{"id":16238,"date":"2013-01-11T09:00:47","date_gmt":"2013-01-11T16:00:47","guid":{"rendered":"https:\/\/gmatclub.com\/blog\/?p=16238"},"modified":"2013-01-04T07:26:20","modified_gmt":"2013-01-04T14:26:20","slug":"gmat-math-the-probability-at-least-question","status":"publish","type":"post","link":"https:\/\/gmatclub.com\/blog\/gmat-math-the-probability-at-least-question\/","title":{"rendered":"GMAT Math: the Probability “At Least” Question"},"content":{"rendered":"

\"\"In the first post in this series, I spoke about the\u00a0AND rule<\/strong>\u00a0and the\u00a0OR rule<\/strong>\u00a0in probability<\/a>.\u00a0 Now, we will focus on probability question involve the words \"at least.\"\u00a0 First, some practice of this genre.<\/p>\n

Set #1 = {A, B, C, D, E}<\/p>\n

Set #2 = {K, L, M, N, O, P}<\/p>\n

1) There are these two sets of letters, and you are going to pick exactly one letter from each set.\u00a0 What is the probability of picking at least one vowel?<\/p>\n

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

    2) Suppose you flip a fair coin six times.\u00a0 What is the probability that, in six flips, you get at least one head?<\/p>\n

      \n
    1. 5\/8<\/li>\n
    2. 13\/16<\/li>\n
    3. 15\/16<\/li>\n
    4. 31\/32<\/li>\n
    5. 63\/64<\/li>\n<\/ol>\n

      3) In a certain game, you pick a card from a standard deck of 52 cards.\u00a0 If the card is a heart, you win.\u00a0 If the card is not a heart, the person replaces the card to the deck, reshuffles, and draws again.\u00a0 The person keeps repeating that process until he picks a heart, and the point is to measure: how many draws did it take before the person picked a heart and won?\u00a0\u00a0 What is the probability that one will have at least three draws before one picks a heart?<\/p>\n

        \n
      1. 1\/2<\/li>\n
      2. 9\/16<\/li>\n
      3. 11\/16<\/li>\n
      4. 13\/16<\/li>\n
      5. 15\/16<\/li>\n<\/ol>\n

         <\/p>\n

        The complement rule<\/h2>\n

        There is a very simple and very important rule relating P(A) and P(not A), linking the probability of any event happening with the probability of that same event not happening.\u00a0\u00a0 For any well-defined event, it's 100% true that either the event happens or it doesn't happen. The GMAT will not ask you probability question about bizarre events in which, for example, you can't tell whether or not the event happened, or complex events which could, in some sense, both happen and not happen.\u00a0 For any event A in a probability question on the GMAT, the two scenarios \"A happens\" and \"A doesn't happen\" exhaust the possibilities that could take place.\u00a0 With certainty, we can say: one of those two will occur.\u00a0 In other words<\/p>\n

        P(A\u00a0OR<\/strong>\u00a0not A) = 1<\/p>\n

        Having a probability of 1 means guaranteed certainty.\u00a0 Obviously, for a variety of deep logical reasons, the events \"A\" and \"not A\" are disjoint and have no overlap.\u00a0 The OR rule, discussed in the last post, implies:<\/p>\n

        P(A) + P(not A) = 1<\/p>\n

        Subtract the first term to isolate P(not A).<\/p>\n

        P(not A) = 1 \u2013 P(A)<\/strong><\/p>\n

        That is known in probability as the\u00a0complement rule<\/strong>, because the probabilistic region in which an event doesn't occur complements the region in which it does occur.\u00a0\u00a0 This is a crucial idea in general, for all GMAT probability questions, and one that will be very important in solving \"at least\" questions in particular.<\/p>\n

         <\/p>\n

        The complement of \"at least\" statements<\/h2>\n

        Suppose event A is a statement involving words \"at least\" --- how would we state what constituted \"not A\"?\u00a0 In other words, how do we negate an \"at least\" statement?\u00a0 Let's be concrete.\u00a0 Suppose there is some event that involves just two outcomes: success and failure.\u00a0 The event could be, for example, making a basketball free throw, or flipping a coin and getting heads.\u00a0 Now, suppose we have a \"contest\" involving ten of these events in a row, and we are counting the number of successes in these ten trials.\u00a0\u00a0 Let A be the event defined as: A = \"there are at least 4 successes in these ten trials.\"\u00a0 What outcomes would constitute \"not A\"?\u00a0 Well, let's think about it.\u00a0 In ten trials, one could get zero successes, exactly one success, exactly two successes, all the way up to ten successes.\u00a0 There are eleven possible outcomes, the numbers from 0 \u2013 10, for the number of successes one could get in 10 trials.\u00a0 Consider the following diagram of the number of possible successes in ten trials.<\/p>\n

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

        The purple numbers are the members of A, the members of \"at least 4 successes\" in ten trials.\u00a0 Therefore, the green numbers are the complement space, the region of \"not A.\"\u00a0 In words, how would we describe the conditions that land you in the green region?\u00a0 We would say: \"not A\" = \"three or fewer success\" in ten trials.\u00a0\u00a0 The negation, the opposite, of \"at least four\" is \"three or fewer.\"<\/p>\n

        Abstracting from this, the negation or opposite of \"at least n\" is the condition \"(n \u2013 1) or fewer.\"\u00a0 One particularly interesting case of this is n = 1:\u00a0the negation or opposite of \"at least one\" is \"none.\"<\/strong>\u00a0\u00a0 That last statement is a hugely important idea, arguably the key to solving most of the \"at least\" questions you will see on the GMAT.<\/p>\n

         <\/p>\n

        Solving an \"at least\" question<\/h2>\n

        The big idea for any \"at least\" question on the GMAT is:\u00a0it is always easier to figure out the complement probability<\/strong>.\u00a0 For example, in the above scenario of ten trials of some sort, calculating \"at least 4\" directly would involve seven different calculations (for the cases from 4 to 10), whereas the calculation of \"three or fewer\" would involve only four separate calculations (for the cases from 0 to 3).\u00a0 In the extreme --- and extremely common ---- case of \"at least one\", the direct approach would involve a calculation for almost case, but the complement calculation simply involves calculating the probability for the \"none\" case, and then subtracting from one.<\/p>\n

        P(not A) = 1 \u2013 P(A)<\/p>\n

        P(at least one success) = 1 \u2013 P(no successes)<\/p>\n

        This is one of the most powerful time-saving shortcuts on the entire GMAT.<\/p>\n

         <\/p>\n

        An example calculation<\/h2>\n

        Consider the following simple question.<\/p>\n

        4)\u00a0A two dice are rolled.\u00a0 What is the probability of rolling a 6 on at least one of them?\u00a0<\/em><\/p>\n

        It turns out, calculating that directly would involve a relatively long calculation --- the probability of exactly one 6, on either die, and the rare probability of both coming up 6's.\u00a0 That calculation easily could take several minutes.<\/p>\n

        Instead, we will use the shortcut defined above:<\/p>\n

        P(not A) = 1 \u2013 P(A)<\/p>\n

        P(at least one 6) = 1 \u2013 P(no 6's)<\/p>\n

        What's the probability of both dice coming up no 6's?\u00a0 Well, first, let's consider one die. The probability of rolling a 6 is 1\/6, so the probability of rolling something other than 6 (\"not 6\") is 5\/6.<\/p>\n

        P(two rolls, no 6's) = P(\"not 6\" on dice #1 AND \"not 6\" on dice #2)<\/p>\n

        As we found in the previous post, the word AND means\u00a0multiply<\/em>.\u00a0 (Clearly, the outcome of each die is independent of the other).\u00a0\u00a0 Thus:<\/p>\n

        P(two rolls, no 6's) =(5\/6)*(5\/6) = 25\/36<\/p>\n

        P(at least one 6) = 1 \u2013 P(no 6's) = 1 \u2013 25\/36 =\u00a011\/36<\/strong><\/p>\n

        What could have been a long calculation becomes remarkably straightforward by means of this shortcut.\u00a0\u00a0 This can be an enormous time-saver on the GMAT!<\/p>\n

         <\/p>\n

        Practice<\/h2>\n

        Having read this post, you may want to take another shot at the three practice questions above before reading the solutions below.\u00a0 Also, here's a free question, with video explanation, on this same topic:<\/p>\n

        5)\u00a0https:\/\/gmat.magoosh.com\/questions\/839<\/a><\/p>\n

         <\/p>\n

        Practice problem explanations<\/h2>\n

        1) P(at least one vowel) = 1 \u2013 P(no vowels)<\/p>\n

        The probability of picking no vowel from the first set is 3\/5.\u00a0 The probability of picking no vowel from the second set is 5\/6.\u00a0 In order to get no vowels at all, we need no vowels from the first set AND no vowels from the second set.\u00a0 According to the AND rule, we multiply those probabilities.<\/p>\n

        P(no vowels) = (3\/5)*(5\/6) = 1\/2<\/p>\n

        P(at least one vowel) = 1 \u2013 P(no vowels) = 1 \u2013 1\/2 =\u00a01\/2<\/strong><\/p>\n

        Answer =\u00a0C<\/strong><\/p>\n

        2)\u00a0 P(at least one H) = 1 \u2013 P(no H's)<\/p>\n

        In one flip, P(\"not H\") = P(T) = 1\/2.\u00a0\u00a0 We would need to have this happen six times --- that is to say, six independent events joined by AND, which means they are multiplied together.<\/p>\n

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

        3) A full deck of 52 cards contains 13 cards from each of the four suits.\u00a0 The probability of drawing a heart from a full deck is 1\/4.\u00a0 Therefore, the probability of \"not heart\" is 3\/4.<\/p>\n

        P(at least three draws to win) = 1 \u2013 P(win in two or fewer draws)<\/p>\n

        Furthermore,<\/p>\n

        P(win in two or fewer draws) = P(win in one draw OR win in two draws)<\/p>\n

        = P(win in one draw) + P(win in two draws)<\/p>\n

        Winning in one draw means: I select one card from a full deck, and it turns out to be a heart.\u00a0 Above, we already said: the probability of this is 1\/4.<\/p>\n

        P(win in one draw) = 1\/4<\/p>\n

        Winning in two draws means: my first draw is \"not heart\", P = 3\/4, AND the second draw is a heart, P = 1\/4.\u00a0 Because we replace and re-shuffle, the draws are independent, so the AND means\u00a0multiply<\/em>.<\/p>\n

        P(win in two draws) =(3\/4)*(1\/4) = 3\/16<\/p>\n

        P(win in two or fewer draws) =P(win in one draw) + P(win in two draws)<\/p>\n

        = 1\/4 + 3\/16 = 7\/16<\/p>\n

        P(at least three draws to win) = 1 \u2013 P(win in two or fewer draws)<\/p>\n

        = 1 \u2013 7\/16 =\u00a09\/16<\/strong><\/p>\n

        Answer =\u00a0B<\/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":"

        In the first post in this series, I spoke about the\u00a0AND rule\u00a0and the\u00a0OR rule\u00a0in probability.\u00a0 Now, we will focus on probability question involve the words “at least.”\u00a0 First, some practice…<\/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-16238","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\/16238","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=16238"}],"version-history":[{"count":2,"href":"https:\/\/gmatclub.com\/blog\/wp-json\/wp\/v2\/posts\/16238\/revisions"}],"predecessor-version":[{"id":16244,"href":"https:\/\/gmatclub.com\/blog\/wp-json\/wp\/v2\/posts\/16238\/revisions\/16244"}],"wp:attachment":[{"href":"https:\/\/gmatclub.com\/blog\/wp-json\/wp\/v2\/media?parent=16238"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/gmatclub.com\/blog\/wp-json\/wp\/v2\/categories?post=16238"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/gmatclub.com\/blog\/wp-json\/wp\/v2\/tags?post=16238"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}