{"id":16434,"date":"2013-01-23T09:00:59","date_gmt":"2013-01-23T16:00:59","guid":{"rendered":"https:\/\/gmatclub.com\/blog\/?p=16434"},"modified":"2013-01-15T09:35:08","modified_gmt":"2013-01-15T16:35:08","slug":"gmat-probability-and-counting-techniques","status":"publish","type":"post","link":"https:\/\/gmatclub.com\/blog\/gmat-probability-and-counting-techniques\/","title":{"rendered":"GMAT Probability and Counting Techniques"},"content":{"rendered":"

\"\"This is the third in a series of probability articles for the GMAT Quantitative Section.\u00a0 In the\u00a0first<\/a>, I discussed the \"AND\" and \"OR\" rules for probability.\u00a0 In the\u00a0second<\/a>, I discussed the \"complement rule\" and how to use this to solve \"at least\" problems in probability.\u00a0 For a warm-up, here are some challenging GMAT Problem Solving problems on probability.<\/p>\n

 <\/p>\n

1) Five children,\u00a0Anaxagoras<\/a>,\u00a0Beatrice<\/a>,\u00a0Childeric<\/a>,\u00a0Desdemona<\/a>, and\u00a0Ethelred<\/a>, sit randomly in five chairs in a row.\u00a0\u00a0 What is the probability that Childeric and Ethelred sit next to each other?<\/p>\n

    \n
  1. 1\/30<\/li>\n
  2. 1\/15<\/li>\n
  3. 1\/5<\/li>\n
  4. 2\/15<\/li>\n
  5. 7\/20<\/li>\n<\/ol>\n

    2) A division of a company consists of seven men and five women.\u00a0 If two of these twelve employees are randomly selected as representatives of the division, what is the probability that both representatives will be female?<\/p>\n

      \n
    1. 1\/6<\/li>\n
    2. 2\/5<\/li>\n
    3. 2\/9<\/li>\n
    4. 5\/12<\/li>\n
    5. 5\/33<\/li>\n<\/ol>\n

      3) John has on his shelf four books of poetry, four novels, and two reference works.\u00a0 Suppose from these ten books, we were to pick two books at random.\u00a0 What is the probability that we pick one novel and one reference work?<\/p>\n

        \n
      1. 1\/2<\/li>\n
      2. 2\/5<\/li>\n
      3. 3\/10<\/li>\n
      4. 7\/20<\/li>\n
      5. 8\/45<\/li>\n<\/ol>\n

        Do these problems make your head spin?\u00a0 Then you have found just the post you need!<\/p>\n

         <\/p>\n

        Probability and counting<\/h2>\n

        Fundamentally, the definition of probability is<\/p>\n

        \"\"<\/a><\/p>\n

        The previous two posts talked about various tricks for calculating probabilities in different scenarios, but in some problems, we just have to count the numbers in the numerator and the denominator.\u00a0 This means, we need to understand\u00a0basic counting techniques<\/a>, including\u00a0combinations and permutations<\/a>.\u00a0 The details of the counting techniques are explained in these posts. \u00a0Any probability problem involving counting is really two counting problems in one ---- we have to calculate the denominator, the number of all possible cases, and the numerator, the number of only those cases that meet the condition specified. \u00a0I'll just mention that often, the most challenging part of any counting problem, especially for special cases (e.g. the numerators of these probability expressions) is\u00a0how we\u00a0frame<\/em><\/strong>\u00a0the problem.\u00a0 Details of this process will be explained below in the solutions to problems.<\/p>\n

         <\/p>\n

        Summary<\/h2>\n

        If counting techniques are unfamiliar to you, read those two other posts.\u00a0 Once you feel confident with counting techniques, give the problems above another try before reading the solutions below.\u00a0 In the last article in this series, I will discuss a special case of probability problems on the GMAT: geometric probability.<\/p>\n

         <\/p>\n

        Practice problem explanations<\/h2>\n

        1) First, we will count all the possible arrangements of the five children on the five seats, all the possible orders.\u00a0\u00a0 This is 5! = 120.\u00a0 That's the denominator.<\/p>\n

        Now, the more challenging part: we have to figure out how many arrangements there are involving C & E sitting together.\u00a0 This is a tricky problem to frame, so I'll demonstrate the steps to follow.\u00a0 First, let's look at the seats these two could be next to each other.\u00a0\u00a0 There are four possible pairs of seats in which they could be next to each other<\/p>\n

        i. X X _ _ _<\/p>\n

        ii. _ X X _ _<\/p>\n

        iii. _ _ X X _<\/p>\n

        iv. _ _ _ X X<\/p>\n

        In each of those four cases, we could have either CE or EC, either order, so that's 4 x 2 = 8 ways we could have just C & E sitting next to each other with the remaining three seats empty.<\/p>\n

        For the final step, we need to consider the other three children, A & B & D.\u00a0 In each of the eight cases, there are three blank seats waiting for those three, and those three could be put in any order in those blank seats.\u00a0 Three elements in any order ---- that's 3! = 6.\u00a0 Thus, the total number of arrangements in which C & E would be next to each other would be 8 x 6 = 48.\u00a0 This is our numerator.<\/p>\n

        The probability would be this number, 48, over the total number of arrangements of the children, 120.<\/p>\n

        \"\"<\/a><\/p>\n

        Answer =\u00a0D<\/strong><\/p>\n

        2) First, the denominator.\u00a0 We have twelve different people, and we want a combination of two selected from these twelve.\u00a0 We will use\u00a0the formula<\/a>:<\/p>\n

        \"\"<\/a><\/p>\n

        which, for profound mathematical reasons we need not address here, is also the formula for the sum of the first (n \u2013 1) positive integers.\u00a0 Here<\/p>\n

        \"\"<\/a><\/p>\n

        That's the total number of pairs we could pick from the twelve employees.\u00a0 That's our denominator.<\/p>\n

        For the numerator, we want every combination of two from the five female employees.\u00a0 That's<\/p>\n

        \"\"<\/a><\/p>\n

        That's the\u00a0 number of pairs of female employees we could pick from the five.\u00a0 That's our numerator.<\/p>\n

        \"\"<\/a><\/p>\n

        Answer =\u00a0E<\/strong><\/p>\n

        3) For the denominator, we are going to pick two books from among ten total: a combination of two from ten.\u00a0 Again, we will use\u00a0the formula<\/a>:<\/p>\n

        \"\"<\/a><\/p>\n

        which, for profound mathematical reasons we need not address here, is also the formula for the sum of the first (n \u2013 1) positive integers.\u00a0 Here<\/p>\n

        \"\"<\/a><\/p>\n

        That's the total number of pairs of books we could pick from the ten on the shelf.\u00a0 That's our denominator.<\/p>\n

        Now, the numerator.\u00a0 We want one novel and one reference work.\u00a0 Well, there are four novels and two reference works, so by\u00a0the FCP<\/a>, the number of ways we can pick this is 4 x 2 = 8.\u00a0 That's the total possible number of pairs involving exactly one of these four novels and exactly one of these two reference works.\u00a0 That's our numerator.<\/p>\n

        \"\"<\/a><\/p>\n

        Answer =\u00a0E<\/strong><\/p>\n

        This post was written by Mike McGarry, GMAT expert at Magoosh<\/a>, and originally posted here<\/a>.<\/p>\n","protected":false},"excerpt":{"rendered":"

        This is the third in a series of probability articles for the GMAT Quantitative Section.\u00a0 In the\u00a0first, I discussed the “AND” and “OR” rules for probability.\u00a0 In the\u00a0second, I discussed…<\/p>\n","protected":false},"author":133,"featured_media":0,"comment_status":"closed","ping_status":"closed","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[9,783,243,736],"tags":[],"class_list":["post-16434","post","type-post","status-publish","format-standard","hentry","category-gmat","category-magoosh-blog","category-blog","category-quant-gmat","entry"],"aioseo_notices":[],"_links":{"self":[{"href":"https:\/\/gmatclub.com\/blog\/wp-json\/wp\/v2\/posts\/16434","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/gmatclub.com\/blog\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/gmatclub.com\/blog\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/gmatclub.com\/blog\/wp-json\/wp\/v2\/users\/133"}],"replies":[{"embeddable":true,"href":"https:\/\/gmatclub.com\/blog\/wp-json\/wp\/v2\/comments?post=16434"}],"version-history":[{"count":3,"href":"https:\/\/gmatclub.com\/blog\/wp-json\/wp\/v2\/posts\/16434\/revisions"}],"predecessor-version":[{"id":16438,"href":"https:\/\/gmatclub.com\/blog\/wp-json\/wp\/v2\/posts\/16434\/revisions\/16438"}],"wp:attachment":[{"href":"https:\/\/gmatclub.com\/blog\/wp-json\/wp\/v2\/media?parent=16434"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/gmatclub.com\/blog\/wp-json\/wp\/v2\/categories?post=16434"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/gmatclub.com\/blog\/wp-json\/wp\/v2\/tags?post=16434"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}