{"id":115877,"date":"2023-08-24T19:44:55","date_gmt":"2023-08-24T19:44:55","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=115877"},"modified":"2023-08-24T19:44:57","modified_gmt":"2023-08-24T19:44:57","slug":"solution-manual-for-a-first-course-in-probability-10th-edition-all-chapters-full-complete-2023-2024","status":"publish","type":"post","link":"https:\/\/www.learnexams.com\/blog\/2023\/08\/24\/solution-manual-for-a-first-course-in-probability-10th-edition-all-chapters-full-complete-2023-2024\/","title":{"rendered":"Solution Manual for A First Course in Probability 10th Edition \/ All Chapters Full Complete 2023 – 2024"},"content":{"rendered":"\n

A First Course in Probability 10th Edition Solution Manual
Problems Chapter 1<\/p>\n\n\n\n

    \n
  1. (a) By the generalized basic principle of counting there are
    26 \uf0d7 26 \uf0d7 10 \uf0d7 10 \uf0d7 10 \uf0d7 10 \uf0d7 10 = 67,600,000
    (b) 26 \uf0d7 25 \uf0d7 10 \uf0d7 9 \uf0d7 8 \uf0d7 7 \uf0d7 6 = 19,656,000<\/li>\n\n\n\n
  2. 6
    4 = 1296<\/li>\n\n\n\n
  3. An assignment is a sequence i1, \u2026, i20 where ij is the job to which person j is assigned. Since
    only one person can be assigned to a job, it follows that the sequence is a permutation of the
    numbers 1, \u2026, 20 and so there are 20! different possible assignments.<\/li>\n\n\n\n
  4. There are 4! possible arrangements. By assigning instruments to Jay, Jack, John and Jim, in
    that order, we see by the generalized basic principle that there are 2 \uf0d7 1 \uf0d7 2 \uf0d7 1 = 4 possibilities.<\/li>\n\n\n\n
  5. There were 8 \uf0d7 2 \uf0d7 9 = 144 possible codes. There were 1 \uf0d7 2 \uf0d7 9 = 18 that started with a 4.<\/li>\n\n\n\n
  6. Each kitten can be identified by a code number i, j, k, l where each of i, j, k, l is any of the
    numbers from 1 to 7. The number i represents which wife is carrying the kitten, j then
    represents which of that wife\u2019s 7 sacks contain the kitten; k represents which of the 7 cats in
    sack j of wife i is the mother of the kitten; and l represents the number of the kitten of cat k in
    sack j of wife i. By the generalized principle there are thus 7 \uf0d7 7 \uf0d7 7 \uf0d7 7 = 2401 kittens<\/li>\n\n\n\n
  7. (a) 6! = 720
    (b) 2 \uf0d7 3! \uf0d7 3! = 72
    (c) 4!3! = 144
    (d) 6 \uf0d7 3 \uf0d7 2 \uf0d7 2 \uf0d7 1 \uf0d7 1 = 72<\/li>\n\n\n\n
  8. (a) 5! = 120
    (b) 7!
    2!2!
    (c) 11!
    = 1260
    = 34,650
    4!4!2!
    (d) 7!
    2!2!
    = 1260<\/li>\n\n\n\n
  9. (12)!
    = 27,720
    6!4!
    1<\/li>\n<\/ol>\n\n\n\n

    2
    5
    \uf0e7
    5
    \uf0f7\uf0e7
    \uf0f7
    5
    5 5
    \uf0e7
    2
    \uf0f7 \uf0e7 \uf0f7 \uf0e7 \uf0f7
    2
    2
    Chapter 1<\/p>\n\n\n\n

      \n
    1. 103 \u2212 10 \uf0d7 9 \uf0d7 8 = 280 numbers have at least 2 equal values. 280 \u2212 10 = 270 have exactly 2
      equal values.<\/li>\n\n\n\n
    2. With ni equal to the number of length i, n1 = 3, n2 = 8, n3 = 12, n4 = 30, n5 = 30, giving the
      answer of 83.<\/li>\n\n\n\n
    3. (a) 305
      (b) 30 \uf0d7 29 \uf0d7 28 \uf0d7 27 \uf0d7 26
      15.
      16.
      \uf0e6 20\uf0f6
      \uf0e7 \uf0f7
      \uf0e8 \uf0f8
      \uf0e652\uf0f6
      \uf0e7 \uf0f7
      \uf0e8 \uf0f8<\/li>\n\n\n\n
    4. There are \uf0e610\uf0f6\uf0e612\uf0f6
      possible choices of the 5 men and 5 women. They can then be paired up
      \uf0e8 \uf0f8\uf0e8 \uf0f8
      in 5! ways, since if we arbitrarily order the men then the first man can be paired with any of
      the 5 women, the next with any of the remaining 4, and so on. Hence, there are
      possible results.
      \uf0e610 \uf0f6\uf0e612 \uf0f6
      5!\uf0e7 \uf0f7\uf0e7 \uf0f7
      \uf0e8 \uf0f8\uf0e8 \uf0f8<\/li>\n\n\n\n
    5. (a)
      \uf0e6 6\uf0f6
      +
      \uf0e67\uf0f6
      +
      \uf0e6 4\uf0f6
      = 42 possibilities.
      \uf0e8 \uf0f8 \uf0e8 \uf0f8 \uf0e8 \uf0f8
      (b) There are 6 \uf0d7 7 choices of a math and a science book, 6 \uf0d7 4 choices of a math and an
      economics book, and 7 \uf0d7 4 choices of a science and an economics book. Hence, there are
      94 possible choices.<\/li>\n\n\n\n
    6. The first gift can go to any of the 10 children, the second to any of the remaining 9 children,
      and so on. Hence, there are 10 \uf0d7 9 \uf0d7 8 \uf0d7 \uf0d7 \uf0d7 5 \uf0d7 4 = 604,800 possibilities.
      2<\/li>\n\n\n\n
    7. (a) 8! = 40,320
      (b)
      (c)
      (d)
      2 \uf0d7 7! = 10,080
      5!4! = 2,880
      4!24 = 384<\/li>\n\n\n\n
    8. (a) 6!
      (b) 3!2!3!
      (c) 3!4!<\/li>\n<\/ol>\n\n\n\n

      \uf0e7
      2
      \uf0f7\uf0e7
      2
      \uf0f7\uf0e7
      3
      \uf0f7
      \uf0e7
      3
      \uf0f7\uf0e7
      3
      \uf0f7 \uf0e7 \uf0f7\uf0e7 \uf0f7\uf0e7 \uf0f7
      3
      1 2
      \uf0e7
      3
      \uf0f7\uf0e7
      3
      \uf0f7 \uf0e7 \uf0f7\uf0e7 \uf0f7\uf0e7 \uf0f7
      1
      2 3
      3 3 3 1 2
      \uf0e7
      3
      \uf0f7\uf0e7
      3
      \uf0f7 \uf0e7 \uf0f7\uf0e7 \uf0f7 \uf0e7 \uf0f7\uf0e7 \uf0f7
      2
      3 3 2
      \uf0e7
      3
      \uf0f7\uf0e7
      3
      \uf0f7
      \uf0e7
      2
      \uf0f7\uf0e7
      3
      \uf0f7
      3 2
      \uf0e7
      5
      \uf0f7 \uf0e7 \uf0f7\uf0e7 \uf0f7 \uf0e7 \uf0f7 \uf0e7 \uf0f7
      1
      4 5 3
      \uf0e8 \uf0f8
      \uf0e8 \uf0f8
      Chapter 1 3<\/p>\n\n\n\n

        \n
      1. \uf0e6 5 \uf0f6\uf0e6 6 \uf0f6\uf0e6 4 \uf0f6
        = 600
        \uf0e8 \uf0f8\uf0e8 \uf0f8\uf0e8 \uf0f8<\/li>\n\n\n\n
      2. (a) There are \uf0e68\uf0f6\uf0e6 4\uf0f6
        +
        \uf0e68\uf0f6\uf0e6 2\uf0f6\uf0e6 4\uf0f6
        \uf0e8 \uf0f8\uf0e8 \uf0f8 \uf0e8 \uf0f8\uf0e8 \uf0f8\uf0e8 \uf0f8
        = 896 possible committees.
        There are
        \uf0e68\uf0f6\uf0e6 4\uf0f6
        that do not contain either of the 2 men, and there are \uf0e68\uf0f6\uf0e6 2\uf0f6\uf0e6 4\uf0f6
        that
        \uf0e7 \uf0f7\uf0e7 \uf0f7
        \uf0e8 \uf0f8\uf0e8 \uf0f8
        contain exactly 1 of them.
        (b) There are \uf0e6 6\uf0f6\uf0e66\uf0f6
        +
        \uf0e6 2\uf0f6\uf0e6 6\uf0f6\uf0e66\uf0f6
        = 1000 possible committees.
        \uf0e8 \uf0f8\uf0e8 \uf0f8 \uf0e8 \uf0f8\uf0e8 \uf0f8\uf0e8 \uf0f8
        \uf0e7 \uf0f7\uf0e7 \uf0f7\uf0e7 \uf0f7
        \uf0e8 \uf0f8\uf0e8 \uf0f8\uf0e8 \uf0f8
        (c) There are
        \uf0e6 7\uf0f6\uf0e65\uf0f6
        +
        \uf0e6 7\uf0f6\uf0e65\uf0f6
        +
        \uf0e6 7\uf0f6\uf0e65 \uf0f6
        \uf0e8 \uf0f8\uf0e8 \uf0f8 \uf0e8 \uf0f8\uf0e8 \uf0f8 \uf0e8 \uf0f8\uf0e8 \uf0f8
        = 910 possible committees. There are \uf0e6 7\uf0f6\uf0e65\uf0f6
        in
        \uf0e8 \uf0f8\uf0e8 \uf0f8
        which neither feuding party serves;
        \uf0e67\uf0f6\uf0e65\uf0f6
        in which the feuding women serves; and
        \uf0e8 \uf0f8\uf0e8 \uf0f8
        \uf0e6 7 \uf0f6\uf0e6 5\uf0f6
        \uf0e7 \uf0f7\uf0e7 \uf0f7
        \uf0e8 \uf0f8\uf0e8 \uf0f8
        in which the feuding man serves.<\/li>\n\n\n\n
      3. \uf0e6 6\uf0f6
        +
        \uf0e6 2\uf0f6\uf0e6 6\uf0f6
        ,
        \uf0e6 6\uf0f6
        +
        \uf0e6 6\uf0f6
        \uf0e8 \uf0f8 \uf0e8 \uf0f8\uf0e8 \uf0f8 \uf0e8 \uf0f8 \uf0e8 \uf0f8<\/li>\n\n\n\n
      4. 7!
        3!4!
        = 35. Each path is a linear arrangement of 4 r\u2019s and 3 u\u2019s (r for right and u for up). For
        instance the arrangement r, r, u, u, r, r, u specifies the path whose first 2 steps are to the right,
        next 2 steps are up, next 2 are to the right, and final step is up.<\/li>\n\n\n\n
      5. There are
        4!
        2!2!
        paths from A to the circled point; and 3!
        2!1!
        paths from the circled point to B.
        Thus, by the basic principle, there are 18 different paths from A to B that go through the
        circled point.<\/li>\n\n\n\n
      6. 3!23<\/li>\n\n\n\n
      7. (a)
        n
        \uf0e6 n \uf0f6
        2
        k = (2 +1)
        n
        \uf0e5k=0
        \uf0e7
        k
        \uf0f7
        (b) \uf0e5
        n \uf0e6 n\uf0f6
        x
        k
        = ( x +1)
        n
        k=0 \uf0e7
        k
        \uf0f7<\/li>\n<\/ol>\n\n\n\n

        \uf0e8 \uf0f8
        \uf0e8 \uf0f8
        \uf0e8 \uf0f8
        \uf0e8 \uf0f8
        \uf0e7
        3
        \uf0f7
        \uf0e7
        3
        \uf0f7
        \uf0e7
        5 \uf0f7
        5 5
        4 Chapter 1<\/p>\n\n\n\n

          \n
        1. \uf0e6 52 \uf0f6
          \uf0e7
          13,13,13,13\uf0f7<\/li>\n<\/ol>\n\n\n\n

          30. \uf0e6 12 \uf0f6<\/h1>\n\n\n\n

          12!
          \uf0e7
          3, 4, 5
          \uf0f7
          3!4!5!<\/p>\n\n\n\n

            \n
          1. Assuming teachers are distinct.
            (a) 4
            8
            (b) \uf0e6<\/li>\n<\/ol>\n\n\n\n

            8 \uf0f6<\/h1>\n\n\n\n

            8! = 2520.
            \uf0e7
            2,2,2,2
            \uf0f7
            (2)4<\/p>\n\n\n\n

              \n
            1. (a) (10)!\/3!4!2!
              (b) 3
              \uf0e63 \uf0f6 7!
              \uf0e7
              2
              \uf0f7
              4!2!<\/li>\n\n\n\n
            2. 2 \uf0d7 9! \u2212 2
              28! since 2 \uf0d7 9! is the number in which the French and English are next to each other
              and 228! the number in which the French and English are next to each other and the U.S. and
              Russian are next to each other.<\/li>\n\n\n\n
            3. (a) number of nonnegative integer solutions of x1 + x2 + x3 + x4 = 8.
              Hence, answer is \uf0e611\uf0f6
              \uf0e8 \uf0f8
              = 165
              (b) here it is the number of positive solutions\u2014hence answer is \uf0e67\uf0f6
              = 35
              \uf0e8 \uf0f8<\/li>\n\n\n\n
            4. (a) number of nonnegative solutions of x1 + \u2026 + x6 = 8
              answer =
              \uf0e613\uf0f6
              \uf0e8 \uf0f8
              (b) (number of solutions of x1 + \u2026 + x6 = 5) \uf0b4 (number of solutions of x1 + \u2026 + x6 = 3) =
              \uf0e610\uf0f6\uf0e68 \uf0f6
              \uf0e7 \uf0f7\uf0e7 \uf0f7
              \uf0e8 \uf0f8\uf0e8 \uf0f8
              <\/li>\n<\/ol>\n","protected":false},"excerpt":{"rendered":"

              A First Course in Probability 10th Edition Solution ManualProblems Chapter 1 25\uf0e75\uf0f7\uf0e7\uf0f755 5\uf0e72\uf0f7 \uf0e7 \uf0f7 \uf0e7 \uf0f722Chapter 1 \uf0e72\uf0f7\uf0e72\uf0f7\uf0e73\uf0f7\uf0e73\uf0f7\uf0e73\uf0f7 \uf0e7 \uf0f7\uf0e7 \uf0f7\uf0e7 \uf0f731 2\uf0e73\uf0f7\uf0e73\uf0f7 \uf0e7 \uf0f7\uf0e7 \uf0f7\uf0e7 \uf0f712 33 3 3 1 2\uf0e73\uf0f7\uf0e73\uf0f7 \uf0e7 \uf0f7\uf0e7 \uf0f7 \uf0e7 \uf0f7\uf0e7 \uf0f723 3 2\uf0e73\uf0f7\uf0e73\uf0f7\uf0e72\uf0f7\uf0e73\uf0f73 2\uf0e75\uf0f7 \uf0e7 \uf0f7\uf0e7 \uf0f7 \uf0e7 \uf0f7 \uf0e7 \uf0f714 5 3\uf0e8 \uf0f8\uf0e8 \uf0f8Chapter 1 3 […]<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"site-sidebar-layout":"default","site-content-layout":"","ast-site-content-layout":"default","site-content-style":"default","site-sidebar-style":"default","ast-global-header-display":"","ast-banner-title-visibility":"","ast-main-header-display":"","ast-hfb-above-header-display":"","ast-hfb-below-header-display":"","ast-hfb-mobile-header-display":"","site-post-title":"","ast-breadcrumbs-content":"","ast-featured-img":"","footer-sml-layout":"","ast-disable-related-posts":"","theme-transparent-header-meta":"","adv-header-id-meta":"","stick-header-meta":"","header-above-stick-meta":"","header-main-stick-meta":"","header-below-stick-meta":"","astra-migrate-meta-layouts":"default","ast-page-background-enabled":"default","ast-page-background-meta":{"desktop":{"background-color":"","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"tablet":{"background-color":"","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"mobile":{"background-color":"","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""}},"ast-content-background-meta":{"desktop":{"background-color":"var(--ast-global-color-5)","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"tablet":{"background-color":"var(--ast-global-color-5)","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"mobile":{"background-color":"var(--ast-global-color-5)","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""}},"footnotes":""},"categories":[25],"tags":[],"class_list":["post-115877","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/115877","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/comments?post=115877"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/115877\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=115877"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=115877"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=115877"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}