{"id":3214,"date":"2024-06-19T12:00:17","date_gmt":"2024-06-19T19:00:17","guid":{"rendered":"https:\/\/magoosh.com\/gmat\/?p=3214"},"modified":"2020-04-07T17:14:12","modified_gmt":"2020-04-08T00:14:12","slug":"gmat-math-the-probability-at-least-question","status":"publish","type":"post","link":"https:\/\/magoosh.com\/gmat\/gmat-math-the-probability-at-least-question\/","title":{"rendered":"GMAT Math: The Probability &#8220;At Least&#8221; Question"},"content":{"rendered":"<p><strong>This post was updated in 2024 for the new GMAT.<\/strong><\/p>\n<p>Ready to refresh your memory on <a title=\"GMAT Math: Probability Rules\" href=\"https:\/\/magoosh.com\/gmat\/gmat-probability\/\" target=\"_blank\" rel=\"noopener\">GMAT probability rules<\/a>? In this post, we will focus on probability questions involving the &#8220;at least&#8221;\u00a0probability. <\/p>\n<p><img decoding=\"async\" src=\"https:\/\/magoosh.com\/gmat\/files\/2012\/12\/GMAT-Probability.jpg\" alt=\"Dice spread across table width=\"1200\" height=\"600\" class=\"aligncenter size-full wp-image-9502\" srcset=\"https:\/\/magoosh.com\/gmat\/files\/2012\/12\/GMAT-Probability.jpg 1200w, https:\/\/magoosh.com\/gmat\/files\/2012\/12\/GMAT-Probability-300x150.jpg 300w, https:\/\/magoosh.com\/gmat\/files\/2012\/12\/GMAT-Probability-600x300.jpg 600w, https:\/\/magoosh.com\/gmat\/files\/2012\/12\/GMAT-Probability-768x384.jpg 768w\" sizes=\"(max-width: 1200px) 100vw, 1200px\" \/><\/p>\n<h2>The complement rule<\/h2>\n<p>There is a very simple and very important rule relating <em>P(A)<\/em> and <em>P(not A)<\/em>, linking the probability of any event happening with the probability of that same event not happening.\u00a0 For any well-defined event, it&#8217;s 100% true that either the event happens or it doesn&#8217;t happen. <\/p>\n<p>The GMAT will not ask you probability question about bizarre events in which, for example, you can&#8217;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 <em>A<\/em> in a probability question on the GMAT, the two scenarios &#8220;<em>A<\/em> happens&#8221; and &#8220;<em>A<\/em> doesn&#8217;t happen&#8221; exhaust the possibilities that could take place.\u00a0 <\/p>\n<p>With certainty, we can say: one of those two will occur.\u00a0 In other words:<\/p>\n<p><em>P(A <strong>OR<\/strong> not A)<\/em> = 1<\/p>\n<p>Having a probability of 1 means guaranteed certainty.\u00a0 Obviously, for a variety of deep logical reasons, the events &#8220;A&#8221; and &#8220;not A&#8221; are disjoint and have no overlap.\u00a0 The OR rule, discussed in the last post, implies:<\/p>\n<p><em>P(A)<\/em> + <em>P(not A)<\/em> = 1<\/p>\n<p>Subtract the first term to isolate <em>P(not A)<\/em>.<\/p>\n<p><strong><em>P(not A)<\/em> = 1 \u2013 <em>P(A)<\/em><\/strong><br \/>\n&nbsp;<\/p>\n<ul class=\"no_bullet\">\n<li class=\"bulb\">That is known in probability as the <strong>complement rule<\/strong>, because the probabilistic region in which an event doesn&#8217;t occur complements the region in which it does occur.<\/li>\n<p>\u00a0\n<\/ul>\n<p>This is a crucial idea in general, for all GMAT probability questions, and one that will be very important in solving &#8220;at least&#8221; questions in particular.<\/p>\n<h2>The complement of &#8220;at least&#8221; statements<\/h2>\n<p>Suppose event <em>A<\/em> is a statement involving words &#8220;at least&#8221;&mdash;how would we state what constituted &#8220;<em>not A<\/em>&#8220;?\u00a0 In other words, how do we negate an &#8220;at least&#8221; statement?\u00a0 <\/p>\n<p>Let&#8217;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 &#8220;contest&#8221; involving ten of these events in a row, and we are counting the number of successes in these ten trials.\u00a0\u00a0 <\/p>\n<p>Let <em>A<\/em> be the event defined as: <em>A<\/em> = &#8220;there are at least 4 successes in these ten trials.&#8221;\u00a0 What outcomes would constitute &#8220;<em>not A<\/em>&#8220;?\u00a0 Well, let&#8217;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 <\/p>\n<p>Consider the following: number of possible successes in ten trials:<\/p>\n<p align=\"center\"> <strong><font style=\"color:#009573;\" > 0 1 2 3 <\/font><font style=\"color:#4d2079;\" align=\"center\" >4 5 6 7 8 9 10 <\/font><\/strong><\/p>\n<p>The purple numbers are the members of <em>A<\/em>, the members of &#8220;at least 4 successes&#8221; in ten trials.\u00a0 Therefore, the green numbers are the complement space, the region of &#8220;<em>not A<\/em>.&#8221;\u00a0 In words, how would we describe the conditions that land you in the green region?\u00a0 We would say: &#8220;<em>not A<\/em>&#8221; = &#8220;three or fewer success&#8221; in ten trials.\u00a0\u00a0 The negation, the opposite, of &#8220;at least four&#8221; is &#8220;three or fewer.&#8221;<\/p>\n<p>Abstracting from this, the negation or opposite of &#8220;at least <em>n<\/em>&#8221; is the condition &#8220;(<em>n<\/em> \u2013 1) or fewer.&#8221;\u00a0 <\/p>\n<ul class=\"no_bullet\">\n<li class=\"bulb\">One particularly interesting case of this is n = 1:<strong> the negation or opposite of &#8220;at least one&#8221; is &#8220;none.&#8221;<\/strong><\/li>\n<\/ul>\n<p>That last statement is a hugely important idea, arguably the key to solving most of the &#8220;at least&#8221; questions you will see on the GMAT.<\/p>\n<h2>Solving an &#8220;at least&#8221; question<\/h2>\n<p>The big idea for any &#8220;at least&#8221; question on the GMAT is: <\/p>\n<ul class=\"no_bullet\">\n<li class=\"bulb\"><strong>It is always easier to figure out the complement probability<\/strong>.<\/li>\n<\/ul>\n<p>\u00a0<br \/>\nFor example, in the above scenario of ten trials of some sort, calculating &#8220;at least 4&#8221; directly would involve seven different calculations (for the cases from 4 to 10), whereas the calculation of &#8220;three or fewer&#8221; would involve only four separate calculations (for the cases from 0 to 3).\u00a0 <\/p>\n<p>In the extreme&mdash;and extremely common&mdash;case of &#8220;the probability of at least one&#8221;, the direct approach would involve a calculation for almost case, but the complement calculation simply involves calculating the probability for the &#8220;none&#8221; case, and then subtracting from one.<\/p>\n<p><em>P(not A)<\/em> = 1 \u2013 <em>P(A)<\/em><\/p>\n<p><em>P(at least one success)<\/em> = 1 \u2013 <em>P(no successes)<\/em><\/p>\n<p>This is one of the most powerful time-saving shortcuts on the entire GMAT.<\/p>\n<p>&nbsp;<\/p>\n<h2>An example calculation<\/h2>\n<p>Consider the following simple question.<\/p>\n<p><em>Two dice are rolled.\u00a0 What is the probability of at least one of the dice rolling a 6? <\/em><\/p>\n<p>It turns out, calculating that directly would involve a relatively long calculation &#8212; the probability of exactly one 6, on either die, and the rare probability of both coming up 6&#8217;s.\u00a0 That calculation easily could take several minutes.<\/p>\n<p>Instead, we will use the shortcut defined above:<\/p>\n<p><em>P(not A)<\/em> = 1 \u2013 <em>P(A)<\/em><\/p>\n<p><em>P(at least one 6)<\/em> = 1 \u2013 <em>P(no 6&#8217;s)<\/em><\/p>\n<p>What&#8217;s the probability of both dice coming up no 6&#8217;s?\u00a0 Well, first, let&#8217;s consider one die. The probability of rolling a 6 is 1\/6, so the probability of rolling something other than 6 (&#8220;not 6&#8221;) is 5\/6.<\/p>\n<p><em>P(two rolls, no 6&#8217;s)<\/em> = <em>P(&#8220;not 6&#8221; on dice #1 AND &#8220;not 6&#8221; on dice #2)<\/em><\/p>\n<p>As we found in the previous post, the word AND means <em>multiply<\/em>.\u00a0 (Clearly, the outcome of each die is independent of the other).\u00a0\u00a0 Thus:<\/p>\n<p><em>P(two rolls, no 6&#8217;s)<\/em> =(5\/6)*(5\/6) = 25\/36<\/p>\n<p><em>P(at least one 6)<\/em> = 1 \u2013 <em>P(no 6&#8217;s)<\/em> = 1 \u2013 25\/36 = <strong>11\/36<\/strong><\/p>\n<p>What could have been a long calculation becomes remarkably straightforward by means of this shortcut.  This can be an enormous time-saver on the GMAT!<\/p>\n<h2>Practice<\/h2>\n<p>Now that you&#8217;ve read this post, take shot at these three practice questions before reading the answers and explanations below.\u00a0 <\/p>\n<p>Set #1 = {A, B, C, D, E}<\/p>\n<p>Set #2 = {K, L, M, N, O, P}<\/p>\n<ol type=\"1\">\n<li>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 at least one vowel being picked?\n<ol type=\"A\">\n<li>1\/6<\/li>\n<li>1\/3<\/li>\n<li>1\/2<\/li>\n<li>2\/3<\/li>\n<li>5\/6<\/li>\n<\/ol>\n<\/li>\n<\/ol>\n<details>\n<summary style=\"color:#4d2079;\">Show answer and explanation<\/summary>\n<p><em>P(at least one vowel)<\/em> = 1 \u2013 <em>P(no vowels)<\/em><\/p>\n<p>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><em>P(no vowels)<\/em> = (3\/5)*(5\/6) = 1\/2<\/p>\n<p><em>P(at least one vowel)<\/em> = 1 \u2013 <em>P(no vowels)<\/em> = 1 \u2013 1\/2 = <strong>1\/2<\/strong><\/p>\n<p>Answer = <strong>C<\/strong><br \/>\n<\/details>\n<p>&nbsp;<\/p>\n<ol start=\"2\" type=\"1\">\n<li>Suppose you flip a fair coin six times.\u00a0 What is the probability of at least one head in six flips?\n<ol type=\"A\">\n<li>5\/8<\/li>\n<li>13\/16<\/li>\n<li>15\/16<\/li>\n<li>31\/32<\/li>\n<li>63\/64<\/li>\n<\/ol>\n<\/li>\n<\/ol>\n<details>\n<summary style=\"color:#4d2079;\">Show answer and explanation<\/summary>\n<p><em>P(at least one H) <\/em>= 1 \u2013 <em>P(no H&#8217;s)<\/em><\/p>\n<p>In one flip, <em>P(&#8220;not H&#8221;)<\/em> = <em>P(T)<\/em> = 1\/2.\u00a0\u00a0 We would need to have this happen six times&mdash;that is to say, six independent events joined by AND, which means they are multiplied together.<\/p>\n<p><a href=\"https:\/\/magoosh.com\/gmat\/files\/2012\/12\/tpatq_img2.png\"><img decoding=\"async\" class=\"alignnone size-full wp-image-3217\" src=\"https:\/\/magoosh.com\/gmat\/files\/2012\/12\/tpatq_img2.png\" alt=\"at least probability\" width=\"288\" height=\"144\" \/><br \/>\n<\/a>Answer = <strong>E<\/strong><br \/>\n<\/details>\n<p>&nbsp;<\/p>\n<ol start=\"3\" type=\"1\">\n<li>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 two &#8220;heartless&#8221; draws on the first two draws, not picking the first heart until at least the third draw?\n<ol type=\"A\">\n<li>1\/2<\/li>\n<li>9\/16<\/li>\n<li>11\/16<\/li>\n<li>13\/16<\/li>\n<li>15\/16<\/li>\n<\/ol>\n<\/li>\n<\/ol>\n<details>\n<summary style=\"color:#4d2079;\">Show answer and explanation<\/summary>\n<p>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 &#8220;not heart&#8221; is 3\/4.<\/p>\n<p><em>P(at least three draws to win)<\/em> = 1 \u2013 <em>P(win in two or fewer draws)<\/em><\/p>\n<p>Furthermore,<\/p>\n<p><em>P(win in two or fewer draws)<\/em> = <em>P(win in one draw OR win in two draws)<\/em><\/p>\n<p>= <em>P(win in one draw)<\/em> + <em>P(win in two draws)<\/em><\/p>\n<p>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><em>P(win in one draw)<\/em> = 1\/4<\/p>\n<p>Winning in two draws means: my first draw is &#8220;not heart&#8221;, <em>P<\/em> = 3\/4, AND the second draw is a heart, <em>P<\/em> = 1\/4.\u00a0 Because we replace and re-shuffle, the draws are independent, so the AND means <em>multiply<\/em>.<\/p>\n<p><em>P(win in two draws)<\/em> =(3\/4)*(1\/4) = 3\/16<\/p>\n<p><em>P(win in two or fewer draws)<\/em> = <em>P(win in one draw)<\/em> + <em>P(win in two draws)<\/em><\/p>\n<p>= 1\/4 + 3\/16 = 7\/16<\/p>\n<p><em>P(at least three draws to win)<\/em> = 1 \u2013 <em>P(win in two or fewer draws)<\/em><\/p>\n<p>= 1 \u2013 7\/16 =<strong> 9\/16<\/strong><\/p>\n<p>Answer = <strong>B<\/strong><br \/>\n<\/details>\n<p>&nbsp;<\/p>\n<p><strong>Bonus Question<\/strong><\/p>\n<blockquote><p>The probability is 0.6 that an \u201cunfair\u201d coin will turn up tails on any given toss. If the coin is tossed 3 times, what is the probability that at least one of the tosses will turn up tails?<\/p><\/blockquote>\n<p><input type=\"radio\" name=\"probabilitycointoss\" value=\"0.064\">0.064<br \/>\n<input type=\"radio\" name=\"probabilitycointoss\" value=\"0.36\">0.36<br \/>\n<input type=\"radio\" name=\"probabilitycointoss\" value=\"0.64\">0.64<br \/>\n<input type=\"radio\" name=\"probabilitycointoss\" value=\"0.784\">0.784<br \/>\n<input type=\"radio\" name=\"probabilitycointoss\" value=\"0.936\">0.936<\/p>\n<p><a href=\"https:\/\/gmat.magoosh.com\/questions\/839?utm_source=gmatblog&#038;utm_medium=blog&#038;utm_campaign=gmatquestions&#038;utm_term=inline&#038;utm_content=gmat-math-the-probability-at-least-question\" target=\"_blank\" rel=\"noopener noreferrer\">Click here for the answer and video explanation!<\/a><\/p>\n<p>If you&#8217;re looking to strengthen your Quant skills, <a href=\"https:\/\/gmat.magoosh.com\/plans?utm_source=gmatblog&#038;utm_medium=blog&#038;utm_campaign=gmatplans&#038;utm_term=inline&#038;utm_content=gmat-math-the-probability-at-least-question\">Magoosh GMAT<\/a> can help! We offer over 800 practice questions, 200 video lessons, plus full-length practice tests to help you get your best score. Try us for free with a 1-week trial!<\/p>\n","protected":false},"excerpt":{"rendered":"<p>This post was updated in 2024 for the new GMAT. Ready to refresh your memory on GMAT probability rules? In this post, we will focus on probability questions involving the &#8220;at least&#8221;\u00a0probability.<\/p>\n","protected":false},"author":26,"featured_media":9605,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[150],"tags":[],"ppma_author":[13209],"class_list":["post-3214","post","type-post","status-publish","format-standard","has-post-thumbnail","hentry","category-basics"],"acf":[],"yoast_head":"<!-- This site is optimized with the Yoast SEO Premium plugin v21.7 (Yoast SEO v21.7) - https:\/\/yoast.com\/wordpress\/plugins\/seo\/ -->\n<title>GMAT Math: The Probability &quot;At Least&quot; Question | Magoosh Study Resources<\/title>\n<meta name=\"description\" content=\"We&#039;ve covered the &quot;and&quot; rule, as well as the &quot;or&quot; rule in probability. 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