{"id":2453,"date":"2017-03-31T19:05:14","date_gmt":"2017-03-31T19:05:14","guid":{"rendered":"https:\/\/courses.lumenlearning.com\/waymakermath4libarts\/?post_type=chapter&#038;p=2453"},"modified":"2021-02-06T00:03:19","modified_gmt":"2021-02-06T00:03:19","slug":"conditional-probability","status":"web-only","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/slcc-mathforliberalartscorequisite\/chapter\/conditional-probability\/","title":{"raw":"Conditional Probability","rendered":"Conditional Probability"},"content":{"raw":"<div class=\"textbox learning-objectives\">\r\n<h3>Learning Outcomes<\/h3>\r\n<ul>\r\n \t<li>Calculate a conditional probability using standard notation<\/li>\r\n<\/ul>\r\n<\/div>\r\n<div class=\"textbox examples\">\r\n<h3>LEARNING PROBABILISTIC PROCESSES<\/h3>\r\nRemember to work through each example in the text and in the EXAMPLE and TRY IT boxes with a pencil on paper, pausing as frequently as needed to digest the process. Watch the videos by working them out on paper, pausing the video as frequently as you need to make sense of the demonstration. Don't be afraid to ask for help -- hard work and willingness to learn translate into success!\r\n\r\n<\/div>\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 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\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 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, 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<\/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\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 Solution[\/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 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>(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<\/h3>\r\n[ohm_question]7110[\/ohm_question]\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 Solution[\/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<\/h3>\r\n[ohm_question]7116[\/ohm_question]\r\n\r\n<\/div>","rendered":"<div class=\"textbox learning-objectives\">\n<h3>Learning Outcomes<\/h3>\n<ul>\n<li>Calculate a conditional probability using standard notation<\/li>\n<\/ul>\n<\/div>\n<div class=\"textbox examples\">\n<h3>LEARNING PROBABILISTIC PROCESSES<\/h3>\n<p>Remember to work through each example in the text and in the EXAMPLE and TRY IT boxes with a pencil on paper, pausing as frequently as needed to digest the process. Watch the videos by working them out on paper, pausing the video as frequently as you need to make sense of the demonstration. Don&#8217;t be afraid to ask for help &#8212; hard work and willingness to learn translate into success!<\/p>\n<\/div>\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 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>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 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, 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<\/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>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 Solution<\/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-1\" 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 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>(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-2\" 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<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 Solution<\/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-3\" 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<\/h3>\n<p><iframe loading=\"lazy\" id=\"ohm7116\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=7116&theme=oea&iframe_resize_id=ohm7116&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-2453\">\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>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>: Lippman, David. <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>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>\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":13,"template":"","meta":{"_candela_citation":"[{\"type\":\"cc\",\"description\":\"Math in Society\",\"author\":\"Lippman, David\",\"organization\":\"\",\"url\":\"http:\/\/www.opentextbookstore.com\/mathinsociety\/\",\"project\":\"\",\"license\":\"cc-by-sa\",\"license_terms\":\"\"},{\"type\":\"cc\",\"description\":\"Statistics, Describing Data, and Probability \",\"author\":\"Jeff Eldridge\",\"organization\":\"\",\"url\":\"\",\"project\":\"\",\"license\":\"cc-by-sa\",\"license_terms\":\"\"},{\"type\":\"cc\",\"description\":\"Question ID 7118\",\"author\":\"WebWork-Rochester\",\"organization\":\"\",\"url\":\"\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"IMathAS Community License CC-BY + GPL\"},{\"type\":\"original\",\"description\":\"Revision and Adaptation\",\"author\":\"\",\"organization\":\"Lumen 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