec}{\mathbf m}\) \(\newcommand{\zerovec}{\mathbf 0}\) \(\newcommand{\onevec}{\mathbf 1}\) \(\newcommand{\real}{\mathbb R}\) \(\newcommand{\twovec}[2]{\left[\begin{array}{r}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\ctwovec}[2]{\left[\begin{array}{c}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\threevec}[3]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\cthreevec}[3]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\fourvec}[4]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\cfourvec}[4]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\fivevec}[5]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\cfivevec}[5]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\mattwo}[4]{\left[\begin{array}{rr}#1 \amp #2 \\ #3 \amp #4 \\ \end{array}\right]}\) \(\newcommand{\laspan}[1]{\text{Span}\{#1\}}\) \(\newcommand{\bcal}{\cal B}\) \(\newcommand{\ccal}{\cal C}\) \(\newcommand{\scal}{\cal S}\) \(\newcommand{\wcal}{\cal W}\) \(\newcommand{\ecal}{\cal E}\) \(\newcommand{\coords}[2]{\left\{#1\right\}_{#2}}\) \(\newcommand{\gray}[1]{\color{gray}{#1}}\) \(\newcommand{\lgray}[1]{\color{lightgray}{#1}}\) \(\newcommand{\rank}{\operatorname{rank}}\) \(\newcommand{\row}{\text{Row}}\) \(\newcommand{\col}{\text{Col}}\) \(\renewcommand{\row}{\text{Row}}\) \(\newcommand{\nul}{\text{Nul}}\) \(\newcommand{\var}{\text{Var}}\) \(\newcommand{\corr}{\text{corr}}\) \(\newcommand{\len}[1]{\left|#1\right|}\) \(\newcommand{\bbar}{\overline{\bvec}}\) \(\newcommand{\bhat}{\widehat{\bvec}}\) \(\newcommand{\bperp}{\bvec^\perp}\) \(\newcommand{\xhat}{\widehat{\xvec}}\) \(\newcommand{\vhat}{\widehat{\vvec}}\) \(\newcommand{\uhat}{\widehat{\uvec}}\) \(\newcommand{\what}{\widehat{\wvec}}\) \(\newcommand{\Sighat}{\widehat{\Sigma}}\) \(\newcommand{\lt}{<}\) \(\newcommand{\gt}{>}\) \(\newcommand{\amp}{&}\) \(\definecolor{fillinmathshade}{gray}{0.9}\)Capítulo 10: Probabilidad
1. a.\(\dfrac{6}{13}\)
b.\(\dfrac{2}{13}\)
3. \(\dfrac{150}{335} = 44.8 \%\)
5. \(\dfrac{1}{6}\)
7. \(\dfrac{26}{65}\)
9. \(\dfrac{3}{6} = \dfrac{1}{2} \)
11. \(\dfrac{4}{52} = \dfrac{1}{4}\)
13. \(1 - \dfrac{1}{12} = \dfrac{11}{12}\)
15. \(1 - \dfrac{25}{65} = \dfrac{40}{65}\)
17. \(\dfrac{1}{6} \cdot \dfrac{1}{6} = \dfrac{1}{36}\)
19. \(\dfrac{1}{6} \cdot \dfrac{3}{6} = \dfrac{3}{36} = \dfrac{1}{12} \)
21. \(\dfrac{17}{49} \cdot \dfrac{16}{48} = \dfrac{17}{49} \cdot \dfrac{1}{3} = \dfrac{17}{147} \)
23. a.\(\dfrac{4}{52} \cdot \dfrac{4}{52} = \dfrac{16}{2704} = \dfrac{1}{169} \)
b.\(\dfrac{4}{52} \cdot \dfrac{48}{52} = \dfrac{192}{2704} = \dfrac{12}{169} \)
c.\(\dfrac{48}{52} \cdot \dfrac{48}{52} = \dfrac{2304}{2704} = \dfrac{144}{169} \)
d.\(\dfrac{13}{52} \cdot \dfrac{13}{52} = \dfrac{169}{2704} = \dfrac{1}{16} \)
e.\(\dfrac{48}{52} \cdot \dfrac{39}{52} = \dfrac{1872}{2704} = \dfrac{117}{169} \)
25. \(\dfrac{4}{52} \cdot \dfrac{4}{51} = \dfrac{16}{2652}\)
27. a.\(\dfrac{11}{25} \cdot \dfrac{14}{24} = \dfrac{154}{600}\)
b.\(\dfrac{14}{25} \cdot \dfrac{11}{24} = \dfrac{154}{600}\)
c.\(\dfrac{11}{25} \cdot \dfrac{10}{24} = \dfrac{110}{600}\)
d.\(\dfrac{14}{25} \cdot \dfrac{13}{24} = \dfrac{182}{600}\)
e. no machos = dos hembras. Igual que la parte d.
29. \(P(\text{F} \text{ and } \text{A}) = \dfrac{10}{65} \)
31. \(P(\text{red} \text{ or } \text{odd}) = \dfrac{6}{14} + \dfrac{7}{14} - \dfrac{3}{14} = \dfrac{10}{14} \). O los canicas azules\(6\) rojos e\(4\) impares están\(10\) fuera de\(14\).
33. \(P(\text{F} \text{ or } \text{B}) = \dfrac{26}{65} + \dfrac{22}{65} - \dfrac{4}{65} = \dfrac{44}{65}\). O\(P(\text{F} \text{ or } \text{B}) = \dfrac{18+4+10+12}{65} = \dfrac{44}{65} \)
35. \(P(\text{King of Hearts or Queen}) = \dfrac{1}{52} + \dfrac{4}{52} = \dfrac{5}{52} \)
37. a.\(P(\text{even} | \text{red}) = \dfrac{2}{5} \)
b.\(P(\text{even} | \text{red}) = \dfrac{2}{6} \)
39. \(P(\text{Heads on second} | \text{Tails on first}) = \dfrac{1}{2}\). Son eventos independientes.
41. \(P(\text{speak French} | \text{female}) = \dfrac{3}{14}\)
43. Fuera de la\(4,000\) gente,\(10\) tendría la enfermedad. De esos\(10\),\(9\) daría positivo, mientras que falsamente\(1\) daría negativo. De las personas\(3990\) no infectadas, falsamente\(399\) daría positivo, mientras que\(3591\) daría negativo.
a.\(P(\text{virus} | \text{positive}) = \dfrac{9}{9 + 399} = \dfrac{9}{408} = 2.2 \%\)
b.\(P(\text{no virus} | \text{negative}) = \dfrac{3591}{3591 + 1} = \dfrac{3591}{3592} = 99.97 \%\)
45. Fuera de la\(100,000\) gente,\(300\) tendría la enfermedad. De esos, falsamente\(18\) daría negativo, mientras que\(282\) daría positivo. De los\(99,700\) sin la enfermedad, falsamente\(3,988\) daría positivo y el otro\(95,712\) daría negativo. \(P(\text{disease} | \text{positive}) = \dfrac{282}{282 + 3988} = \dfrac{720}{7664} = 6.6 \%\)
47. Fuera de\(100,000\) las mujeres,\(800\) tendría cáncer de mama. De esos, falsamente\(80\) daría negativo, mientras que\(720\) daría positivo. De los\(99,200\) sin cáncer, falsamente\(6,944\) daría positivo. \(P(\text{cancer} | \text{positive}) = \dfrac{720}{720 + 6944} = \dfrac{720}{7664} = 9.4 \%\)
49. \(2 \cdot 3 \cdot 8 \cdot 2 = 96\)atuendos
51. a.\(4 \cdot 4 \cdot 4 = 64\)
b.\(4 \cdot 3 \cdot 2 = 24\)
53. \(26 \cdot 26 \cdot 26 \cdot 10 \cdot 10 \cdot 10 = 17,576,000\)
55. \(_4P_4 \text{ or } 4 \cdot 3 \cdot 2 \cdot 1 = 24\)pedidos posibles
57. Cuestiones de orden. \(_7P_4 = 840\)posibles equipos
59. Cuestiones de orden. \(_{12}P_5 = 95,040\)posibles temas
61. El orden no importa. \(_{12}C_4 = 495\)
63. \(_{50}C_6 = 15,890,700\)
65. \(_{27}C_{11} \cdot 16 = 208,606,320\)
67. Sólo hay\(1\) forma de organizar los\(5\) CD's en orden alfabético. La probabilidad de que los CD estén en orden alfabético es una dividida por el número total de formas de organizar los\(5\) CD's. Dado que el orden alfabético es solo uno de todos los ordenamientos posibles puedes usar permutaciones, o simplemente usar\(5!\). \(P(\text{alphabetical}) = \dfrac{1}{5!} = \dfrac{1}{(_5P_5)} = \dfrac{1}{120}\).
69. Hay boletos\(_{48}C_6\) totales. Para igualar\(5\) del\(6\), un jugador tendría que elegir entre\(5\) esos\(6\),\(_6C_5\), y uno de los números\(42\) no ganadores,\(_{42}C_1\). \(\dfrac{6 \cdot 42}{12271512} = \dfrac{252}{12271512}\)
71. Todas las manos posibles es\(_{52}C_5\). Las manos serán todos los corazones es\(_{13}C_5\). \(\dfrac{1287}{2598960}\).
73. \( $3 \left(\dfrac{3}{37} \right) + $2 \left(\dfrac{6}{37} \right) + (-$1) \left(\dfrac{3}{37} \right) = -$ \dfrac{7}{37} + = -$0.19\)
75. Hay\(_{23}C_6 = 100,947\) posibles boletos.
Valor esperado =\($29,999 \left(\dfrac{1}{100947} \right) + (-$1) \left(\dfrac{100946}{100947} \right) = -$0.70\)
77. \($48 (0.993) + (−$302)(0.007) = $45.55\)