{"id":401,"date":"2023-06-05T15:31:03","date_gmt":"2023-06-05T15:31:03","guid":{"rendered":"https:\/\/courses.lumenlearning.com\/tulsacc-math1473\/chapter\/working-with-events\/"},"modified":"2026-04-20T17:00:35","modified_gmt":"2026-04-20T17:00:35","slug":"working-with-events","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/tulsacc-math1473\/chapter\/working-with-events\/","title":{"raw":"Types of Events and Their Probabilities","rendered":"Types of Events and Their Probabilities"},"content":{"raw":"<div class=\"textbox learning-objectives\">\r\n<h3>Learning Outcomes<\/h3>\r\n<ul>\r\n \t<li>Calculate the probability of a complementary event.<\/li>\r\n \t<li>Identify if events are mutually exclusive.<\/li>\r\n \t<li>Use the Addition Rule to compute the probability of events combined with \"or\".<\/li>\r\n \t<li>Determine if events are independent.<\/li>\r\n \t<li>Use the Multiplication Rule to compute the probability of joint (simultaneous) events.<\/li>\r\n<\/ul>\r\n<\/div>\r\n<div class=\"textbox examples\">\r\n<h3>Recall operations on fractions<\/h3>\r\nAdding and subtracting fractions with common denominators\r\n\r\n[latex]\\dfrac{a}{c}\\pm \\dfrac{b}{c}=\\dfrac{a\\pm b}{c}[\/latex]\r\n\r\nIn the two equations below, note that this relationship is described in both directions.\r\n\r\nThat is, it is also true that\r\n\r\n[latex]\\dfrac{a\\pm b}{c}=\\dfrac{a}{c}\\pm \\dfrac{b}{c}[\/latex]\r\n\r\nThe second equation furthermore includes the fact that\r\n\r\n[latex]\\dfrac{a}{a}=1[\/latex]\r\n\r\n<\/div>\r\n<h2>Complementary Events<\/h2>\r\nNow let us examine the probability that an event does <strong>not<\/strong> happen. As in the previous section, consider the situation of rolling a six-sided die and first compute the probability of rolling a six: the answer is <em>P<\/em>(six) =1\/6. Now consider the probability that we do <em>not<\/em> roll a six: there are 5 outcomes that are not a six, so the answer is <em>P<\/em>(not a six) = [latex]\\frac{5}{6}[\/latex]. Notice that\r\n<p style=\"text-align: center;\">[latex]P(\\text{six})+P(\\text{not a six})=\\frac{1}{6}+\\frac{5}{6}=\\frac{6}{6}=1[\/latex]<\/p>\r\nThis is not a coincidence.\u00a0 Consider a generic situation with <em>n<\/em> possible outcomes and an event <em>E<\/em> that corresponds to <em>m<\/em> of these outcomes. Then the remaining <em>n<\/em> - <em>m<\/em> outcomes correspond to <em>E<\/em> not happening, thus\r\n<p style=\"text-align: center;\">[latex]P(\\text{not}E)=\\frac{n-m}{n}=\\frac{n}{n}-\\frac{m}{n}=1-\\frac{m}{n}=1-P(E)[\/latex]<\/p>\r\n<p style=\"text-align: center;\"><a href=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/1141\/2016\/12\/28183109\/ace-167052_640.jpg\"><img class=\"aligncenter size-full wp-image-991\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/1141\/2016\/12\/28183109\/ace-167052_640.jpg\" alt=\"scattered playing cards on a table. The Ace of Spades is on top.\" width=\"640\" height=\"426\" \/><\/a><\/p>\r\n\r\n<div class=\"textbox\">\r\n<h3>Complement of an Event<\/h3>\r\nThe <strong>complement<\/strong> of an event is the event \u201c<em>E<\/em> doesn\u2019t happen\u201d\r\n<ul>\r\n \t<li>The notation [latex]\\bar{E}[\/latex] is used for the complement of event <em>E<\/em>.<\/li>\r\n \t<li>We can compute the probability of the complement using [latex]P\\left({\\bar{E}}\\right)=1-P(E)[\/latex]<\/li>\r\n \t<li>Notice also that [latex]P(E)=1-P\\left({\\bar{E}}\\right)[\/latex]<\/li>\r\n<\/ul>\r\n<\/div>\r\n<div class=\"textbox exercises\">\r\n<h3>example<\/h3>\r\nIf you pull a random card from a deck of playing cards, what is the probability it is not a heart?\r\n[reveal-answer q=\"926985\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"926985\"]\r\n\r\nThere are 13 hearts in the deck, so [latex]P(\\text{heart})=\\frac{13}{52}=\\frac{1}{4}[\/latex].\r\n\r\nThe probability of <em>not<\/em> drawing a heart is the complement: [latex]P(\\text{not heart})=1-P(\\text{heart})=1-\\frac{1}{4}=\\frac{3}{4}[\/latex]\r\n\r\n[\/hidden-answer]\r\n\r\nThis situation is explained in the following video.\r\n\r\nhttps:\/\/youtu.be\/RnljiW6ZM6A\r\n\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Try It<\/h3>\r\n[ohm_question]7097[\/ohm_question]\r\n\r\n<\/div>\r\n<h2>Probability of either of two events occurring<\/h2>\r\n<div class=\"textbox exercises\">\r\n<h3>example<\/h3>\r\nSuppose we flipped a coin and rolled a die, and wanted to know the probability of getting a head on the coin <em>or<\/em> a 6 on the die.\r\n[reveal-answer q=\"443646\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"443646\"]\r\n\r\nHere, there are still 12 possible outcomes: {H1,H2,H3,H4,H5,H6,T1,T2,T3,T4,T5,T6}\r\n\r\nBy simply counting, we can see that 7 of the outcomes have a head on the coin <em>or<\/em> a 6 on the die <em>or<\/em> both \u2013 we use <em>or<\/em> inclusively here (these 7 outcomes are H1, H2, H3, H4, H5, H6, T6), so the probability is [latex]\\frac{7}{12}[\/latex]. How could we have found this from the individual probabilities?\r\n\r\nAs we would expect, [latex]\\frac{1}{2}[\/latex] of these outcomes have a head, and [latex]\\frac{1}{6}[\/latex] of these outcomes have a 6 on the die. If we add these, [latex]\\frac{1}{2}+\\frac{1}{6}=\\frac{6}{12}+\\frac{2}{12}=\\frac{8}{12}[\/latex], which is not the correct probability. Looking at the outcomes we can see why: the outcome H6 would have been counted twice, since it contains both a head and a 6; the probability of both a head <em>and<\/em> rolling a 6 is [latex]\\frac{1}{12}[\/latex].\r\n\r\nIf we subtract out this double count, we have the correct probability: [latex]\\frac{8}{12}-\\frac{1}{12}=\\frac{7}{12}[\/latex].\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox\">\r\n<h3><em>ADDITION RULE OF PROBABILITY:\u00a0 P<\/em>(<em>A<\/em> or <em>B<\/em>)<\/h3>\r\nThe probability of either <em>A<\/em> or <em>B <\/em>occurring (or both) is\r\n<p style=\"text-align: center;\">[latex]P(A\\text{ or }B)=P(A)+P(B)\u2013P(A\\text{ and }B)[\/latex]<\/p>\r\n\r\n<\/div>\r\n<div class=\"textbox exercises\">\r\n<h3>example<\/h3>\r\nSuppose we draw one card from a standard deck. What is the probability that we get a Queen or a King?\r\n[reveal-answer q=\"657503\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"657503\"]\r\n\r\nThere are 4 Queens and 4 Kings in the deck, hence 8 outcomes corresponding to a Queen or King out of 52 possible outcomes. Thus the probability of drawing a Queen or a King is:\r\n<p style=\"text-align: center;\">[latex]P(\\text{King or Queen})=\\frac{8}{52}[\/latex]<\/p>\r\nNote that in this case, there are no cards that are both a Queen and a King, so [latex]P(\\text{King and Queen})=0[\/latex]. Using our probability rule, we could have said:\r\n<p style=\"text-align: center;\">[latex]P(\\text{King or Queen})=P(\\text{King})+P(\\text{Queen})-P(\\text{King and Queen})=\\frac{4}{52}+\\frac{4}{52}-0=\\frac{8}{52}[\/latex]<\/p>\r\n[\/hidden-answer]\r\n\r\nSee more about this example and the previous one in the following video.\r\n\r\nhttps:\/\/youtu.be\/klbPZeH1np4\r\n\r\n<\/div>\r\nIn the last example, the events were mutually exclusive (cannot happen at the same time), so <strong><em>P<\/em>(<em>A<\/em> or <em>B<\/em>) = <em>P<\/em>(<em>A<\/em>) + <em>P<\/em>(<em>B<\/em>)<\/strong>.\r\n<div class=\"textbox\">\r\n<h3>Mutually Exclusive Events<\/h3>\r\nEvents A and B are mutually exclusive events if Event B and Event A <strong><em>cannot<\/em> occur at the same time<\/strong>.\r\n\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Try It<\/h3>\r\n[ohm_question]7113[\/ohm_question]\r\n\r\n<\/div>\r\n<div class=\"textbox exercises\">\r\n<h3>example<\/h3>\r\nSuppose we draw one card from a standard deck. What is the probability that we get a red card or a King?\r\n[reveal-answer q=\"649692\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"649692\"]\r\n\r\nHalf the cards are red, so [latex]P(\\text{red})=\\frac{26}{52}[\/latex]\r\n\r\nThere are four kings, so [latex]P(\\text{King})=\\frac{4}{52}[\/latex]\r\n\r\nThere are two red kings, so [latex]P(\\text{Red and King})=\\frac{2}{52}[\/latex]\r\n\r\nWe can then calculate\r\n<p style=\"text-align: center;\">[latex]P(\\text{Red or King})=P(\\text{Red})+P(\\text{King})-P(\\text{Red and King})=\\frac{26}{52}+\\frac{4}{52}-\\frac{2}{52}=\\frac{28}{52}[\/latex]<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Try It<\/h3>\r\nIn your drawer you have 10 pairs of socks, 6 of which are white, and 7 tee shirts, 3 of which are white. If you reach in and randomly grab a pair of socks and a tee shirt, what is the probability at least one is white?\r\n\r\n<\/div>\r\n<h2>Probability of two events occurring simultaneously<\/h2>\r\nThe previous examples looked at the probability of <em>either<\/em> event occurring. Now we will look at the probability of <em>both<\/em> events occurring.\r\n<div class=\"textbox exercises\">\r\n<h3>example<\/h3>\r\nSuppose we flipped a coin and rolled a die, and wanted to know the probability of getting a head on the coin and a 6 on the die.\r\n[reveal-answer q=\"453669\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"453669\"]\r\n\r\nWe could list all possible outcomes: \u00a0{H1,H2,H3,H4,H5,H6,T1,T2,T3,T4,T5,T6}.\r\n\r\nNotice there are [latex]2\\cdot6=12[\/latex] total outcomes. Out of these, only 1 is the desired outcome, so the probability is [latex]\\frac{1}{12}[\/latex].\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\nThe prior example contained\u00a0two <em>independent<\/em> events. Getting a certain outcome from rolling a die had no influence on the outcome from flipping the coin.\r\n<div class=\"textbox\">\r\n<h3>Independent Events<\/h3>\r\nEvents A and B are <strong>independent events<\/strong> if the probability of Event B occurring is the same whether or not Event A occurs.\r\n\r\n<\/div>\r\n<div class=\"textbox exercises\">\r\n<h3>example<\/h3>\r\nAre these events independent?\r\n<ol>\r\n \t<li>A fair coin is tossed two times. The two events are (1) first toss is a head and (2) second toss is a head.<\/li>\r\n \t<li>The two events (1) \"It will rain tomorrow in Houston\" and (2) \"It will rain tomorrow in Galveston\u201d (a city near Houston).<\/li>\r\n \t<li>You draw a card from a deck, then draw a second card without replacing the first.<\/li>\r\n<\/ol>\r\n[reveal-answer q=\"621789\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"621789\"]\r\n<ol>\r\n \t<li>The probability that a head comes up on the second toss is 1\/2 regardless of whether or not a head came up on the first toss, so these events are independent.<\/li>\r\n \t<li>These events are not independent because it is more likely that it will rain in Galveston on days it rains in Houston than on days it does not.<\/li>\r\n \t<li>The probability of the second card being red depends on whether the first card is red or not, so these events are not independent.<\/li>\r\n<\/ol>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\nWhen two events are independent, the probability of both occurring is the product of the probabilities of the individual events.\r\n<div class=\"textbox examples\">\r\n<h3>recall multiplying fractions<\/h3>\r\nTo multiply fractions, place the product of the numerators over the product of the denominators.\r\n\r\n[latex]\\dfrac{a}{b} \\cdot \\dfrac{c}{d} = \\dfrac{ac}{bd}[\/latex]\r\n\r\n<\/div>\r\n<div class=\"textbox\">\r\n<h3><em>MULTIPLICATION RULE OF PROBABILITY:\u00a0 P<\/em>(<em>A<\/em> and <em>B<\/em>) for independent events<\/h3>\r\nIf events <em>A<\/em> and <em>B<\/em> are independent, then the probability of both <em>A<\/em> and <em>B <\/em>occurring is\r\n<p style=\"text-align: center;\">[latex]P\\left(A\\text{ and }B\\right)=P\\left(A\\right)\\cdot{P}\\left(B\\right)[\/latex]<\/p>\r\nwhere <em>P<\/em>(<em>A<\/em> and <em>B<\/em>) is the probability of events <em>A<\/em> and <em>B<\/em> both occurring, <em>P<\/em>(<em>A<\/em>) is the probability of event <em>A<\/em> occurring, and <em>P<\/em>(<em>B<\/em>) is the probability of event <em>B<\/em> occurring\r\n\r\n<\/div>\r\nIf you look back at the coin and die example from earlier, you can see how the number of outcomes of the first event multiplied by the number of outcomes in the second event multiplied to equal the total number of possible outcomes in the combined event.\r\n<div class=\"textbox examples\">\r\n<h3><span style=\"text-align: initial; background-color: #dacacc;\">Recall fraction reduction<\/span><\/h3>\r\nTo write a fraction in reduced terms, first take the prime factorization of the numerator and denominator, then cancel out factors that are common in the numerator and the denominator.\r\n\r\nEx. [latex]\\dfrac{12}{18}=\\dfrac{\\cancel{2}\\cdot 2\\cdot \\cancel{3}}{\\cancel{2}\\cdot 3\\cdot \\cancel{3}}=\\dfrac{2}{3}[\/latex]\r\n\r\n<\/div>\r\n<div class=\"textbox exercises\">\r\n<h3>example<\/h3>\r\nIn your drawer you have 10 pairs of socks, 6 of which are white, and 7 tee shirts, 3 of which are white. If you randomly reach in and pull out a pair of socks and a tee shirt, what is the probability both are white?\r\n[reveal-answer q=\"410325\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"410325\"]\r\n\r\nThe probability of choosing a white pair of socks is [latex]\\frac{6}{10}[\/latex].\r\n\r\nThe probability of choosing a white tee shirt is [latex]\\frac{3}{7}[\/latex].\r\n\r\nThe probability of both being white is [latex]\\frac{6}{10}\\cdot\\frac{3}{7}=\\frac{18}{70}=\\frac{9}{35}[\/latex]\r\n\r\n[\/hidden-answer]\r\n\r\nExamples of joint probabilities are discussed in this video.\r\n\r\nhttps:\/\/youtu.be\/6F17WLp-EL8\r\n\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Try It<\/h3>\r\n[ohm_question]7111[\/ohm_question]\r\n\r\n<\/div>\r\n<div class=\"textbox exercises\">\r\n<h3>Example<\/h3>\r\nThe table below shows the number of survey subjects who have received and not received a speeding ticket in the last year, and the color of their car. Find the probability that a randomly chosen person:\r\n<ol>\r\n \t<li>Has a red car <em>and<\/em> got a speeding ticket<\/li>\r\n \t<li>Has a red car <em>or<\/em> got a speeding ticket.<\/li>\r\n<\/ol>\r\n<table>\r\n<tbody>\r\n<tr>\r\n<td><\/td>\r\n<td>Speeding ticket<\/td>\r\n<td>No speeding ticket<\/td>\r\n<td>Total<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>Red car<\/td>\r\n<td>15<\/td>\r\n<td>135<\/td>\r\n<td>150<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>Not red car<\/td>\r\n<td>45<\/td>\r\n<td>470<\/td>\r\n<td>515<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>Total<\/td>\r\n<td>60<\/td>\r\n<td>605<\/td>\r\n<td>665<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n[reveal-answer q=\"167092\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"167092\"]\r\n\r\nWe can see that 15 people of the 665 surveyed had both a red car and got a speeding ticket, so the probability is [latex]\\frac{15}{665}\\approx0.0226[\/latex].\r\n\r\nNotice that having a red car and getting a speeding ticket are not independent events, so the probability of both of them occurring is not simply the product of probabilities of each one occurring.\r\n\r\nWe could answer this question by simply adding up the numbers: 15 people with red cars and speeding tickets + 135 with red cars but no ticket + 45 with a ticket but no red car = 195 people. So the probability is [latex]\\frac{195}{665}\\approx0.2932[\/latex].\r\n\r\nWe also could have found this probability by:\r\n<p style=\"text-align: center;\">P(had a red car) + P(got a speeding ticket) \u2013 P(had a red car and got a speeding ticket)<\/p>\r\n<p style=\"text-align: center;\">= [latex]\\frac{150}{665}+\\frac{60}{665}-\\frac{15}{665}=\\frac{195}{665}[\/latex].<\/p>\r\n[\/hidden-answer]\r\n\r\nThis table example is detailed in the following explanatory video.\r\n\r\nhttps:\/\/youtu.be\/HWrGoM1yRaU\r\n\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Try It<\/h3>\r\n[ohm_question]7114[\/ohm_question]\r\n\r\n[ohm_question]7115[\/ohm_question]\r\n\r\n<\/div>\r\nNow, let's consider the situation where the events are not independent. When two events are dependent, the probability of both occurring is the product of the first event happening and the second event happening assuming the first one happened.\r\n\r\nFor example, if you draw a card from a deck, then the sample space for the next card drawn has changed, because you are now working with a deck of 51 cards. In the following example we will show you how the computations for events like this are different from the computations we did for independent events.\r\n<div class=\"textbox exercises\">\r\n<h3>example<\/h3>\r\nWhat is the probability that two cards drawn at random from a deck of playing cards will both be aces?\r\n[reveal-answer q=\"284277\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"284277\"]\r\n\r\nIt might seem that you could use the formula for the probability of two independent events and simply multiply [latex]\\frac{4}{52}\\cdot\\frac{4}{52}=\\frac{1}{169}[\/latex]. This would be incorrect, however, because the two events are not independent. If the first card drawn is an ace, then the probability that the second card is also an ace would be lower because there would only be three aces left in the deck.\r\n\r\nWe assume the first card chosen is an ace, then the probability that the second card chosen is also an ace is dependent because there are now 3 aces in the remaining 51 cards. Symbolically, we write this as:\r\n\r\n<em>P<\/em>(ace on second draw | an ace on the first draw).\r\n\r\nThe vertical bar \"|\" is read as \"given,\" so the above expression is short for \"The probability that an ace is drawn on the second draw given that an ace was drawn on the first draw.\"\r\n\r\nThus, the probability of both cards being aces is [latex]\\frac{4}{52}\\cdot\\frac{3}{51}=\\frac{12}{2652}=\\frac{1}{221}[\/latex].\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox\">\r\n<h3><em>MULTIPLICATION RULE OF PROBABILITY:\u00a0\u00a0<\/em><em>P<\/em>(<em>A<\/em> and <em>B<\/em>) for dependent events<\/h3>\r\nIf Events <em>A<\/em> and <em>B<\/em> are not independent, then\u00a0<em>P<\/em>(<em>A<\/em> and <em>B<\/em>) = <em>P<\/em>(<em>A<\/em>) \u00b7 <em>P<\/em>(<em>B<\/em> | <em>A<\/em>)\r\n\r\n<\/div>\r\n<div class=\"textbox examples\">\r\n<h3>Converting a fraction to decimal form<\/h3>\r\nProbabilities can be expressed in fraction or decimal form. To convert a fraction to a decimal, use a calculator to divide the numerator by the denominator.\r\n\r\nEx. [latex]\\dfrac{19}{51}=19 \\div 51 \\approx 0.3725[\/latex]\r\n\r\n<\/div>\r\n<div class=\"textbox exercises\">\r\n<h3>example<\/h3>\r\nIf you pull 2 cards out of a deck, what is the probability that both are spades?\r\n[reveal-answer q=\"400876\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"400876\"]\r\n\r\nThe probability that the first card is a spade is [latex]\\frac{13}{52}[\/latex].\r\n\r\nThe probability that the second card is a spade, is dependent on the first being a spade, is [latex]\\frac{12}{51}[\/latex], since there is one less spade in the deck, and one less total cards.\r\n\r\nThe probability that both cards are spades is [latex]\\frac{13}{52}\\cdot\\frac{12}{51}=\\frac{156}{2652}\\approx0.0588[\/latex]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Try It<\/h3>\r\n[ohm_question]7118[\/ohm_question]\r\n\r\n<\/div>\r\n<div class=\"textbox exercises\">\r\n<h3>Example<\/h3>\r\nIf you draw two cards from a deck, what is the probability that you will get the Ace of Diamonds and a black card?\r\n[reveal-answer q=\"774421\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"774421\"]\r\n\r\nYou can satisfy this condition by having Case A or Case B, as follows:\r\n\r\nCase A) you can get the Ace of Diamonds first and then a black card or\r\n\r\nCase B) you can get a black card first and then the Ace of Diamonds.\r\n\r\nLet's calculate the probability of Case A. The probability that the first card is the Ace of Diamonds is [latex]\\frac{1}{52}[\/latex]. The probability that the second card is black given that the first card is the Ace of Diamonds is [latex]\\frac{26}{51}[\/latex] because 26 of the remaining 51 cards are black. The probability is therefore [latex]\\frac{1}{52}\\cdot\\frac{26}{51}=\\frac{1}{102}[\/latex].\r\n\r\nNow for Case B: the probability that the first card is black is [latex]\\frac{26}{52}=\\frac{1}{2}[\/latex]. The probability that the second card is the Ace of Diamonds given that the first card is black is [latex]\\frac{1}{51}[\/latex]. The probability of Case B is therefore [latex]\\frac{1}{2}\\cdot\\frac{1}{51}=\\frac{1}{102}[\/latex], the same as the probability of Case 1.\r\n\r\nRecall that the probability of A or B is <em>P<\/em>(A) + <em>P<\/em>(B) - <em>P<\/em>(A and B). In this problem, <em>P<\/em>(case A and case B) = 0 since the first card cannot be the Ace of Diamonds and be a black card. Therefore, the probability of Case A or Case B is [latex]\\frac{1}{102\r\n}+\\frac{1}{102}=\\frac{2}{102}=\\frac{1}{51}[\/latex]. The probability that you will get the Ace of Diamonds and a black card when drawing two cards from a deck is [latex]\\frac{1}{51}[\/latex].\r\n\r\n[\/hidden-answer]\r\n\r\nThese two playing card scenarios are discussed further in the following video.\r\n\r\nhttps:\/\/youtu.be\/ngyGsgV4_0U\r\n\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Try It<\/h3>\r\n[ohm_question]7110[\/ohm_question]\r\n\r\n<\/div>","rendered":"<div class=\"textbox learning-objectives\">\n<h3>Learning Outcomes<\/h3>\n<ul>\n<li>Calculate the probability of a complementary event.<\/li>\n<li>Identify if events are mutually exclusive.<\/li>\n<li>Use the Addition Rule to compute the probability of events combined with &#8220;or&#8221;.<\/li>\n<li>Determine if events are independent.<\/li>\n<li>Use the Multiplication Rule to compute the probability of joint (simultaneous) events.<\/li>\n<\/ul>\n<\/div>\n<div class=\"textbox examples\">\n<h3>Recall operations on fractions<\/h3>\n<p>Adding and subtracting fractions with common denominators<\/p>\n<p>[latex]\\dfrac{a}{c}\\pm \\dfrac{b}{c}=\\dfrac{a\\pm b}{c}[\/latex]<\/p>\n<p>In the two equations below, note that this relationship is described in both directions.<\/p>\n<p>That is, it is also true that<\/p>\n<p>[latex]\\dfrac{a\\pm b}{c}=\\dfrac{a}{c}\\pm \\dfrac{b}{c}[\/latex]<\/p>\n<p>The second equation furthermore includes the fact that<\/p>\n<p>[latex]\\dfrac{a}{a}=1[\/latex]<\/p>\n<\/div>\n<h2>Complementary Events<\/h2>\n<p>Now let us examine the probability that an event does <strong>not<\/strong> happen. As in the previous section, consider the situation of rolling a six-sided die and first compute the probability of rolling a six: the answer is <em>P<\/em>(six) =1\/6. Now consider the probability that we do <em>not<\/em> roll a six: there are 5 outcomes that are not a six, so the answer is <em>P<\/em>(not a six) = [latex]\\frac{5}{6}[\/latex]. Notice that<\/p>\n<p style=\"text-align: center;\">[latex]P(\\text{six})+P(\\text{not a six})=\\frac{1}{6}+\\frac{5}{6}=\\frac{6}{6}=1[\/latex]<\/p>\n<p>This is not a coincidence.\u00a0 Consider a generic situation with <em>n<\/em> possible outcomes and an event <em>E<\/em> that corresponds to <em>m<\/em> of these outcomes. Then the remaining <em>n<\/em> &#8211; <em>m<\/em> outcomes correspond to <em>E<\/em> not happening, thus<\/p>\n<p style=\"text-align: center;\">[latex]P(\\text{not}E)=\\frac{n-m}{n}=\\frac{n}{n}-\\frac{m}{n}=1-\\frac{m}{n}=1-P(E)[\/latex]<\/p>\n<p style=\"text-align: center;\"><a href=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/1141\/2016\/12\/28183109\/ace-167052_640.jpg\"><img loading=\"lazy\" decoding=\"async\" class=\"aligncenter size-full wp-image-991\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/1141\/2016\/12\/28183109\/ace-167052_640.jpg\" alt=\"scattered playing cards on a table. The Ace of Spades is on top.\" width=\"640\" height=\"426\" \/><\/a><\/p>\n<div class=\"textbox\">\n<h3>Complement of an Event<\/h3>\n<p>The <strong>complement<\/strong> of an event is the event \u201c<em>E<\/em> doesn\u2019t happen\u201d<\/p>\n<ul>\n<li>The notation [latex]\\bar{E}[\/latex] is used for the complement of event <em>E<\/em>.<\/li>\n<li>We can compute the probability of the complement using [latex]P\\left({\\bar{E}}\\right)=1-P(E)[\/latex]<\/li>\n<li>Notice also that [latex]P(E)=1-P\\left({\\bar{E}}\\right)[\/latex]<\/li>\n<\/ul>\n<\/div>\n<div class=\"textbox exercises\">\n<h3>example<\/h3>\n<p>If you pull a random card from a deck of playing cards, what is the probability it is not a heart?<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q926985\">Show Solution<\/span><\/p>\n<div id=\"q926985\" class=\"hidden-answer\" style=\"display: none\">\n<p>There are 13 hearts in the deck, so [latex]P(\\text{heart})=\\frac{13}{52}=\\frac{1}{4}[\/latex].<\/p>\n<p>The probability of <em>not<\/em> drawing a heart is the complement: [latex]P(\\text{not heart})=1-P(\\text{heart})=1-\\frac{1}{4}=\\frac{3}{4}[\/latex]<\/p>\n<\/div>\n<\/div>\n<p>This situation is explained in the following video.<\/p>\n<p><iframe loading=\"lazy\" id=\"oembed-1\" title=\"Probability - Complements\" width=\"500\" height=\"281\" src=\"https:\/\/www.youtube.com\/embed\/RnljiW6ZM6A?feature=oembed&#38;rel=0\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/p>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p><iframe loading=\"lazy\" id=\"ohm7097\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=7097&theme=oea&iframe_resize_id=ohm7097&show_question_numbers\" width=\"100%\" height=\"150\"><\/iframe><\/p>\n<\/div>\n<h2>Probability of either of two events occurring<\/h2>\n<div class=\"textbox exercises\">\n<h3>example<\/h3>\n<p>Suppose we flipped a coin and rolled a die, and wanted to know the probability of getting a head on the coin <em>or<\/em> a 6 on the die.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q443646\">Show Solution<\/span><\/p>\n<div id=\"q443646\" class=\"hidden-answer\" style=\"display: none\">\n<p>Here, there are still 12 possible outcomes: {H1,H2,H3,H4,H5,H6,T1,T2,T3,T4,T5,T6}<\/p>\n<p>By simply counting, we can see that 7 of the outcomes have a head on the coin <em>or<\/em> a 6 on the die <em>or<\/em> both \u2013 we use <em>or<\/em> inclusively here (these 7 outcomes are H1, H2, H3, H4, H5, H6, T6), so the probability is [latex]\\frac{7}{12}[\/latex]. How could we have found this from the individual probabilities?<\/p>\n<p>As we would expect, [latex]\\frac{1}{2}[\/latex] of these outcomes have a head, and [latex]\\frac{1}{6}[\/latex] of these outcomes have a 6 on the die. If we add these, [latex]\\frac{1}{2}+\\frac{1}{6}=\\frac{6}{12}+\\frac{2}{12}=\\frac{8}{12}[\/latex], which is not the correct probability. Looking at the outcomes we can see why: the outcome H6 would have been counted twice, since it contains both a head and a 6; the probability of both a head <em>and<\/em> rolling a 6 is [latex]\\frac{1}{12}[\/latex].<\/p>\n<p>If we subtract out this double count, we have the correct probability: [latex]\\frac{8}{12}-\\frac{1}{12}=\\frac{7}{12}[\/latex].<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox\">\n<h3><em>ADDITION RULE OF PROBABILITY:\u00a0 P<\/em>(<em>A<\/em> or <em>B<\/em>)<\/h3>\n<p>The probability of either <em>A<\/em> or <em>B <\/em>occurring (or both) is<\/p>\n<p style=\"text-align: center;\">[latex]P(A\\text{ or }B)=P(A)+P(B)\u2013P(A\\text{ and }B)[\/latex]<\/p>\n<\/div>\n<div class=\"textbox exercises\">\n<h3>example<\/h3>\n<p>Suppose we draw one card from a standard deck. What is the probability that we get a Queen or a King?<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q657503\">Show Solution<\/span><\/p>\n<div id=\"q657503\" class=\"hidden-answer\" style=\"display: none\">\n<p>There are 4 Queens and 4 Kings in the deck, hence 8 outcomes corresponding to a Queen or King out of 52 possible outcomes. Thus the probability of drawing a Queen or a King is:<\/p>\n<p style=\"text-align: center;\">[latex]P(\\text{King or Queen})=\\frac{8}{52}[\/latex]<\/p>\n<p>Note that in this case, there are no cards that are both a Queen and a King, so [latex]P(\\text{King and Queen})=0[\/latex]. Using our probability rule, we could have said:<\/p>\n<p style=\"text-align: center;\">[latex]P(\\text{King or Queen})=P(\\text{King})+P(\\text{Queen})-P(\\text{King and Queen})=\\frac{4}{52}+\\frac{4}{52}-0=\\frac{8}{52}[\/latex]<\/p>\n<\/div>\n<\/div>\n<p>See more about this example and the previous one in the following video.<\/p>\n<p><iframe loading=\"lazy\" id=\"oembed-2\" title=\"Probability of two events: P(A or B)\" width=\"500\" height=\"281\" src=\"https:\/\/www.youtube.com\/embed\/klbPZeH1np4?feature=oembed&#38;rel=0\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/p>\n<\/div>\n<p>In the last example, the events were mutually exclusive (cannot happen at the same time), so <strong><em>P<\/em>(<em>A<\/em> or <em>B<\/em>) = <em>P<\/em>(<em>A<\/em>) + <em>P<\/em>(<em>B<\/em>)<\/strong>.<\/p>\n<div class=\"textbox\">\n<h3>Mutually Exclusive Events<\/h3>\n<p>Events A and B are mutually exclusive events if Event B and Event A <strong><em>cannot<\/em> occur at the same time<\/strong>.<\/p>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p><iframe loading=\"lazy\" id=\"ohm7113\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=7113&theme=oea&iframe_resize_id=ohm7113&show_question_numbers\" width=\"100%\" height=\"150\"><\/iframe><\/p>\n<\/div>\n<div class=\"textbox exercises\">\n<h3>example<\/h3>\n<p>Suppose we draw one card from a standard deck. What is the probability that we get a red card or a King?<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q649692\">Show Solution<\/span><\/p>\n<div id=\"q649692\" class=\"hidden-answer\" style=\"display: none\">\n<p>Half the cards are red, so [latex]P(\\text{red})=\\frac{26}{52}[\/latex]<\/p>\n<p>There are four kings, so [latex]P(\\text{King})=\\frac{4}{52}[\/latex]<\/p>\n<p>There are two red kings, so [latex]P(\\text{Red and King})=\\frac{2}{52}[\/latex]<\/p>\n<p>We can then calculate<\/p>\n<p style=\"text-align: center;\">[latex]P(\\text{Red or King})=P(\\text{Red})+P(\\text{King})-P(\\text{Red and King})=\\frac{26}{52}+\\frac{4}{52}-\\frac{2}{52}=\\frac{28}{52}[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p>In your drawer you have 10 pairs of socks, 6 of which are white, and 7 tee shirts, 3 of which are white. If you reach in and randomly grab a pair of socks and a tee shirt, what is the probability at least one is white?<\/p>\n<\/div>\n<h2>Probability of two events occurring simultaneously<\/h2>\n<p>The previous examples looked at the probability of <em>either<\/em> event occurring. Now we will look at the probability of <em>both<\/em> events occurring.<\/p>\n<div class=\"textbox exercises\">\n<h3>example<\/h3>\n<p>Suppose we flipped a coin and rolled a die, and wanted to know the probability of getting a head on the coin and a 6 on the die.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q453669\">Show Solution<\/span><\/p>\n<div id=\"q453669\" class=\"hidden-answer\" style=\"display: none\">\n<p>We could list all possible outcomes: \u00a0{H1,H2,H3,H4,H5,H6,T1,T2,T3,T4,T5,T6}.<\/p>\n<p>Notice there are [latex]2\\cdot6=12[\/latex] total outcomes. Out of these, only 1 is the desired outcome, so the probability is [latex]\\frac{1}{12}[\/latex].<\/p>\n<\/div>\n<\/div>\n<\/div>\n<p>The prior example contained\u00a0two <em>independent<\/em> events. Getting a certain outcome from rolling a die had no influence on the outcome from flipping the coin.<\/p>\n<div class=\"textbox\">\n<h3>Independent Events<\/h3>\n<p>Events A and B are <strong>independent events<\/strong> if the probability of Event B occurring is the same whether or not Event A occurs.<\/p>\n<\/div>\n<div class=\"textbox exercises\">\n<h3>example<\/h3>\n<p>Are these events independent?<\/p>\n<ol>\n<li>A fair coin is tossed two times. The two events are (1) first toss is a head and (2) second toss is a head.<\/li>\n<li>The two events (1) &#8220;It will rain tomorrow in Houston&#8221; and (2) &#8220;It will rain tomorrow in Galveston\u201d (a city near Houston).<\/li>\n<li>You draw a card from a deck, then draw a second card without replacing the first.<\/li>\n<\/ol>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q621789\">Show Solution<\/span><\/p>\n<div id=\"q621789\" class=\"hidden-answer\" style=\"display: none\">\n<ol>\n<li>The probability that a head comes up on the second toss is 1\/2 regardless of whether or not a head came up on the first toss, so these events are independent.<\/li>\n<li>These events are not independent because it is more likely that it will rain in Galveston on days it rains in Houston than on days it does not.<\/li>\n<li>The probability of the second card being red depends on whether the first card is red or not, so these events are not independent.<\/li>\n<\/ol>\n<\/div>\n<\/div>\n<\/div>\n<p>When two events are independent, the probability of both occurring is the product of the probabilities of the individual events.<\/p>\n<div class=\"textbox examples\">\n<h3>recall multiplying fractions<\/h3>\n<p>To multiply fractions, place the product of the numerators over the product of the denominators.<\/p>\n<p>[latex]\\dfrac{a}{b} \\cdot \\dfrac{c}{d} = \\dfrac{ac}{bd}[\/latex]<\/p>\n<\/div>\n<div class=\"textbox\">\n<h3><em>MULTIPLICATION RULE OF PROBABILITY:\u00a0 P<\/em>(<em>A<\/em> and <em>B<\/em>) for independent events<\/h3>\n<p>If events <em>A<\/em> and <em>B<\/em> are independent, then the probability of both <em>A<\/em> and <em>B <\/em>occurring is<\/p>\n<p style=\"text-align: center;\">[latex]P\\left(A\\text{ and }B\\right)=P\\left(A\\right)\\cdot{P}\\left(B\\right)[\/latex]<\/p>\n<p>where <em>P<\/em>(<em>A<\/em> and <em>B<\/em>) is the probability of events <em>A<\/em> and <em>B<\/em> both occurring, <em>P<\/em>(<em>A<\/em>) is the probability of event <em>A<\/em> occurring, and <em>P<\/em>(<em>B<\/em>) is the probability of event <em>B<\/em> occurring<\/p>\n<\/div>\n<p>If you look back at the coin and die example from earlier, you can see how the number of outcomes of the first event multiplied by the number of outcomes in the second event multiplied to equal the total number of possible outcomes in the combined event.<\/p>\n<div class=\"textbox examples\">\n<h3><span style=\"text-align: initial; background-color: #dacacc;\">Recall fraction reduction<\/span><\/h3>\n<p>To write a fraction in reduced terms, first take the prime factorization of the numerator and denominator, then cancel out factors that are common in the numerator and the denominator.<\/p>\n<p>Ex. [latex]\\dfrac{12}{18}=\\dfrac{\\cancel{2}\\cdot 2\\cdot \\cancel{3}}{\\cancel{2}\\cdot 3\\cdot \\cancel{3}}=\\dfrac{2}{3}[\/latex]<\/p>\n<\/div>\n<div class=\"textbox exercises\">\n<h3>example<\/h3>\n<p>In your drawer you have 10 pairs of socks, 6 of which are white, and 7 tee shirts, 3 of which are white. If you randomly reach in and pull out a pair of socks and a tee shirt, what is the probability both are white?<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q410325\">Show Solution<\/span><\/p>\n<div id=\"q410325\" class=\"hidden-answer\" style=\"display: none\">\n<p>The probability of choosing a white pair of socks is [latex]\\frac{6}{10}[\/latex].<\/p>\n<p>The probability of choosing a white tee shirt is [latex]\\frac{3}{7}[\/latex].<\/p>\n<p>The probability of both being white is [latex]\\frac{6}{10}\\cdot\\frac{3}{7}=\\frac{18}{70}=\\frac{9}{35}[\/latex]<\/p>\n<\/div>\n<\/div>\n<p>Examples of joint probabilities are discussed in this video.<\/p>\n<p><iframe loading=\"lazy\" id=\"oembed-3\" title=\"Joint probabilities of independent events: P(A and B)\" width=\"500\" height=\"281\" src=\"https:\/\/www.youtube.com\/embed\/6F17WLp-EL8?feature=oembed&#38;rel=0\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/p>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p><iframe loading=\"lazy\" id=\"ohm7111\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=7111&theme=oea&iframe_resize_id=ohm7111&show_question_numbers\" width=\"100%\" height=\"150\"><\/iframe><\/p>\n<\/div>\n<div class=\"textbox exercises\">\n<h3>Example<\/h3>\n<p>The table below shows the number of survey subjects who have received and not received a speeding ticket in the last year, and the color of their car. Find the probability that a randomly chosen person:<\/p>\n<ol>\n<li>Has a red car <em>and<\/em> got a speeding ticket<\/li>\n<li>Has a red car <em>or<\/em> got a speeding ticket.<\/li>\n<\/ol>\n<table>\n<tbody>\n<tr>\n<td><\/td>\n<td>Speeding ticket<\/td>\n<td>No speeding ticket<\/td>\n<td>Total<\/td>\n<\/tr>\n<tr>\n<td>Red car<\/td>\n<td>15<\/td>\n<td>135<\/td>\n<td>150<\/td>\n<\/tr>\n<tr>\n<td>Not red car<\/td>\n<td>45<\/td>\n<td>470<\/td>\n<td>515<\/td>\n<\/tr>\n<tr>\n<td>Total<\/td>\n<td>60<\/td>\n<td>605<\/td>\n<td>665<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q167092\">Show Solution<\/span><\/p>\n<div id=\"q167092\" class=\"hidden-answer\" style=\"display: none\">\n<p>We can see that 15 people of the 665 surveyed had both a red car and got a speeding ticket, so the probability is [latex]\\frac{15}{665}\\approx0.0226[\/latex].<\/p>\n<p>Notice that having a red car and getting a speeding ticket are not independent events, so the probability of both of them occurring is not simply the product of probabilities of each one occurring.<\/p>\n<p>We could answer this question by simply adding up the numbers: 15 people with red cars and speeding tickets + 135 with red cars but no ticket + 45 with a ticket but no red car = 195 people. So the probability is [latex]\\frac{195}{665}\\approx0.2932[\/latex].<\/p>\n<p>We also could have found this probability by:<\/p>\n<p style=\"text-align: center;\">P(had a red car) + P(got a speeding ticket) \u2013 P(had a red car and got a speeding ticket)<\/p>\n<p style=\"text-align: center;\">= [latex]\\frac{150}{665}+\\frac{60}{665}-\\frac{15}{665}=\\frac{195}{665}[\/latex].<\/p>\n<\/div>\n<\/div>\n<p>This table example is detailed in the following explanatory video.<\/p>\n<p><iframe loading=\"lazy\" id=\"oembed-4\" title=\"Probabilities from a table: AND and OR\" width=\"500\" height=\"281\" src=\"https:\/\/www.youtube.com\/embed\/HWrGoM1yRaU?feature=oembed&#38;rel=0\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/p>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p><iframe loading=\"lazy\" id=\"ohm7114\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=7114&theme=oea&iframe_resize_id=ohm7114&show_question_numbers\" width=\"100%\" height=\"150\"><\/iframe><\/p>\n<p><iframe loading=\"lazy\" id=\"ohm7115\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=7115&theme=oea&iframe_resize_id=ohm7115&show_question_numbers\" width=\"100%\" height=\"150\"><\/iframe><\/p>\n<\/div>\n<p>Now, let&#8217;s consider the situation where the events are not independent. When two events are dependent, the probability of both occurring is the product of the first event happening and the second event happening assuming the first one happened.<\/p>\n<p>For example, if you draw a card from a deck, then the sample space for the next card drawn has changed, because you are now working with a deck of 51 cards. In the following example we will show you how the computations for events like this are different from the computations we did for independent events.<\/p>\n<div class=\"textbox exercises\">\n<h3>example<\/h3>\n<p>What is the probability that two cards drawn at random from a deck of playing cards will both be aces?<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q284277\">Show Solution<\/span><\/p>\n<div id=\"q284277\" class=\"hidden-answer\" style=\"display: none\">\n<p>It might seem that you could use the formula for the probability of two independent events and simply multiply [latex]\\frac{4}{52}\\cdot\\frac{4}{52}=\\frac{1}{169}[\/latex]. This would be incorrect, however, because the two events are not independent. If the first card drawn is an ace, then the probability that the second card is also an ace would be lower because there would only be three aces left in the deck.<\/p>\n<p>We assume the first card chosen is an ace, then the probability that the second card chosen is also an ace is dependent because there are now 3 aces in the remaining 51 cards. Symbolically, we write this as:<\/p>\n<p><em>P<\/em>(ace on second draw | an ace on the first draw).<\/p>\n<p>The vertical bar &#8220;|&#8221; is read as &#8220;given,&#8221; so the above expression is short for &#8220;The probability that an ace is drawn on the second draw given that an ace was drawn on the first draw.&#8221;<\/p>\n<p>Thus, the probability of both cards being aces is [latex]\\frac{4}{52}\\cdot\\frac{3}{51}=\\frac{12}{2652}=\\frac{1}{221}[\/latex].<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox\">\n<h3><em>MULTIPLICATION RULE OF PROBABILITY:\u00a0\u00a0<\/em><em>P<\/em>(<em>A<\/em> and <em>B<\/em>) for dependent events<\/h3>\n<p>If Events <em>A<\/em> and <em>B<\/em> are not independent, then\u00a0<em>P<\/em>(<em>A<\/em> and <em>B<\/em>) = <em>P<\/em>(<em>A<\/em>) \u00b7 <em>P<\/em>(<em>B<\/em> | <em>A<\/em>)<\/p>\n<\/div>\n<div class=\"textbox examples\">\n<h3>Converting a fraction to decimal form<\/h3>\n<p>Probabilities can be expressed in fraction or decimal form. To convert a fraction to a decimal, use a calculator to divide the numerator by the denominator.<\/p>\n<p>Ex. [latex]\\dfrac{19}{51}=19 \\div 51 \\approx 0.3725[\/latex]<\/p>\n<\/div>\n<div class=\"textbox exercises\">\n<h3>example<\/h3>\n<p>If you pull 2 cards out of a deck, what is the probability that both are spades?<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q400876\">Show Solution<\/span><\/p>\n<div id=\"q400876\" class=\"hidden-answer\" style=\"display: none\">\n<p>The probability that the first card is a spade is [latex]\\frac{13}{52}[\/latex].<\/p>\n<p>The probability that the second card is a spade, is dependent on the first being a spade, is [latex]\\frac{12}{51}[\/latex], since there is one less spade in the deck, and one less total cards.<\/p>\n<p>The probability that both cards are spades is [latex]\\frac{13}{52}\\cdot\\frac{12}{51}=\\frac{156}{2652}\\approx0.0588[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p><iframe loading=\"lazy\" id=\"ohm7118\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=7118&theme=oea&iframe_resize_id=ohm7118&show_question_numbers\" width=\"100%\" height=\"150\"><\/iframe><\/p>\n<\/div>\n<div class=\"textbox exercises\">\n<h3>Example<\/h3>\n<p>If you draw two cards from a deck, what is the probability that you will get the Ace of Diamonds and a black card?<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q774421\">Show Solution<\/span><\/p>\n<div id=\"q774421\" class=\"hidden-answer\" style=\"display: none\">\n<p>You can satisfy this condition by having Case A or Case B, as follows:<\/p>\n<p>Case A) you can get the Ace of Diamonds first and then a black card or<\/p>\n<p>Case B) you can get a black card first and then the Ace of Diamonds.<\/p>\n<p>Let&#8217;s calculate the probability of Case A. The probability that the first card is the Ace of Diamonds is [latex]\\frac{1}{52}[\/latex]. The probability that the second card is black given that the first card is the Ace of Diamonds is [latex]\\frac{26}{51}[\/latex] because 26 of the remaining 51 cards are black. The probability is therefore [latex]\\frac{1}{52}\\cdot\\frac{26}{51}=\\frac{1}{102}[\/latex].<\/p>\n<p>Now for Case B: the probability that the first card is black is [latex]\\frac{26}{52}=\\frac{1}{2}[\/latex]. The probability that the second card is the Ace of Diamonds given that the first card is black is [latex]\\frac{1}{51}[\/latex]. The probability of Case B is therefore [latex]\\frac{1}{2}\\cdot\\frac{1}{51}=\\frac{1}{102}[\/latex], the same as the probability of Case 1.<\/p>\n<p>Recall that the probability of A or B is <em>P<\/em>(A) + <em>P<\/em>(B) &#8211; <em>P<\/em>(A and B). In this problem, <em>P<\/em>(case A and case B) = 0 since the first card cannot be the Ace of Diamonds and be a black card. Therefore, the probability of Case A or Case B is [latex]\\frac{1}{102  }+\\frac{1}{102}=\\frac{2}{102}=\\frac{1}{51}[\/latex]. The probability that you will get the Ace of Diamonds and a black card when drawing two cards from a deck is [latex]\\frac{1}{51}[\/latex].<\/p>\n<\/div>\n<\/div>\n<p>These two playing card scenarios are discussed further in the following video.<\/p>\n<p><iframe loading=\"lazy\" id=\"oembed-5\" title=\"Conditional probability with cards\" width=\"500\" height=\"281\" src=\"https:\/\/www.youtube.com\/embed\/ngyGsgV4_0U?feature=oembed&#38;rel=0\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/p>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p><iframe loading=\"lazy\" id=\"ohm7110\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=7110&theme=oea&iframe_resize_id=ohm7110&show_question_numbers\" width=\"100%\" height=\"150\"><\/iframe><\/p>\n<\/div>\n\n\t\t\t <section class=\"citations-section\" role=\"contentinfo\">\n\t\t\t <h3>Candela Citations<\/h3>\n\t\t\t\t\t <div>\n\t\t\t\t\t\t <div id=\"citation-list-401\">\n\t\t\t\t\t\t\t <div class=\"licensing\"><div class=\"license-attribution-dropdown-subheading\">CC licensed content, Original<\/div><ul class=\"citation-list\"><li>Revisoin and Adaptation. <strong>Provided by<\/strong>: Lumen Learning. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em><\/li><\/ul><div class=\"license-attribution-dropdown-subheading\">CC licensed content, Shared previously<\/div><ul class=\"citation-list\"><li>Working With Events. <strong>Authored by<\/strong>: David Lippman. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"http:\/\/www.opentextbookstore.com\/mathinsociety\/\">http:\/\/www.opentextbookstore.com\/mathinsociety\/<\/a>. <strong>Project<\/strong>: Math in Society. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by-sa\/4.0\/\">CC BY-SA: Attribution-ShareAlike<\/a><\/em><\/li><li>ace-playing-cards-deck-spades. <strong>Authored by<\/strong>: PDPics. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"https:\/\/pixabay.com\/en\/ace-playing-cards-deck-spades-167052\/\">https:\/\/pixabay.com\/en\/ace-playing-cards-deck-spades-167052\/<\/a>. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/about\/cc0\">CC0: No Rights Reserved<\/a><\/em><\/li><li>Probability - Complements. <strong>Authored by<\/strong>: OCLPhase2&#039;s channel. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"https:\/\/youtu.be\/RnljiW6ZM6A\">https:\/\/youtu.be\/RnljiW6ZM6A<\/a>. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em><\/li><li>Probability of two events: P(A or B). <strong>Authored by<\/strong>: OCLPhase2&#039;s channel. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"https:\/\/youtu.be\/klbPZeH1np4\">https:\/\/youtu.be\/klbPZeH1np4<\/a>. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em><\/li><li>Joint probabilities of independent events: P(A and B). <strong>Authored by<\/strong>: OCLPhase2&#039;s channel. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"https:\/\/youtu.be\/6F17WLp-EL8\">https:\/\/youtu.be\/6F17WLp-EL8<\/a>. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em><\/li><li>Probabilities from a table: AND and OR. <strong>Authored by<\/strong>: OCLPhase2&#039;s channel. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"https:\/\/youtu.be\/HWrGoM1yRaU\">https:\/\/youtu.be\/HWrGoM1yRaU<\/a>. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em><\/li><li>Basic conditional probability. <strong>Authored by<\/strong>: OCLPhase2&#039;s channel. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"https:\/\/youtu.be\/b6tstekMlb8\">https:\/\/youtu.be\/b6tstekMlb8<\/a>. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em><\/li><li>Conditional probability with cards. <strong>Authored by<\/strong>: OCLPhase2&#039;s channel. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"https:\/\/youtu.be\/ngyGsgV4_0U\">https:\/\/youtu.be\/ngyGsgV4_0U<\/a>. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em><\/li><li>Conditional probability from a table. <strong>Authored by<\/strong>: OCLPhase2&#039;s channel. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"https:\/\/youtu.be\/LH0cuHS9Ez0\">https:\/\/youtu.be\/LH0cuHS9Ez0<\/a>. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em><\/li><li>Question ID 7111, 7113, 7114, 7115. <strong>Authored by<\/strong>: unknown. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em>. <strong>License Terms<\/strong>: IMathAS Community License CC-BY + GPL<\/li><\/ul><\/div>\n\t\t\t\t\t\t <\/div>\n\t\t\t\t\t <\/div>\n\t\t\t <\/section>","protected":false},"author":503070,"menu_order":18,"template":"","meta":{"_candela_citation":"[{\"type\":\"cc\",\"description\":\"Working With Events\",\"author\":\"David Lippman\",\"organization\":\"\",\"url\":\"http:\/\/www.opentextbookstore.com\/mathinsociety\/\",\"project\":\"Math in Society\",\"license\":\"cc-by-sa\",\"license_terms\":\"\"},{\"type\":\"cc\",\"description\":\"ace-playing-cards-deck-spades\",\"author\":\"PDPics\",\"organization\":\"\",\"url\":\"https:\/\/pixabay.com\/en\/ace-playing-cards-deck-spades-167052\/\",\"project\":\"\",\"license\":\"cc0\",\"license_terms\":\"\"},{\"type\":\"cc\",\"description\":\"Probability - 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