{"id":859,"date":"2017-06-13T23:14:29","date_gmt":"2017-06-13T23:14:29","guid":{"rendered":"https:\/\/courses.lumenlearning.com\/suny-hccc-ma-124-1\/?post_type=chapter&#038;p=859"},"modified":"2018-08-14T20:34:45","modified_gmt":"2018-08-14T20:34:45","slug":"computing-the-probability-of-an-event","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/suny-hccc-ma-124-1\/chapter\/computing-the-probability-of-an-event\/","title":{"raw":"Computing the Probability of an Event","rendered":"Computing the Probability of an Event"},"content":{"raw":"<div class=\"textbox learning-objectives\">\r\n<h3>Learning Objectives<\/h3>\r\nThe learning outcomes for this section include:\r\n<ul>\r\n \t<li>Describe a sample space and simple and compound events in it using standard notation<\/li>\r\n \t<li>Calculate the probability of an event using standard notation<\/li>\r\n \t<li>Calculate the probability of two independent events using standard notation<\/li>\r\n \t<li>Recognize when two events are mutually exclusive<\/li>\r\n \t<li>Calculate a conditional probability using standard notation<\/li>\r\n<\/ul>\r\n<\/div>\r\nProbability is the likelihood of a particular outcome or event happening. Statisticians and actuaries use probability to make predictions about events. \u00a0An actuary that works for a car insurance company would, for example, be interested in how likely a 17 year old male would be to get in a car accident. \u00a0They would use data from past events to make predictions about future events using the characteristics of probabilities, then use this information to calculate an insurance rate.\r\n\r\nIn this section, we will explore the definition of an event, and learn how to calculate the probability of it's occurance. \u00a0We will also practice using standard mathematical notation to calculate and describe different kinds of probabilities.\r\n\r\n<a href=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/1141\/2017\/02\/01222858\/Car_crash_1.jpg\"><img class=\"wp-image-1263 aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/1141\/2017\/02\/01222858\/Car_crash_1-300x225.jpg\" alt=\"\" width=\"613\" height=\"462\" \/><\/a>\r\n<h2>Basic Concepts<\/h2>\r\nIf you roll a die, pick a card from deck of playing cards, or randomly select a person and observe their hair color, we are executing an experiment or procedure. In probability, we look at the likelihood of different outcomes.\r\n<a href=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/1141\/2016\/12\/28180953\/dice-219263_640.jpg\"><img class=\"aligncenter size-full wp-image-987\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/1141\/2016\/12\/28180953\/dice-219263_640.jpg\" alt=\"Five dice, red and white, on a marble surface\" width=\"640\" height=\"480\" \/><\/a>\r\n\r\nWe begin with some terminology.\r\n<div class=\"textbox\">\r\n<h3>Events and Outcomes<\/h3>\r\n<ul>\r\n \t<li>The result of an experiment is called an <strong>outcome<\/strong>.<\/li>\r\n \t<li>An <strong>event<\/strong> is any particular outcome or group of outcomes.<\/li>\r\n \t<li>A <strong>simple event <\/strong>is an event that cannot be broken down further<\/li>\r\n \t<li>The <strong>sample space<\/strong> is the set of all possible simple events.<\/li>\r\n<\/ul>\r\n<\/div>\r\n<div class=\"textbox exercises\">\r\n<h3>example<\/h3>\r\nIf we roll a standard 6-sided die, describe the sample space and some simple events.\r\n\r\n[reveal-answer q=\"997648\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"997648\"]\r\n\r\nThe sample space is the set of all possible simple events: {1,2,3,4,5,6}\r\n\r\nSome examples of simple events:\r\n<ul>\r\n \t<li>We roll a 1<\/li>\r\n \t<li>We roll a 5<\/li>\r\n<\/ul>\r\nSome compound events:\r\n<ul>\r\n \t<li>We roll a number bigger than 4<\/li>\r\n \t<li>We roll an even number<\/li>\r\n<\/ul>\r\n[\/hidden-answer]\r\n\r\n&nbsp;\r\n\r\n<\/div>\r\n<div class=\"textbox\">\r\n<h3>Basic Probability<\/h3>\r\nGiven that all outcomes are equally likely, we can compute the probability of an event <em>E<\/em> using this formula:\r\n<p style=\"text-align: center;\">[latex]P(E)=\\frac{\\text{Number of outcomes corresponding to the event E}}{\\text{Total number of equally-likely outcomes}}[\/latex]<\/p>\r\n\r\n<\/div>\r\n<div class=\"textbox exercises\">\r\n<h3>examples<\/h3>\r\nIf we roll a 6-sided die, calculate\r\n<ol>\r\n \t<li>P(rolling a 1)<\/li>\r\n \t<li>P(rolling a number bigger than 4)<\/li>\r\n<\/ol>\r\n[reveal-answer q=\"417705\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"417705\"]\r\n\r\nRecall that the sample space is {1,2,3,4,5,6}\r\n<ol>\r\n \t<li>There is one outcome corresponding to \u201crolling a 1,\u201d so the probability is [latex]\\frac{1}{6}[\/latex]<\/li>\r\n \t<li>There are two outcomes bigger than a 4, so the probability is [latex]\\frac{2}{6}=\\frac{1}{3}[\/latex]<\/li>\r\n<\/ol>\r\nProbabilities are essentially fractions, and can be reduced to lower terms like fractions.\r\n\r\n[\/hidden-answer]\r\n\r\nThis video describes this example and the previous one in detail.\r\n\r\nhttps:\/\/youtu.be\/37P01dt0zsE\r\n\r\nLet's say you have a bag with 20 cherries, 14 sweet and 6 sour. If you pick a cherry at random, what is the probability that it will be sweet?\r\n[reveal-answer q=\"9680\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"9680\"]\r\n\r\nThere are 20 possible cherries that could be picked, so the number of possible outcomes is 20. Of these 20 possible outcomes, 14 are favorable (sweet), so the probability that the cherry will be sweet is [latex]\\frac{14}{20}=\\frac{7}{10}[\/latex].\r\nThere is one potential complication to this example, however. It must be assumed that the probability of picking any of the cherries is the same as the probability of picking any other. This wouldn't be true if (let us imagine) the sweet cherries are smaller than the sour ones. (The sour cherries would come to hand more readily when you sampled from the bag.) Let us keep in mind, therefore, that when we assess probabilities in terms of the ratio of favorable to all potential cases, we rely heavily on the assumption of equal probability for all outcomes.\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Try It Now<\/h3>\r\nAt some random moment, you look at your clock and note the minutes reading.\r\n\r\na. What is probability the minutes reading is 15?\r\n\r\nb. What is the probability the minutes reading is 15 or less?\r\n\r\n<\/div>\r\n<div class=\"textbox\">\r\n<h3>Cards<\/h3>\r\nA standard deck of 52 playing cards consists of four <strong>suits<\/strong> (hearts, spades, diamonds and clubs). Spades and clubs are black while hearts and diamonds are red. Each suit contains 13 cards, each of a different <strong>rank<\/strong>: an Ace (which in many games functions as both a low card and a high card), cards numbered 2 through 10, a Jack, a Queen and a King.\r\n\r\n<\/div>\r\n<div class=\"textbox exercises\">\r\n<h3>example<\/h3>\r\nCompute the probability of randomly drawing one card from a deck and getting an Ace.\r\n[reveal-answer q=\"652517\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"652517\"]\r\n\r\nThere are 52 cards in the deck and 4 Aces so\u00a0[latex]P(Ace)=\\frac{4}{52}=\\frac{1}{13}\\approx 0.0769[\/latex]\r\n\r\nWe can also think of probabilities as percents: There is a 7.69% chance that a randomly selected card will be an Ace.\r\n\r\nNotice that the smallest possible probability is 0 \u2013 if there are no outcomes that correspond with the event. The largest possible probability is 1 \u2013 if all possible outcomes correspond with the event.\r\n\r\n[\/hidden-answer]\r\n\r\nThis video demonstrates both this example and the previous cherry example\u00a0on the page.\r\n\r\nhttps:\/\/youtu.be\/EBqj_R3dzd4\r\n\r\n<\/div>\r\n<div class=\"textbox\">\r\n<h3>Certain and Impossible events<\/h3>\r\n<ul>\r\n \t<li>An impossible event has a probability of 0.<\/li>\r\n \t<li>A certain event has a probability of 1.<\/li>\r\n \t<li>The probability of any event must be [latex]0\\le P(E)\\le 1[\/latex]<\/li>\r\n<\/ul>\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Try it now<\/h3>\r\n<iframe id=\"mom1\" class=\"resizable\" src=\"https:\/\/www.myopenmath.com\/multiembedq.php?id=7130&amp;theme=oea&amp;iframe_resize_id=mom1\" width=\"100%\" height=\"400\"><\/iframe>\r\n\r\n<\/div>\r\nIn the course of this section, <strong>if you compute a probability and get an answer that is negative or greater than 1, you have made a mistake and should check your work<\/strong>.\r\n<h2>Types of Events<\/h2>\r\n<h3>Complementary Events<\/h3>\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 Answer[\/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 now<\/h3>\r\n<iframe id=\"mom19\" class=\"resizable\" src=\"https:\/\/www.myopenmath.com\/multiembedq.php?id=7097&amp;theme=oea&amp;iframe_resize_id=mom19\" width=\"100%\" height=\"300\"><\/iframe>\r\n\r\n<\/div>\r\n<h3>Probability of two independent events<\/h3>\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 Answer[\/reveal-answer]\r\n[hidden-answer a=\"453669\"]\r\n\r\nWe could list all possible outcomes: \u00a0{H1,H2,h2,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 Answer[\/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\">\r\n<h3><em>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 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 Answer[\/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 now<\/h3>\r\n<iframe id=\"mom100\" class=\"resizable\" src=\"https:\/\/www.myopenmath.com\/multiembedq.php?id=7111&amp;theme=oea&amp;iframe_resize_id=mom100\" width=\"100%\" height=\"300\"><\/iframe>\r\n\r\n<\/div>\r\nThe previous examples looked at the probability of <em>both<\/em> events occurring. Now we will look at the probability of <em>either<\/em> event 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 <em>or<\/em> a 6 on the die.\r\n[reveal-answer q=\"443646\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"443646\"]\r\n\r\nHere, there are still 12 possible outcomes: {H1,H2,h2,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, h2, 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>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 Answer[\/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 <strong>mutually exclusive<\/strong>, so <em>P<\/em>(<em>A<\/em> or <em>B<\/em>) = <em>P<\/em>(<em>A<\/em>) + <em>P<\/em>(<em>B<\/em>).\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Try it now<\/h3>\r\n<iframe id=\"mom10\" class=\"resizable\" src=\"https:\/\/www.myopenmath.com\/multiembedq.php?id=7113&amp;theme=oea&amp;iframe_resize_id=mom10\" width=\"100%\" height=\"250\"><\/iframe>\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 Answer[\/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 Now<\/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 the probability at least one is white?\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 Answer[\/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 now<\/h3>\r\n<iframe id=\"mom15\" class=\"resizable\" src=\"https:\/\/www.myopenmath.com\/multiembedq.php?id=7114&amp;theme=oea&amp;iframe_resize_id=mom15\" width=\"100%\" height=\"300\"><\/iframe>\r\n<iframe id=\"mom1\" class=\"resizable\" src=\"https:\/\/www.myopenmath.com\/multiembedq.php?id=7115&amp;theme=oea&amp;iframe_resize_id=mom1\" width=\"100%\" height=\"300\"><\/iframe>\r\n\r\n<\/div>\r\n<h2>Conditional Probability<\/h2>\r\nIn the previous section we computed the probabilities of events that were independent of each other. We saw that getting a certain outcome from rolling a die had no influence on the outcome from flipping a coin, even though we were computing a probability based on doing them at the same time.\r\n\r\nIn this section, we will consider events that\u00a0<em>are\u00a0<\/em>dependent on each other, called <strong>conditional probabilities<\/strong>.\r\n<div class=\"textbox\">\r\n<h3>Conditional Probability<\/h3>\r\nThe probability the event <em>B<\/em> occurs, given that event <em>A<\/em> has happened, is represented as\r\n<p style=\"text-align: center;\"><em>P<\/em>(<em>B<\/em> | <em>A<\/em>)<\/p>\r\nThis is read as \u201cthe probability of <em>B<\/em> given <em>A<\/em>\u201d\r\n\r\n<\/div>\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 in the last section.\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 Answer[\/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\nOnce the first card chosen is an ace, the probability that the second card chosen is also an ace is called the <strong>conditional probability<\/strong> of drawing an ace. In this case the \"condition\" is that the first card is an ace. 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.\" What is this probability? After an ace is drawn on the first draw, there are 3 aces out of 51 total cards left. This means that the conditional probability of drawing an ace after one ace has already been drawn is [latex]\\frac{3}{51}=\\frac{1}{17}[\/latex].\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>Conditional Probability Formula<\/h3>\r\nIf Events <em>A<\/em> and <em>B<\/em> are not independent, then\r\n\r\n<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 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 Answer[\/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, given the first was 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 now<\/h3>\r\n<iframe id=\"mom11\" class=\"resizable\" src=\"https:\/\/www.myopenmath.com\/multiembedq.php?id=7118&amp;theme=oea&amp;iframe_resize_id=mom11\" width=\"100%\" height=\"350\"><\/iframe>\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 speeding ticket <em>given<\/em> they have a red car<\/li>\r\n \t<li>has a red car <em>given<\/em> they have a speeding ticket<\/li>\r\n<\/ol>\r\n<table>\r\n<tbody>\r\n<tr>\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=\"849340\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"849340\"]\r\n<ol>\r\n \t<li>Since we know the person has a red car, we are only considering the 150 people in the first row of the table. Of those, 15 have a speeding ticket, so P(ticket | red car) = [latex]\\frac{15}{150}=\\frac{1}{10}=0.1[\/latex]<\/li>\r\n \t<li>Since we know the person has a speeding ticket, we are only considering the 60 people in the first column of the table. Of those, 15 have a red car, so P(red car | ticket) = [latex]\\frac{15}{60}=\\frac{1}{4}=0.25[\/latex].<\/li>\r\n<\/ol>\r\nNotice from the last example that P(B | A) is <strong>not<\/strong> equal to P(A | B).\r\n\r\n[\/hidden-answer]\r\n\r\nThese kinds of conditional probabilities are what insurance companies use to determine your insurance rates. They look at the conditional probability of you having accident, given your age, your car, your car color, your driving history, etc., and price your policy based on that likelihood.\r\n\r\nView more about conditional probability in the following video.\r\n\r\nhttps:\/\/youtu.be\/b6tstekMlb8\r\n\r\n<\/div>\r\n&nbsp;\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 Answer[\/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>(A and 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}{101}+\\frac{1}{101}=\\frac{2}{101}[\/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{2}{101}[\/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 now<\/h3>\r\n<iframe id=\"mom1\" class=\"resizable\" src=\"https:\/\/www.myopenmath.com\/multiembedq.php?id=7110&amp;theme=oea&amp;iframe_resize_id=mom1\" width=\"100%\" height=\"350\"><\/iframe>\r\n\r\n<\/div>\r\n<div class=\"textbox exercises\">\r\n<h3>Example<\/h3>\r\nA home pregnancy test was given to women, then pregnancy was verified through blood tests. \u00a0The following table shows the home pregnancy test results.\r\n\r\nFind\r\n<ol>\r\n \t<li><em>P<\/em>(not pregnant | positive test result)<\/li>\r\n \t<li><em>P<\/em>(positive test result | not pregnant)<\/li>\r\n<\/ol>\r\n<table>\r\n<tbody>\r\n<tr>\r\n<td><\/td>\r\n<td>Positive test<\/td>\r\n<td>Negative test<\/td>\r\n<td>Total<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>Pregnant<\/td>\r\n<td>70<\/td>\r\n<td>4<\/td>\r\n<td>74<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>Not Pregnant<\/td>\r\n<td>5<\/td>\r\n<td>14<\/td>\r\n<td>19<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>Total<\/td>\r\n<td>75<\/td>\r\n<td>18<\/td>\r\n<td>93<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n[reveal-answer q=\"968710\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"968710\"]\r\n<ol>\r\n \t<li>Since we know the test result was positive, we\u2019re limited to the 75 women in the first column, of which 5 were not pregnant. <em>P<\/em>(not pregnant | positive test result) = [latex]\\frac{5}{75}\\approx0.067[\/latex].<\/li>\r\n \t<li>Since we know the woman is not pregnant, we are limited to the 19 women in the second row, of which 5 had a positive test.\u00a0<em>P<\/em>(positive test result | not pregnant) = [latex]\\frac{5}{19}\\approx0.263[\/latex]<\/li>\r\n<\/ol>\r\nThe second result is what is usually called a false positive: A positive result when the woman is not actually pregnant.\r\n\r\n[\/hidden-answer]\r\n\r\nSee more about this example here.\r\n\r\nhttps:\/\/youtu.be\/LH0cuHS9Ez0\r\n\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Try it now<\/h3>\r\n<iframe id=\"mom10\" class=\"resizable\" src=\"https:\/\/www.myopenmath.com\/multiembedq.php?id=7116&amp;theme=oea&amp;iframe_resize_id=mom10\" width=\"100%\" height=\"550\"><\/iframe>\r\n\r\n<\/div>","rendered":"<div class=\"textbox learning-objectives\">\n<h3>Learning Objectives<\/h3>\n<p>The learning outcomes for this section include:<\/p>\n<ul>\n<li>Describe a sample space and simple and compound events in it using standard notation<\/li>\n<li>Calculate the probability of an event using standard notation<\/li>\n<li>Calculate the probability of two independent events using standard notation<\/li>\n<li>Recognize when two events are mutually exclusive<\/li>\n<li>Calculate a conditional probability using standard notation<\/li>\n<\/ul>\n<\/div>\n<p>Probability is the likelihood of a particular outcome or event happening. Statisticians and actuaries use probability to make predictions about events. \u00a0An actuary that works for a car insurance company would, for example, be interested in how likely a 17 year old male would be to get in a car accident. \u00a0They would use data from past events to make predictions about future events using the characteristics of probabilities, then use this information to calculate an insurance rate.<\/p>\n<p>In this section, we will explore the definition of an event, and learn how to calculate the probability of it&#8217;s occurance. \u00a0We will also practice using standard mathematical notation to calculate and describe different kinds of probabilities.<\/p>\n<p><a href=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/1141\/2017\/02\/01222858\/Car_crash_1.jpg\"><img loading=\"lazy\" decoding=\"async\" class=\"wp-image-1263 aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/1141\/2017\/02\/01222858\/Car_crash_1-300x225.jpg\" alt=\"\" width=\"613\" height=\"462\" \/><\/a><\/p>\n<h2>Basic Concepts<\/h2>\n<p>If you roll a die, pick a card from deck of playing cards, or randomly select a person and observe their hair color, we are executing an experiment or procedure. In probability, we look at the likelihood of different outcomes.<br \/>\n<a href=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/1141\/2016\/12\/28180953\/dice-219263_640.jpg\"><img loading=\"lazy\" decoding=\"async\" class=\"aligncenter size-full wp-image-987\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/1141\/2016\/12\/28180953\/dice-219263_640.jpg\" alt=\"Five dice, red and white, on a marble surface\" width=\"640\" height=\"480\" \/><\/a><\/p>\n<p>We begin with some terminology.<\/p>\n<div class=\"textbox\">\n<h3>Events and Outcomes<\/h3>\n<ul>\n<li>The result of an experiment is called an <strong>outcome<\/strong>.<\/li>\n<li>An <strong>event<\/strong> is any particular outcome or group of outcomes.<\/li>\n<li>A <strong>simple event <\/strong>is an event that cannot be broken down further<\/li>\n<li>The <strong>sample space<\/strong> is the set of all possible simple events.<\/li>\n<\/ul>\n<\/div>\n<div class=\"textbox exercises\">\n<h3>example<\/h3>\n<p>If we roll a standard 6-sided die, describe the sample space and some simple events.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q997648\">Show Answer<\/span><\/p>\n<div id=\"q997648\" class=\"hidden-answer\" style=\"display: none\">\n<p>The sample space is the set of all possible simple events: {1,2,3,4,5,6}<\/p>\n<p>Some examples of simple events:<\/p>\n<ul>\n<li>We roll a 1<\/li>\n<li>We roll a 5<\/li>\n<\/ul>\n<p>Some compound events:<\/p>\n<ul>\n<li>We roll a number bigger than 4<\/li>\n<li>We roll an even number<\/li>\n<\/ul>\n<\/div>\n<\/div>\n<p>&nbsp;<\/p>\n<\/div>\n<div class=\"textbox\">\n<h3>Basic Probability<\/h3>\n<p>Given that all outcomes are equally likely, we can compute the probability of an event <em>E<\/em> using this formula:<\/p>\n<p style=\"text-align: center;\">[latex]P(E)=\\frac{\\text{Number of outcomes corresponding to the event E}}{\\text{Total number of equally-likely outcomes}}[\/latex]<\/p>\n<\/div>\n<div class=\"textbox exercises\">\n<h3>examples<\/h3>\n<p>If we roll a 6-sided die, calculate<\/p>\n<ol>\n<li>P(rolling a 1)<\/li>\n<li>P(rolling a number bigger than 4)<\/li>\n<\/ol>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q417705\">Show Answer<\/span><\/p>\n<div id=\"q417705\" class=\"hidden-answer\" style=\"display: none\">\n<p>Recall that the sample space is {1,2,3,4,5,6}<\/p>\n<ol>\n<li>There is one outcome corresponding to \u201crolling a 1,\u201d so the probability is [latex]\\frac{1}{6}[\/latex]<\/li>\n<li>There are two outcomes bigger than a 4, so the probability is [latex]\\frac{2}{6}=\\frac{1}{3}[\/latex]<\/li>\n<\/ol>\n<p>Probabilities are essentially fractions, and can be reduced to lower terms like fractions.<\/p>\n<\/div>\n<\/div>\n<p>This video describes this example and the previous one in detail.<\/p>\n<p><iframe loading=\"lazy\" id=\"oembed-1\" title=\"Basics of Probability - events and outcomes\" width=\"500\" height=\"281\" src=\"https:\/\/www.youtube.com\/embed\/37P01dt0zsE?feature=oembed&#38;rel=0\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/p>\n<p>Let&#8217;s say you have a bag with 20 cherries, 14 sweet and 6 sour. If you pick a cherry at random, what is the probability that it will be sweet?<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q9680\">Show Answer<\/span><\/p>\n<div id=\"q9680\" class=\"hidden-answer\" style=\"display: none\">\n<p>There are 20 possible cherries that could be picked, so the number of possible outcomes is 20. Of these 20 possible outcomes, 14 are favorable (sweet), so the probability that the cherry will be sweet is [latex]\\frac{14}{20}=\\frac{7}{10}[\/latex].<br \/>\nThere is one potential complication to this example, however. It must be assumed that the probability of picking any of the cherries is the same as the probability of picking any other. This wouldn&#8217;t be true if (let us imagine) the sweet cherries are smaller than the sour ones. (The sour cherries would come to hand more readily when you sampled from the bag.) Let us keep in mind, therefore, that when we assess probabilities in terms of the ratio of favorable to all potential cases, we rely heavily on the assumption of equal probability for all outcomes.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try It Now<\/h3>\n<p>At some random moment, you look at your clock and note the minutes reading.<\/p>\n<p>a. What is probability the minutes reading is 15?<\/p>\n<p>b. What is the probability the minutes reading is 15 or less?<\/p>\n<\/div>\n<div class=\"textbox\">\n<h3>Cards<\/h3>\n<p>A standard deck of 52 playing cards consists of four <strong>suits<\/strong> (hearts, spades, diamonds and clubs). Spades and clubs are black while hearts and diamonds are red. Each suit contains 13 cards, each of a different <strong>rank<\/strong>: an Ace (which in many games functions as both a low card and a high card), cards numbered 2 through 10, a Jack, a Queen and a King.<\/p>\n<\/div>\n<div class=\"textbox exercises\">\n<h3>example<\/h3>\n<p>Compute the probability of randomly drawing one card from a deck and getting an Ace.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q652517\">Show Answer<\/span><\/p>\n<div id=\"q652517\" class=\"hidden-answer\" style=\"display: none\">\n<p>There are 52 cards in the deck and 4 Aces so\u00a0[latex]P(Ace)=\\frac{4}{52}=\\frac{1}{13}\\approx 0.0769[\/latex]<\/p>\n<p>We can also think of probabilities as percents: There is a 7.69% chance that a randomly selected card will be an Ace.<\/p>\n<p>Notice that the smallest possible probability is 0 \u2013 if there are no outcomes that correspond with the event. The largest possible probability is 1 \u2013 if all possible outcomes correspond with the event.<\/p>\n<\/div>\n<\/div>\n<p>This video demonstrates both this example and the previous cherry example\u00a0on the page.<\/p>\n<p><iframe loading=\"lazy\" id=\"oembed-2\" title=\"Basic Probabilities\" width=\"500\" height=\"281\" src=\"https:\/\/www.youtube.com\/embed\/EBqj_R3dzd4?feature=oembed&#38;rel=0\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/p>\n<\/div>\n<div class=\"textbox\">\n<h3>Certain and Impossible events<\/h3>\n<ul>\n<li>An impossible event has a probability of 0.<\/li>\n<li>A certain event has a probability of 1.<\/li>\n<li>The probability of any event must be [latex]0\\le P(E)\\le 1[\/latex]<\/li>\n<\/ul>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try it now<\/h3>\n<p><iframe loading=\"lazy\" id=\"mom1\" class=\"resizable\" src=\"https:\/\/www.myopenmath.com\/multiembedq.php?id=7130&amp;theme=oea&amp;iframe_resize_id=mom1\" width=\"100%\" height=\"400\"><\/iframe><\/p>\n<\/div>\n<p>In the course of this section, <strong>if you compute a probability and get an answer that is negative or greater than 1, you have made a mistake and should check your work<\/strong>.<\/p>\n<h2>Types of Events<\/h2>\n<h3>Complementary Events<\/h3>\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 Answer<\/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-3\" 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 now<\/h3>\n<p><iframe loading=\"lazy\" id=\"mom19\" class=\"resizable\" src=\"https:\/\/www.myopenmath.com\/multiembedq.php?id=7097&amp;theme=oea&amp;iframe_resize_id=mom19\" width=\"100%\" height=\"300\"><\/iframe><\/p>\n<\/div>\n<h3>Probability of two independent events<\/h3>\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 Answer<\/span><\/p>\n<div id=\"q453669\" class=\"hidden-answer\" style=\"display: none\">\n<p>We could list all possible outcomes: \u00a0{H1,H2,h2,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 Answer<\/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\">\n<h3><em>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 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 Answer<\/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-4\" 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 now<\/h3>\n<p><iframe loading=\"lazy\" id=\"mom100\" class=\"resizable\" src=\"https:\/\/www.myopenmath.com\/multiembedq.php?id=7111&amp;theme=oea&amp;iframe_resize_id=mom100\" width=\"100%\" height=\"300\"><\/iframe><\/p>\n<\/div>\n<p>The previous examples looked at the probability of <em>both<\/em> events occurring. Now we will look at the probability of <em>either<\/em> event 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 <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 Answer<\/span><\/p>\n<div id=\"q443646\" class=\"hidden-answer\" style=\"display: none\">\n<p>Here, there are still 12 possible outcomes: {H1,H2,h2,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, h2, 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>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 Answer<\/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-5\" 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 <strong>mutually exclusive<\/strong>, so <em>P<\/em>(<em>A<\/em> or <em>B<\/em>) = <em>P<\/em>(<em>A<\/em>) + <em>P<\/em>(<em>B<\/em>).<\/p>\n<div class=\"textbox key-takeaways\">\n<h3>Try it now<\/h3>\n<p><iframe loading=\"lazy\" id=\"mom10\" class=\"resizable\" src=\"https:\/\/www.myopenmath.com\/multiembedq.php?id=7113&amp;theme=oea&amp;iframe_resize_id=mom10\" width=\"100%\" height=\"250\"><\/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 Answer<\/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 Now<\/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 the probability at least one is white?<\/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 Answer<\/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-6\" 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 now<\/h3>\n<p><iframe loading=\"lazy\" id=\"mom15\" class=\"resizable\" src=\"https:\/\/www.myopenmath.com\/multiembedq.php?id=7114&amp;theme=oea&amp;iframe_resize_id=mom15\" width=\"100%\" height=\"300\"><\/iframe><br \/>\n<iframe loading=\"lazy\" id=\"mom1\" class=\"resizable\" src=\"https:\/\/www.myopenmath.com\/multiembedq.php?id=7115&amp;theme=oea&amp;iframe_resize_id=mom1\" width=\"100%\" height=\"300\"><\/iframe><\/p>\n<\/div>\n<h2>Conditional Probability<\/h2>\n<p>In the previous section we computed the probabilities of events that were independent of each other. We saw that getting a certain outcome from rolling a die had no influence on the outcome from flipping a coin, even though we were computing a probability based on doing them at the same time.<\/p>\n<p>In this section, we will consider events that\u00a0<em>are\u00a0<\/em>dependent on each other, called <strong>conditional probabilities<\/strong>.<\/p>\n<div class=\"textbox\">\n<h3>Conditional Probability<\/h3>\n<p>The probability the event <em>B<\/em> occurs, given that event <em>A<\/em> has happened, is represented as<\/p>\n<p style=\"text-align: center;\"><em>P<\/em>(<em>B<\/em> | <em>A<\/em>)<\/p>\n<p>This is read as \u201cthe probability of <em>B<\/em> given <em>A<\/em>\u201d<\/p>\n<\/div>\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 in the last section.<\/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 Answer<\/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>Once the first card chosen is an ace, the probability that the second card chosen is also an ace is called the <strong>conditional probability<\/strong> of drawing an ace. In this case the &#8220;condition&#8221; is that the first card is an ace. 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; What is this probability? After an ace is drawn on the first draw, there are 3 aces out of 51 total cards left. This means that the conditional probability of drawing an ace after one ace has already been drawn is [latex]\\frac{3}{51}=\\frac{1}{17}[\/latex].<\/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>Conditional Probability Formula<\/h3>\n<p>If Events <em>A<\/em> and <em>B<\/em> are not independent, then<\/p>\n<p><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 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 Answer<\/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, given the first was 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 now<\/h3>\n<p><iframe loading=\"lazy\" id=\"mom11\" class=\"resizable\" src=\"https:\/\/www.myopenmath.com\/multiembedq.php?id=7118&amp;theme=oea&amp;iframe_resize_id=mom11\" width=\"100%\" height=\"350\"><\/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 speeding ticket <em>given<\/em> they have a red car<\/li>\n<li>has a red car <em>given<\/em> they have a speeding ticket<\/li>\n<\/ol>\n<table>\n<tbody>\n<tr>\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=\"q849340\">Show Answer<\/span><\/p>\n<div id=\"q849340\" class=\"hidden-answer\" style=\"display: none\">\n<ol>\n<li>Since we know the person has a red car, we are only considering the 150 people in the first row of the table. Of those, 15 have a speeding ticket, so P(ticket | red car) = [latex]\\frac{15}{150}=\\frac{1}{10}=0.1[\/latex]<\/li>\n<li>Since we know the person has a speeding ticket, we are only considering the 60 people in the first column of the table. Of those, 15 have a red car, so P(red car | ticket) = [latex]\\frac{15}{60}=\\frac{1}{4}=0.25[\/latex].<\/li>\n<\/ol>\n<p>Notice from the last example that P(B | A) is <strong>not<\/strong> equal to P(A | B).<\/p>\n<\/div>\n<\/div>\n<p>These kinds of conditional probabilities are what insurance companies use to determine your insurance rates. They look at the conditional probability of you having accident, given your age, your car, your car color, your driving history, etc., and price your policy based on that likelihood.<\/p>\n<p>View more about conditional probability in the following video.<\/p>\n<p><iframe loading=\"lazy\" id=\"oembed-7\" title=\"Basic conditional probability\" width=\"500\" height=\"281\" src=\"https:\/\/www.youtube.com\/embed\/b6tstekMlb8?feature=oembed&#38;rel=0\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/p>\n<\/div>\n<p>&nbsp;<\/p>\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 Answer<\/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>(A and 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}{101}+\\frac{1}{101}=\\frac{2}{101}[\/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{2}{101}[\/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-8\" 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 now<\/h3>\n<p><iframe loading=\"lazy\" id=\"mom1\" class=\"resizable\" src=\"https:\/\/www.myopenmath.com\/multiembedq.php?id=7110&amp;theme=oea&amp;iframe_resize_id=mom1\" width=\"100%\" height=\"350\"><\/iframe><\/p>\n<\/div>\n<div class=\"textbox exercises\">\n<h3>Example<\/h3>\n<p>A home pregnancy test was given to women, then pregnancy was verified through blood tests. \u00a0The following table shows the home pregnancy test results.<\/p>\n<p>Find<\/p>\n<ol>\n<li><em>P<\/em>(not pregnant | positive test result)<\/li>\n<li><em>P<\/em>(positive test result | not pregnant)<\/li>\n<\/ol>\n<table>\n<tbody>\n<tr>\n<td><\/td>\n<td>Positive test<\/td>\n<td>Negative test<\/td>\n<td>Total<\/td>\n<\/tr>\n<tr>\n<td>Pregnant<\/td>\n<td>70<\/td>\n<td>4<\/td>\n<td>74<\/td>\n<\/tr>\n<tr>\n<td>Not Pregnant<\/td>\n<td>5<\/td>\n<td>14<\/td>\n<td>19<\/td>\n<\/tr>\n<tr>\n<td>Total<\/td>\n<td>75<\/td>\n<td>18<\/td>\n<td>93<\/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=\"q968710\">Show Answer<\/span><\/p>\n<div id=\"q968710\" class=\"hidden-answer\" style=\"display: none\">\n<ol>\n<li>Since we know the test result was positive, we\u2019re limited to the 75 women in the first column, of which 5 were not pregnant. <em>P<\/em>(not pregnant | positive test result) = [latex]\\frac{5}{75}\\approx0.067[\/latex].<\/li>\n<li>Since we know the woman is not pregnant, we are limited to the 19 women in the second row, of which 5 had a positive test.\u00a0<em>P<\/em>(positive test result | not pregnant) = [latex]\\frac{5}{19}\\approx0.263[\/latex]<\/li>\n<\/ol>\n<p>The second result is what is usually called a false positive: A positive result when the woman is not actually pregnant.<\/p>\n<\/div>\n<\/div>\n<p>See more about this example here.<\/p>\n<p><iframe loading=\"lazy\" id=\"oembed-9\" title=\"Conditional probability from a table\" width=\"500\" height=\"281\" src=\"https:\/\/www.youtube.com\/embed\/LH0cuHS9Ez0?feature=oembed&#38;rel=0\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/p>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try it now<\/h3>\n<p><iframe loading=\"lazy\" id=\"mom10\" class=\"resizable\" src=\"https:\/\/www.myopenmath.com\/multiembedq.php?id=7116&amp;theme=oea&amp;iframe_resize_id=mom10\" width=\"100%\" height=\"550\"><\/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-859\">\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>Introduction and outcomes. <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><li>Revision 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>Math in Society. <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>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>dice-die-probability-fortune-luck. <strong>Authored by<\/strong>: jodylehigh. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"https:\/\/pixabay.com\/en\/dice-die-probability-fortune-luck-219263\/\">https:\/\/pixabay.com\/en\/dice-die-probability-fortune-luck-219263\/<\/a>. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/about\/cc0\">CC0: No Rights Reserved<\/a><\/em><\/li><li>Basics of Probability - events and outcomes. <strong>Authored by<\/strong>: OCLPhase2&#039;s channel. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"https:\/\/youtu.be\/37P01dt0zsE\">https:\/\/youtu.be\/37P01dt0zsE<\/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 Probabilities. <strong>Authored by<\/strong>: OCLPhase2&#039;s channel. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"https:\/\/youtu.be\/EBqj_R3dzd4\">https:\/\/youtu.be\/EBqj_R3dzd4<\/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 7130. <strong>Authored by<\/strong>: WebWork-Rochester, mb Lippman,David. <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><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><li>Statistics, Describing Data, and Probability . <strong>Authored by<\/strong>: Jeff Eldridge. <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>Question ID 7118. <strong>Authored by<\/strong>: WebWork-Rochester. <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 class=\"license-attribution-dropdown-subheading\">Public domain content<\/div><ul class=\"citation-list\"><li> A car crash on Jagtvej, a road in Copenhagen, Denmark. <strong>Authored by<\/strong>: By Thue (Own work) . <strong>Provided by<\/strong>: Wikimedia Commons. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"https:\/\/commons.wikimedia.org\/wiki\/File%3ACar_crash_1.jpg,%20https:\/\/upload.wikimedia.org\/wikipedia\/commons\/e\/e1\/Car_crash_1.jpg\">https:\/\/commons.wikimedia.org\/wiki\/File%3ACar_crash_1.jpg,%20https:\/\/upload.wikimedia.org\/wikipedia\/commons\/e\/e1\/Car_crash_1.jpg<\/a>. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/about\/pdm\">Public Domain: No Known Copyright<\/a><\/em><\/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":21,"menu_order":2,"template":"","meta":{"_candela_citation":"[{\"type\":\"pd\",\"description\":\" A car crash on Jagtvej, a road in Copenhagen, Denmark\",\"author\":\"By Thue (Own work) \",\"organization\":\"Wikimedia Commons\",\"url\":\"https:\/\/commons.wikimedia.org\/wiki\/File%3ACar_crash_1.jpg, https:\/\/upload.wikimedia.org\/wikipedia\/commons\/e\/e1\/Car_crash_1.jpg\",\"project\":\"\",\"license\":\"pd\",\"license_terms\":\"\"},{\"type\":\"original\",\"description\":\"Introduction and outcomes\",\"author\":\"\",\"organization\":\"Lumen Learning\",\"url\":\"\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"\"},{\"type\":\"cc\",\"description\":\"Math in Society\",\"author\":\"David 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