{"id":17024,"date":"2013-02-27T09:00:22","date_gmt":"2013-02-27T16:00:22","guid":{"rendered":"https:\/\/gmatclub.com\/blog\/?p=17024"},"modified":"2013-02-17T07:06:42","modified_gmt":"2013-02-17T14:06:42","slug":"geometric-probability-on-the-gmat","status":"publish","type":"post","link":"https:\/\/gmatclub.com\/blog\/geometric-probability-on-the-gmat\/","title":{"rendered":"Geometric Probability on the GMAT"},"content":{"rendered":"

\"\"<\/a>In the first three articles in this series on GMAT probability questions, I discussed the\u00a0AND and OR<\/a>\u00a0probability rules, \u00a0\"at least<\/a>\" probability questions, and\u00a0probability questions that involve counting<\/a>.\u00a0 This post covers relatively rare kind of probability question not covered in the first three posts:\u00a0geometric probability<\/strong>\u00a0questions.\u00a0 First, a few challenging practice GMAT questions on this topic.<\/p>\n

\"gp_img1\"<\/a><\/p>\n

1) In the diagram above, the sides of rectangle ABCD have a ratio AB:BC = 1:2, and the circle is tangent to three sides of the rectangle.\u00a0 If a point is chosen at random inside the rectangle, what is the probability that it is\u00a0not<\/em>\u00a0inside the circle?<\/p>\n

\"gp_img2\"<\/a><\/p>\n

2) Region R is a square in the x-y plane with vertices J = (\u20131, \u20132), K = (\u20131, 4), L = (5, 4), and M = (5, \u20132).\u00a0 What is the probability that a randomly selected point in region R lies below the line 3x \u2013 5y = 10?<\/p>\n

    \n
  1. 5\/12<\/li>\n
  2. 5\/18<\/li>\n
  3. 5\/24<\/li>\n
  4. 5\/36<\/li>\n
  5. 5\/72<\/li>\n<\/ol>\n

    \"gp_img3\"<\/a><\/p>\n

    3) In the diagram above, WZ = XZ, and circular arc XY has a center at W.\u00a0 If a point is selected from anywhere within this figure, what is the probability that it is selected from the shaded region?<\/p>\n

    \"gp_img4\"<\/a><\/p>\n

    These three are very challenging problems.\u00a0 If they seem easy to you, you are probably a GMAT Quantitative expert!\u00a0 If these make your head spin, please read on!<\/p>\n

     <\/p>\n

    Probability and area<\/h2>\n

    As anti-intuitive as this may seem, there is a deep mathematical link between probability and area.\u00a0\u00a0 For example, if you are dealing with regions under the\u00a0Normal Distribution<\/a>\u00a0or any other distribution, the probability of landing in those regions is measured by finding the area under the curve.\u00a0\u00a0 This connection between area and probability because quite apparent if you happen to study calculus-based statistics.\u00a0 If you haven't ventured into such regions, I'll just say, remember: there's an important link between probability and area. \u00a0In particular, if a diagram is given or described in a GMAT Quantitative question, and you are asked for a probability, then you almost always will be finding a ratio of two areas.\u00a0\u00a0 Thus, any geometric probability problem just becomes a 2-in1 find-the-area problem.<\/p>\n

     <\/p>\n

    Finding area<\/h2>\n

    Know your basic geometry.\u00a0 You should know how to find the area of rectangles and triangles: if you don't remember, say, the formula for the area of a trapezoid, just break it into triangles & rectangles and find the area piece-by-piece.\u00a0 You should know\u00a0Archimedes<\/a>' remarkable formula for the area of a circle:<\/p>\n

    \"gp_img5\"<\/a><\/p>\n

    You should also be comfortable with setting up the proportion necessary to find the\u00a0area of a circular sector<\/a>\u00a0(i.e. a \"slice of the pie\").<\/p>\n

     <\/p>\n

    Visualize<\/h2>\n

    If you are given a diagram as part of the problem, that's great.\u00a0 If you are not given a diagram, sketch one on the notepad.\u00a0 In any GMAT problem involving Geometry, it's very important to have your right-brain visual skills engaged.<\/p>\n

     <\/p>\n

    A ratio of areas<\/h2>\n

    The most general formula for probability is<\/p>\n

    \"gp_img6\"<\/a><\/p>\n

    In the case of geometry probability, each of these terms, the numerator and the denominator, is an area.<\/p>\n

    \"gp_img7\"<\/a><\/p>\n

    Many times in geometric probability problems, as in #1 and #3 above, no absolute lengths are given --- we know the relative ratios, but we don't know in absolute terms how big anything is.\u00a0 The information in the problem does not allow us to compute the numerical value of any area.\u00a0 That's fine.\u00a0 Just pick\u00a0any convenient number<\/em>\u00a0for one of the lengths, and figure out everything from there.\u00a0\u00a0It doesn't\u00a0 matter what value you pick<\/em>, because in the probability ratio, the value you pick will cancel.\u00a0 All that matters are the ratios.<\/p>\n

     <\/p>\n

    Summary<\/h2>\n

    With these recommendations, go back and look at those problems again before reading the explanations below.\u00a0 Here's another practice question from inside the product:<\/p>\n

    4)\u00a0https:\/\/gmat.magoosh.com\/questions\/829<\/a><\/p>\n

     <\/p>\n

    Practice question explanations<\/h2>\n

    1) We aren't given any absolute lengths.\u00a0 For convenience, I am going to assume that the radius of the circle is r = 1.\u00a0 That's very easy.\u00a0 Right away, we know the area of the circle is <m>pi<\/m>.\u00a0 Notice, the height of the triangle is equal to the diameter of the circle, so h = AB = 2.\u00a0 We are told the ratio of AB:BC = 1:2, so BC, the width, must equal w = BC = 4.\u00a0 Area of the rectangle is h*w = (AB)*(BC) = 8.\u00a0 That, right there, is our \"denominator area\".\u00a0\u00a0 Now, for the numerator area, the area of the rectangle that\u00a0does not include<\/em>\u00a0the circle, subtract the circle from the rectangle: A = <m>8 - pi<\/m>.\u00a0 That's our numerator.<\/p>\n

    \"gp_img8\"<\/a><\/p>\n

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

     <\/p>\n

    2)\u00a0 Here, we are not given a diagram, so we will sketch one.\u00a0 In the real test, just a rough diagram will be enough to visualize things.\u00a0 Here's an accurate diagram.<\/p>\n

    \"gp_img9\"<\/a><\/p>\n

    Notice the square JKLM has an area of 6 x 6 = 36 ---- right there, that's our denominator-area.<\/p>\n

    Now, let's think about this line.\u00a0 Let's solve for y, to put the line into\u00a0slope-intercept form<\/a>\u00a0(i.e. y = mx + b form).<\/p>\n

    \"gp_img10\"<\/a><\/p>\n

    First of all, putting it in slope-intercept form makes clear --- the y-intercept of this line is point S, (0, \u20132), which is on one side of the square.\u00a0 The slope is 3\/5, so over five and up three puts us at point T, (5, 1), also on a side of the square.\u00a0 Triangle STM has a base of 5 and height of three, so the area is<\/p>\n

    \"gp_img11\"<\/a><\/p>\n

    That's the numerator-area, so just divide<\/p>\n

    \"gp_img12\"<\/a><\/p>\n

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

     <\/p>\n

    3)\u00a0 Again, we are not given any absolute lengths, so we will pick something convenient.\u00a0 Here, I am going to pick WX = XZ = 1.\u00a0\u00a0 Triangle WXZ is an\u00a0Isosceles Right Triangle<\/a>, i.e. a 45-45-90 triangle.\u00a0 Its area is A= (1\/2)bh, and since both base and height are 1, the area of the triangle is 1\/2.<\/p>\n

    Both WX and WY are radii of the circle.\u00a0 From the proportions in the 45-45-90 triangle, we know<\/p>\n

    \"gp_img13\"<\/a><\/p>\n

    Now, we can find the area of the circle, using Archimedes' formula:<\/p>\n

    \"gp_img14\"<\/a><\/p>\n

    Because triangle WXZ is a 45-45-90 triangle, we know \u2220W = 45\u00b0.\u00a0\u00a0 Well, a 45\u00b0 angle must be half of 90\u00b0 ---- which means a 45\u00b0 is 1\/4 of 180\u00b0 ---- which means a 45\u00b0 is 1\/8 of 360\u00b0.\u00a0 Thus, a 45\u00b0 sector would be 1\/8 of a circle and have 1\/8 of the entire circle's area.<\/p>\n

    \"gp_img15\"<\/a><\/p>\n

    First of all, that's the denominator-area in our probability.\u00a0 Also, we can use this to figure out the area of the shaded region.<\/p>\n

    Area of shaded region = (area of sector) \u2013 (area of triangle)<\/p>\n

    \"gp_img16\"<\/a><\/p>\n

    That's our numerator-area.\u00a0 Now, we can put together the probability --- we will just have to simply the complex fraction by multiplying numerator and denominator by 4.<\/p>\n

    \"gp_img17\"<\/a><\/p>\n

    Answer =\u00a0A<\/strong><\/p>\n

     <\/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 three articles in this series on GMAT probability questions, I discussed the\u00a0AND and OR\u00a0probability rules, \u00a0“at least” probability questions, and\u00a0probability questions that involve counting.\u00a0 This post covers…<\/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],"tags":[],"class_list":["post-17024","post","type-post","status-publish","format-standard","hentry","category-gmat","category-magoosh-blog","category-blog","entry"],"aioseo_notices":[],"_links":{"self":[{"href":"https:\/\/gmatclub.com\/blog\/wp-json\/wp\/v2\/posts\/17024","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=17024"}],"version-history":[{"count":3,"href":"https:\/\/gmatclub.com\/blog\/wp-json\/wp\/v2\/posts\/17024\/revisions"}],"predecessor-version":[{"id":17033,"href":"https:\/\/gmatclub.com\/blog\/wp-json\/wp\/v2\/posts\/17024\/revisions\/17033"}],"wp:attachment":[{"href":"https:\/\/gmatclub.com\/blog\/wp-json\/wp\/v2\/media?parent=17024"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/gmatclub.com\/blog\/wp-json\/wp\/v2\/categories?post=17024"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/gmatclub.com\/blog\/wp-json\/wp\/v2\/tags?post=17024"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}