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First, some practice questions.\u00a0 The scenario below is relevant to questions #1-#3.<\/p>\nFirst, some practice questions.\u00a0 The scenario below is relevant to questions #1-#3.<\/p>\n
There are two sets of letters, and you are going to pick exactly one letter from each set.<\/p>\n
Set #1 = {A, B, C, D, E}<\/p>\n
Set #2 = {K, L, M, N, O, P}<\/p>\n
1) What is the probability of picking a C\u00a0and\u00a0an M?<\/p>\n
2) What is the probability of picking a C\u00a0or\u00a0an M?<\/p>\n
3) What is the probability of picking two vowels?<\/p>\n
_______________________________________________________________________________<\/p>\n
4) In a certain corporation, there are 300 male employees and 100 female employees.\u00a0 It is known that 20% of the male employees have advanced degrees and 40% of the females have advanced degrees.\u00a0 If one of the 400 employees is chosen at random, what is the probability this employee has an advanced degree\u00a0and\u00a0is female?<\/p>\n
5) In a certain corporation, there are 300 male employees and 100 female employees.\u00a0 It is known that 20% of the male employees have advanced degrees and 40% of the females have advanced degrees.\u00a0 If one of the 400 employees is chosen at random, what is the probability this employee has an advanced degree\u00a0or\u00a0is female?<\/p>\n
<\/p>\n
Here is the absolute bare minimum you need to know for probability calculations on the GMAT:<\/p>\n
\"AND\" means MULTIPLY<\/strong><\/p>\n \"OR\" means ADD<\/strong><\/p>\n Is this the whole story? Well, not exactly.\u00a0 But if you can't remember or don't understand anything else about probability, at least know these two bare-bones rules, because just this will put ahead of so many people.\u00a0 Just this is enough to solve the problems #1 and #3, although these alone could lead to problems on the others.\u00a0 Before we qualify these simplistic rules, we need to discuss two distinctions.<\/p>\n <\/p>\n Two events are disjoint if they are mutually exclusive.\u00a0 In other words, two events are disjoint if the probability of their simultaneous occurrence is zero, that is, it is absolutely impossible to have them both happen at the same time.\u00a0\u00a0 For example, different faces of a single die are disjoint: under ordinary circumstances, if you roll one die once, you can't simultaneously get, say, both a 3 and a 5.\u00a0\u00a0 Those two are disjoint.\u00a0 Suppose we are picking random people and classifying them by their current age.\u00a0 In this process, being in the category \"teenager\" and being in the category \"senior citizen\" are disjoint: there is no one we could pick who is simultaneously in both categories.\u00a0\u00a0 Most categories involving human beings are too messy for the distinction \"disjoint\" to apply.<\/p>\n If events\u00a0A<\/strong>\u00a0and\u00a0B<\/strong>\u00a0are\u00a0disjoint<\/strong>, then we can use the\u00a0simplified OR rule<\/strong>:<\/p>\n P(A or B) = P(A) + P(B)<\/strong><\/p>\n That's the case in which the simplified rule, OR means ADD, works perfectly.\u00a0 If events A and B are\u00a0not<\/strong>\u00a0disjoint<\/strong>, then we have to use the\u00a0generalized OR rule<\/strong>:<\/p>\n P(A or B) = P(A) + P(B) \u2013 P(A and B)<\/strong><\/p>\n The reason for that final term: we need to subtract the overlap.\u00a0 The events in the region \"A and B\" are included in region A and also in region B, so if we add those two regions, the overlap gets counted twice.\u00a0 We need to subtract it, so it is only counted once like everything else.<\/p>\n <\/p>\n Two events are independent if whether one happens has absolutely no influence on whether the other happens.\u00a0 In other words, knowing about the outcome of one event gives absolutely no information about how the other event will turn out.\u00a0\u00a0 For example, if I roll two ordinary dice, the outcome of each die is independent of the other die.\u00a0 If I tell you I rolled two dice, and the first die was a 4, then knowing that give you no clue about what the number on the other die might be.\u00a0\u00a0 On any given day, what the weather is in the San Francisco Bay Area and how the Dow Jones Industrial Average performs are independent: knowing one gives us absolutely no information about how the other turned out.<\/p>\n We have to be careful.\u00a0 If I shuffle a deck of cards, draw one, replace it, re-shuffle, and draw another, then the two cards are independent.\u00a0 BUT, if I shuffle the deck, draw one card, and then without replacement draw a second card, then they are not independent.\u00a0 For example, if the first card is the 7 of Hearts, then it is less likely that the second card would be either a 7 or a Heart, because there are fewer of those options among the remaining 51 cards.<\/p>\n Also, notice that there are many human situations which\u00a0would be<\/em>\u00a0independent in a perfect just ideal world, but regrettably are not independent in a real world full of inequities.\u00a0\u00a0 In a perfect world, gender and corporate promotion would be independent, but in practice, they are not.\u00a0 In a perfect world, race and criminal conviction would be independent, but in practice, they are not.<\/p>\n If events\u00a0A<\/strong>\u00a0and\u00a0B<\/strong>\u00a0are\u00a0independent<\/strong>, then we can use the\u00a0simplified AND rule<\/strong>:<\/p>\n P(A and B) = P(A)*P(B)<\/strong><\/p>\n That's the case in which the simplified rule, AND means MULTIPLY, works perfectly.\u00a0 If events A and B are\u00a0not<\/strong>\u00a0disjoint<\/strong>, then things get complicated.\u00a0 Technically, the \"generalized AND rule\" formula would involve a concept known as \"conditional probability\", which would lead into realms of probability theory that are rarely, if at all, tested on the GMAT.\u00a0 For GMAT purposes, if you find yourself in this situation, you will have to think through the particulars of the situation (as I will demonstrate below): just know that you can't simply rely on a formula in this case.<\/p>\n <\/p>\n Having read this post, take another look at those practice questions, and see if you understand them better, before simply reading the explanations below.\u00a0\u00a0 Be patient with yourself as you work through probability: it takes time to internalize these distinctions, such as \"disjoint\" and \"independent\".\u00a0 There will be more information in the\u00a0next post<\/a>\u00a0in this sequence.<\/p>\n <\/p>\n 1) Whatever we pick from the first set is independent with whatever we pick from the second set, so we can use the simplified AND rule.<\/p>\n P(first pick = C) = 1\/5<\/p>\n P(second pick = M) = 1\/6<\/p>\n P(C and M) = P(C)*P(M) = (1\/5)*(1\/6) = 1\/30<\/p>\n Answer =\u00a0A<\/strong><\/p>\n 2) Picking an M is not disjoint with picking a C --- they both could happen on the same round of the game.\u00a0\u00a0 We have to use the generalized OR rule for this:<\/p>\n P(C or M) = P(C) + P(M) \u2013 P(C and M)<\/p>\n Fortunately, we know the first two, and we calculated the value of the third term already in #1.<\/p>\n P(C or M) = P(C) + P(M) \u2013 P(C and M)<\/p>\n Answer =\u00a0E<\/strong><\/p>\n 3) On the first pick, two of the five letters are vowels --- A & E --- so the probability of picking a vowel on the first pick is 2\/5.\u00a0\u00a0 On the second pick, only one letter out of the six is a vowel --- O --- so the probability of picking a vowel on the second pick is 1\/6.\u00a0 The two picks are independent: what one selects from one set has absolutely no bearing on what one picks from the other set.\u00a0\u00a0 Therefore, we can use the generalized AND rule.<\/p>\n P(two vowels) = P(vowel on first pick)*P(vowel on second pick) =(2\/5)*(1\/6) = 2\/30 = 1\/15<\/p>\n Answer =\u00a0B<\/strong><\/p>\n 4) Here we have an AND question, and the parameters --- gender and advanced degree --- are not independent.\u00a0 If I tell you the gender of a certain employee, then that give me information about how likely it is that this employee has an advanced degree.\u00a0\u00a0 One parameter gives information about the other, which means they are not independent.\u00a0 Therefore, we cannot use the simplified AND rule.\u00a0 Fortunately, it is relatively easy here to calculate everything directly.<\/p>\n There are 100 female employees, and we know 40% of them have advanced degrees, so there are 40 employees who both are female and have an advanced degree.\u00a0 That's the number of employees in the AND region.\u00a0 Well, there are 400 employees altogether.\u00a0 Of these 400 total employees, the probability of picking someone in this AND region is<\/p>\n P = 40\/400 = 1\/10<\/p>\n Answer =\u00a0B<\/strong><\/p>\n 5)\u00a0 In this corporation, there are 400 total employees.\u00a0 There are 100 women.\u00a0 Of the 300 men, 20% have advanced degrees ----10% of 300 must be 30, so 20% of 300 must be 60.\u00a0 Add the women and the men with advanced degrees: 100 + 60 = 160.\u00a0\u00a0 This is the OR region, full set of individuals that satisfy the condition \"has an advanced degree or is female.\"\u00a0 Of the 400 employees, what's the probability of picking one of the 160 in this particular group?<\/p>\n P = 160\/400 = 16\/40 = 4\/10 = 2\/5<\/p>\n Answer =\u00a0D<\/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":" First, some practice questions.\u00a0 The scenario below is relevant to questions #1-#3. There are two sets of letters, and you are going to pick exactly one letter from each set.…<\/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-16229","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\/16229","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=16229"}],"version-history":[{"count":4,"href":"https:\/\/gmatclub.com\/blog\/wp-json\/wp\/v2\/posts\/16229\/revisions"}],"predecessor-version":[{"id":16237,"href":"https:\/\/gmatclub.com\/blog\/wp-json\/wp\/v2\/posts\/16229\/revisions\/16237"}],"wp:attachment":[{"href":"https:\/\/gmatclub.com\/blog\/wp-json\/wp\/v2\/media?parent=16229"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/gmatclub.com\/blog\/wp-json\/wp\/v2\/categories?post=16229"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/gmatclub.com\/blog\/wp-json\/wp\/v2\/tags?post=16229"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}Disjoint<\/h2>\n
Independent<\/h2>\n
Practice<\/h2>\n
Practice problem solutions<\/h2>\n
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<\/a><\/p>\n