{"id":6050,"date":"2021-03-04T23:01:57","date_gmt":"2021-03-04T23:01:57","guid":{"rendered":"https:\/\/courses.lumenlearning.com\/frontrange-mathforliberalartscorequisite1\/?post_type=chapter&#038;p=6050"},"modified":"2022-01-07T00:30:44","modified_gmt":"2022-01-07T00:30:44","slug":"4-4-the-two-basic-rules-of-probability","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/frontrange-mathforliberalartscorequisite1\/chapter\/4-4-the-two-basic-rules-of-probability\/","title":{"raw":"4.4  The Two Basic Rules of Probability","rendered":"4.4  The Two Basic Rules of Probability"},"content":{"raw":"<div class=\"textbox learning-objectives\">\r\n<h3>Learning Objectives<\/h3>\r\n1.\u00a0 The Multiplication Rule (AND)\r\n<ul>\r\n \t<li>Calculate the probability of two independent events both occurring together<\/li>\r\n<\/ul>\r\n2.\u00a0 The Addition Rule (OR)\r\n<ul>\r\n \t<li>Calculate the probability of two mutually exclusive events<\/li>\r\n \t<li>Calculate the probability of two events that are not mutually exclusive<\/li>\r\n<\/ul>\r\n<\/div>\r\nWhen calculating probability, there are two rules to consider:\u00a0 <strong>The Multiplication Rule<\/strong> and <strong>The Addition Rule.<\/strong>\r\n\r\nBefore we talk about the Multiplication Rule, we need to define <strong>independent events<\/strong>.\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>\r\n<div class=\"textbox examples\">\r\n<h3>Examples<\/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) \u201cIt will rain tomorrow in Houston\u201d and (2) \u201cIt 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=\"513139\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"513139\"]\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\n<\/div>\r\nIn order to use the Multiplication Rule, we will be multiplying fractions, so let's review a couple of things.\r\n<div class=\"textbox examples\">\r\n<h3>recall multiplying fractions<\/h3>\r\nTo multiply fractions, place the product of the numerators over the product of the denominators.\r\n\r\n[latex]\\dfrac{a}{b} \\cdot \\dfrac{c}{d} = \\dfrac{ac}{bd}[\/latex]\r\n\r\n<\/div>\r\n<div class=\"textbox examples\">\r\n<h3><span style=\"text-align: initial; background-color: #dacacc;\">Recall fraction SIMPLIFICATION<\/span><\/h3>\r\nTo write a fraction in simplified terms, first take the prime factorization of the numerator and denominator, then cancel out factors that are common in the numerator and the denominator.\r\n\r\nEx. [latex]\\dfrac{12}{18}=\\dfrac{\\cancel{2}\\cdot 2\\cdot \\cancel{3}}{\\cancel{2}\\cdot 3\\cdot \\cancel{3}}=\\dfrac{2}{3}[\/latex]\r\n\r\n<\/div>\r\nWe are now ready for the Multiplication Rule.\r\n<div class=\"textbox tryit\">\r\n<h3>multiplication rule (and)<\/h3>\r\nAssume A and B are events in the sample space (all possible outcomes) of a probability experiment. Also, <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<strong>P(A and B) for Independent Events<\/strong>\r\n\r\nIf events <em>A<\/em> and <em>B<\/em> are independent, then\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\n<strong>P(A and B) for Events that are NOT Independent\r\n<\/strong>\r\n\r\nIf\u00a0<em>A<\/em> and <em>B<\/em> are <strong>not<\/strong> independent events, then\r\n<p style=\"text-align: center;\">[latex]P\\left(A\\text{ and }B\\right)=P\\left(A\\right)\\cdot{P}\\left(B \\right |A)[\/latex]<\/p>\r\n<p style=\"text-align: center;\">[latex]{P}\\left(B\\right |A)[\/latex] means the probability of [latex]B[\/latex] given that [latex]A[\/latex] has already occurred).<\/p>\r\n\r\n<\/div>\r\n<div><\/div>\r\n<div class=\"textbox exercises\">\r\n<h3>example (Multiplication rule with independent events)<\/h3>\r\nSuppose we flipped a coin and rolled a die, and wanted to know the probability of getting a head on the coin and a 6 on the die.\r\n[reveal-answer q=\"453669\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"453669\"]\r\n\r\nWe could list all possible outcomes: \u00a0{H1, H2, H3, H4, H5, H6, T1, T2, T3, T4, T5, T6}.\r\n\r\nNotice there are [latex]2\\cdot6=12[\/latex] total outcomes. Out of these, only 1 is the desired outcome, so the probability is [latex]\\frac{1}{12}[\/latex].\r\n\r\nThese are independent events (because a certain outcome from rolling a die has no influence on the outcome from flipping the coin), so we use the formula:\r\n\r\n[latex]P\\left(A\\text{ and }B\\right)=P\\left(A\\right)\\cdot{P}\\left(B\\right)[\/latex], where [latex]P\\left(A\\right)=\\frac{1}{2}[\/latex] (the probability of getting a head) and [latex]P\\left(B\\right)=\\frac{1}{6}[\/latex] (the probability of getting a 6).\r\n<p style=\"text-align: center;\">[latex]P\\left(A\\text{ and }B\\right)=\\frac{1}{2}\\cdot\\frac{1}{6}[\/latex].<\/p>\r\nMultiplying and simplifying, we get\r\n<p style=\"text-align: center;\">[latex]P\\left(A\\text{ and }B\\right)=\\frac{1}{12}[\/latex].[\/hidden-answer]<\/p>\r\n\r\n<\/div>\r\n&nbsp;\r\n<div class=\"textbox exercises\">\r\n<h3>example (Multiplication rule with events that are not independent)<\/h3>\r\nSuppose we wanted to eat two cookies from a cookie jar containing 4 chocolate chip cookies, 6 oatmeal raisin cookies, and 10 peanut butter cookies. Suppose we reach our hand into the cookie jar and choose one cookie and eat it. Then we reach our hand in and choose a second cookie and eat it. Find the probability that we would get a peanut butter cookie first and then a chocolate chip cookie.\r\n\r\n[reveal-answer q=\"268578\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"268578\"]\r\n\r\nThere are a total of 20 cookies, or 20 possible outcomes if we reach our hand into the cookie jar. There are 10 peanut butter cookies, so the probability of picking a peanut butter cookie first is [latex]\\frac{10}{20}[\/latex] or [latex]\\frac{1}{2}[\/latex]. Because the problem said we eat the first cookie, there are now only 19 cookies in the jar. This is called \"without replacement.\" The next cookie we pick will only be out of 19. These events are not independent because not replacing the first cookie changes the total number of cookies for the second pick. The probability of picking a chocolate chip cookie next will be [latex]\\frac{4}{19}[\/latex].\r\n\r\nUsing the multiplication formula for events that are not independent:\r\n<p style=\"text-align: center;\">[latex]P\\left(A\\text{ and }B\\right)=P\\left(A\\right)\\cdot{P}\\left(B \\right |A)[\/latex]<\/p>\r\nwe get\r\n\r\n[latex]P[\/latex](first cookie is peanut butter AND second cookie is chocolate chip) = [latex]P[\/latex](second cookie is chocolate chip given that the first is peanut butter), or\r\n<p style=\"text-align: center;\">[latex]P\\left(A\\text{ and }B\\right)=\\frac{1}{2}\\cdot\\frac{4}{19}[\/latex].<\/p>\r\nMultiplying and simplifying, we get\r\n<p style=\"text-align: center;\">[latex]P\\left(A\\text{ and }B\\right)=\\frac{2}{19}[\/latex].[\/hidden-answer]<\/p>\r\n\r\n<\/div>\r\n&nbsp;\r\n\r\nNow we will look at the probability of <em>either<\/em> event occurring.\r\n<div class=\"textbox tryit\">\r\n<h3>ADDITION RULE (OR)<\/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&nbsp;\r\n<div class=\"textbox exercises\">\r\n<h3>example (ADDITION RULE)<\/h3>\r\nSuppose we flipped a coin and rolled a die, and wanted to know the probability of getting a head on the coin <em>or<\/em> a 6 on the die.\r\n[reveal-answer q=\"443646\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"443646\"]\r\n\r\nHere, there are still 12 possible outcomes: {H1, H2, H3, H4, H5, H6, T1, T2, T3, T4, T5, T6}\r\n\r\nBy simply counting, we can see that 7 of the outcomes have a head on the coin <em>or<\/em> a 6 on the die <em>or<\/em> both \u2013 we use <em>or<\/em> inclusively here (these 7 outcomes are H1, H2, H3, H4, H5, H6, T6), so the probability is [latex]\\frac{7}{12}[\/latex]. How could we have found this from the individual probabilities?\r\n\r\nAs we would expect, [latex]\\frac{1}{2}[\/latex] of these outcomes have a head, and [latex]\\frac{1}{6}[\/latex] of these outcomes have a 6 on the die. If we add these, [latex]\\frac{1}{2}+\\frac{1}{6}=\\frac{6}{12}+\\frac{2}{12}=\\frac{8}{12}[\/latex], which is not the correct probability. Looking at the outcomes we can see why: the outcome H6 would have been counted twice, since it contains both a head and a 6; the probability of both a head <em>and<\/em> rolling a 6 is [latex]\\frac{1}{12}[\/latex].\r\n\r\nIf we subtract out this double count, we have the correct probability: [latex]\\frac{8}{12}-\\frac{1}{12}=\\frac{7}{12}[\/latex].\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox exercises\">\r\n<h3>example<\/h3>\r\n<a href=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/5548\/2016\/10\/04191241\/Deck-of-Cards.png\"><img class=\"alignnone size-medium wp-image-5985\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/5548\/2016\/10\/04191241\/Deck-of-Cards-223x300.png\" alt=\"\" width=\"223\" height=\"300\" \/><\/a>\r\n\r\nSuppose we draw one card from a standard deck. What is the probability that we get a Queen or a King?\r\n[reveal-answer q=\"657503\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"657503\"]\r\n\r\nThere are 4 Queens and 4 Kings in the deck, hence 8 outcomes corresponding to a Queen or King out of 52 possible outcomes. Thus the probability of drawing a Queen or a King is:\r\n<p style=\"text-align: center;\">[latex]P(\\text{King or Queen})=\\frac{8}{52}[\/latex]<\/p>\r\nNote that in this case, there are no cards that are both a Queen and a King, so [latex]P(\\text{King and Queen})=0[\/latex]. Using our probability rule, we could have said:\r\n<p style=\"text-align: center;\">[latex]P(\\text{King or Queen})=P(\\text{King})+P(\\text{Queen})-P(\\text{King and Queen})=\\frac{4}{52}+\\frac{4}{52}-0=\\frac{8}{52}=\\frac{2}{13}[\/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\r\nWhen the events are <strong>not mutually exclusive<\/strong>, it is useful to draw a picture or use a diagram.\u00a0 For instance, let's use a picture of a deck of cards to calculate probability.\r\n\r\n<a href=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/5548\/2016\/10\/04191241\/Deck-of-Cards.png\"><img class=\"size-medium wp-image-5985 aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/5548\/2016\/10\/04191241\/Deck-of-Cards-223x300.png\" alt=\"\" width=\"223\" height=\"300\" \/><\/a>\r\n<div class=\"textbox examples\">\r\n<h3>Example<\/h3>\r\nUsing the picture of a standard deck of cards above, calculate the following probabilities.\r\n\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\r\n<a href=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/5548\/2016\/10\/04193205\/Red-Card-and-King1.png\"><img class=\"size-medium wp-image-5989 aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/5548\/2016\/10\/04193205\/Red-Card-and-King1-201x300.png\" alt=\"\" width=\"201\" height=\"300\" \/><\/a>\r\n\r\nFrom the picture, we can determine that the number of cards that are red or a King is the 26 red cards plus the King of spades and the King of clubs for a total of 28 cards.\u00a0 Therefore the P(red card if a King) = 28\/52 which simplifies to 7\/13.\r\n\r\nLet's look at the same example using the formula.\r\n\r\nP(A or B)\u00a0 =\u00a0 P(A) + P(B) - P(A and B)\r\n\r\nP(red or King) = P(red) + P(King) - P(red and King)\r\n\r\nP(red or King) = 26\/52\u00a0 +\u00a0 \u00a04\/52\u00a0 \u00a0 -\u00a0 \u00a0 2\/52\r\n\r\nP(red or King) = 28\/52\r\n\r\nP(red or King) = 7\/13\r\n\r\n<\/div>\r\n<a href=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/5548\/2016\/10\/04220345\/red-and-blue-die.png\"><img class=\"alignnone size-full wp-image-6037\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/5548\/2016\/10\/04220345\/red-and-blue-die.png\" alt=\"\" width=\"95\" height=\"55\" \/><\/a>Let's calculate probability when rolling 2 dice.\u00a0 How many different outcomes are possible when rolling 2 dice, one red and one blue?\u00a0 Start with drawing a picture of the possible outcomes.\r\n\r\n<a href=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/5548\/2016\/10\/04220607\/outcome-for-2-standard-dice.png\"><img class=\"size-medium wp-image-6038 aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/5548\/2016\/10\/04220607\/outcome-for-2-standard-dice-300x232.png\" alt=\"\" width=\"300\" height=\"232\" \/><\/a>\r\n<div class=\"textbox examples\">\r\n<h3>Example<\/h3>\r\nRoll 2 dice.\u00a0 What's the probability that the sum of the 2 dice is either a 7 or 11?\u00a0 That is, P(sum is 7 or 11)\r\n\r\n<a href=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/5548\/2016\/10\/04221246\/sum-of-7-or-11.png\"><img class=\"size-medium wp-image-6039 aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/5548\/2016\/10\/04221246\/sum-of-7-or-11-300x216.png\" alt=\"\" width=\"300\" height=\"216\" \/><\/a>\r\n\r\nP(sum is 7) =\u00a0 6\/36\r\n\r\nP(sum is 11) = 2\/36\r\n\r\nThese events are mutually exclusive since the sum cannot be 7 and 11.\r\n\r\nP(sum is 7 or 11) = P(sum is 7) + P(sum is 11)\r\n\r\nP(sum is 7 or 11) =\u00a0 \u00a0 \u00a0 6\/36\u00a0 \u00a0 \u00a0 \u00a0+\u00a0 \u00a0 \u00a0 \u00a02\/36\r\n\r\nP(sum is 7 or 11) =\u00a0 \u00a08\/36 which simplifies to 2\/9\r\n\r\n<\/div>\r\n<div class=\"textbox examples\">\r\n<h3>Examples<\/h3>\r\nRoll 2 dice.\u00a0 What's the probability that the sum of the 2 dice is greater than or equal to 10 or the red die shows a 5 on its face?\u00a0 That is, P(sum <span style=\"text-decoration: underline;\">&gt;<\/span> 10 or 5 on red die)\r\n\r\n<a href=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/5548\/2016\/10\/04223433\/dice-problem-2.png\"><img class=\"size-medium wp-image-6044 aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/5548\/2016\/10\/04223433\/dice-problem-2-300x223.png\" alt=\"\" width=\"300\" height=\"223\" \/><\/a>Looking at the picture and counting the outcomes that are shaded the probability is 10\/36 which simplifies to 5\/18.\r\n\r\nNow, let's walk through the <span style=\"text-decoration: underline;\">or<\/span> formula.\r\n\r\nP(sum <span style=\"text-decoration: underline;\">&gt;<\/span> 10) =\u00a0 6\/36\r\n\r\nP(red die is a 5) = 6\/36\r\n\r\nThese events are not mutually exclusive since the sum can have a 5 on the red die and be a sum greater than or equal to 10.\r\n\r\nP(sum <span style=\"text-decoration: underline;\">&gt;<\/span>10 or red 5) = P( <span style=\"text-decoration: underline;\">&gt;<\/span> 10) + P(red 5) - P( <span style=\"text-decoration: underline;\">&gt;<\/span>10&amp;red 5)\r\n\r\nP(sum <span style=\"text-decoration: underline;\">&gt;<\/span>10 or red 5) =\u00a0 \u00a0 \u00a0 6\/36\u00a0 \u00a0+\u00a0 \u00a06\/36\u00a0 \u00a0-\u00a0 \u00a02\/36\r\n\r\nP(sum &gt;10 or red 5) =\u00a0 \u00a010\/36 which simplifies to 5\/18.\r\n\r\n<\/div>\r\n<p style=\"text-align: center;\"><span style=\"color: #ff0000;\"><strong>This is the end of the section. Close this tab and proceed to the corresponding assignment.<\/strong><\/span><\/p>","rendered":"<div class=\"textbox learning-objectives\">\n<h3>Learning Objectives<\/h3>\n<p>1.\u00a0 The Multiplication Rule (AND)<\/p>\n<ul>\n<li>Calculate the probability of two independent events both occurring together<\/li>\n<\/ul>\n<p>2.\u00a0 The Addition Rule (OR)<\/p>\n<ul>\n<li>Calculate the probability of two mutually exclusive events<\/li>\n<li>Calculate the probability of two events that are not mutually exclusive<\/li>\n<\/ul>\n<\/div>\n<p>When calculating probability, there are two rules to consider:\u00a0 <strong>The Multiplication Rule<\/strong> and <strong>The Addition Rule.<\/strong><\/p>\n<p>Before we talk about the Multiplication Rule, we need to define <strong>independent events<\/strong>.<\/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>\n<div class=\"textbox examples\">\n<h3>Examples<\/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) \u201cIt will rain tomorrow in Houston\u201d and (2) \u201cIt 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=\"q513139\">Show Answer<\/span><\/p>\n<div id=\"q513139\" 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<\/div>\n<p>In order to use the Multiplication Rule, we will be multiplying fractions, so let&#8217;s review a couple of things.<\/p>\n<div class=\"textbox examples\">\n<h3>recall multiplying fractions<\/h3>\n<p>To multiply fractions, place the product of the numerators over the product of the denominators.<\/p>\n<p>[latex]\\dfrac{a}{b} \\cdot \\dfrac{c}{d} = \\dfrac{ac}{bd}[\/latex]<\/p>\n<\/div>\n<div class=\"textbox examples\">\n<h3><span style=\"text-align: initial; background-color: #dacacc;\">Recall fraction SIMPLIFICATION<\/span><\/h3>\n<p>To write a fraction in simplified terms, first take the prime factorization of the numerator and denominator, then cancel out factors that are common in the numerator and the denominator.<\/p>\n<p>Ex. [latex]\\dfrac{12}{18}=\\dfrac{\\cancel{2}\\cdot 2\\cdot \\cancel{3}}{\\cancel{2}\\cdot 3\\cdot \\cancel{3}}=\\dfrac{2}{3}[\/latex]<\/p>\n<\/div>\n<p>We are now ready for the Multiplication Rule.<\/p>\n<div class=\"textbox tryit\">\n<h3>multiplication rule (and)<\/h3>\n<p>Assume A and B are events in the sample space (all possible outcomes) of a probability experiment. Also, <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<p><strong>P(A and B) for Independent Events<\/strong><\/p>\n<p>If events <em>A<\/em> and <em>B<\/em> are independent, then<\/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><strong>P(A and B) for Events that are NOT Independent<br \/>\n<\/strong><\/p>\n<p>If\u00a0<em>A<\/em> and <em>B<\/em> are <strong>not<\/strong> independent events, then<\/p>\n<p style=\"text-align: center;\">[latex]P\\left(A\\text{ and }B\\right)=P\\left(A\\right)\\cdot{P}\\left(B \\right |A)[\/latex]<\/p>\n<p style=\"text-align: center;\">[latex]{P}\\left(B\\right |A)[\/latex] means the probability of [latex]B[\/latex] given that [latex]A[\/latex] has already occurred).<\/p>\n<\/div>\n<div><\/div>\n<div class=\"textbox exercises\">\n<h3>example (Multiplication rule with independent events)<\/h3>\n<p>Suppose we flipped a coin and rolled a die, and wanted to know the probability of getting a head on the coin and a 6 on the die.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q453669\">Show Solution<\/span><\/p>\n<div id=\"q453669\" class=\"hidden-answer\" style=\"display: none\">\n<p>We could list all possible outcomes: \u00a0{H1, H2, H3, H4, H5, H6, T1, T2, T3, T4, T5, T6}.<\/p>\n<p>Notice there are [latex]2\\cdot6=12[\/latex] total outcomes. Out of these, only 1 is the desired outcome, so the probability is [latex]\\frac{1}{12}[\/latex].<\/p>\n<p>These are independent events (because a certain outcome from rolling a die has no influence on the outcome from flipping the coin), so we use the formula:<\/p>\n<p>[latex]P\\left(A\\text{ and }B\\right)=P\\left(A\\right)\\cdot{P}\\left(B\\right)[\/latex], where [latex]P\\left(A\\right)=\\frac{1}{2}[\/latex] (the probability of getting a head) and [latex]P\\left(B\\right)=\\frac{1}{6}[\/latex] (the probability of getting a 6).<\/p>\n<p style=\"text-align: center;\">[latex]P\\left(A\\text{ and }B\\right)=\\frac{1}{2}\\cdot\\frac{1}{6}[\/latex].<\/p>\n<p>Multiplying and simplifying, we get<\/p>\n<p style=\"text-align: center;\">[latex]P\\left(A\\text{ and }B\\right)=\\frac{1}{12}[\/latex].<\/div>\n<\/div>\n<\/div>\n<p>&nbsp;<\/p>\n<div class=\"textbox exercises\">\n<h3>example (Multiplication rule with events that are not independent)<\/h3>\n<p>Suppose we wanted to eat two cookies from a cookie jar containing 4 chocolate chip cookies, 6 oatmeal raisin cookies, and 10 peanut butter cookies. Suppose we reach our hand into the cookie jar and choose one cookie and eat it. Then we reach our hand in and choose a second cookie and eat it. Find the probability that we would get a peanut butter cookie first and then a chocolate chip cookie.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q268578\">Show Answer<\/span><\/p>\n<div id=\"q268578\" class=\"hidden-answer\" style=\"display: none\">\n<p>There are a total of 20 cookies, or 20 possible outcomes if we reach our hand into the cookie jar. There are 10 peanut butter cookies, so the probability of picking a peanut butter cookie first is [latex]\\frac{10}{20}[\/latex] or [latex]\\frac{1}{2}[\/latex]. Because the problem said we eat the first cookie, there are now only 19 cookies in the jar. This is called &#8220;without replacement.&#8221; The next cookie we pick will only be out of 19. These events are not independent because not replacing the first cookie changes the total number of cookies for the second pick. The probability of picking a chocolate chip cookie next will be [latex]\\frac{4}{19}[\/latex].<\/p>\n<p>Using the multiplication formula for events that are not independent:<\/p>\n<p style=\"text-align: center;\">[latex]P\\left(A\\text{ and }B\\right)=P\\left(A\\right)\\cdot{P}\\left(B \\right |A)[\/latex]<\/p>\n<p>we get<\/p>\n<p>[latex]P[\/latex](first cookie is peanut butter AND second cookie is chocolate chip) = [latex]P[\/latex](second cookie is chocolate chip given that the first is peanut butter), or<\/p>\n<p style=\"text-align: center;\">[latex]P\\left(A\\text{ and }B\\right)=\\frac{1}{2}\\cdot\\frac{4}{19}[\/latex].<\/p>\n<p>Multiplying and simplifying, we get<\/p>\n<p style=\"text-align: center;\">[latex]P\\left(A\\text{ and }B\\right)=\\frac{2}{19}[\/latex].<\/div>\n<\/div>\n<\/div>\n<p>&nbsp;<\/p>\n<p>Now we will look at the probability of <em>either<\/em> event occurring.<\/p>\n<div class=\"textbox tryit\">\n<h3>ADDITION RULE (OR)<\/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<p>&nbsp;<\/p>\n<div class=\"textbox exercises\">\n<h3>example (ADDITION RULE)<\/h3>\n<p>Suppose we flipped a coin and rolled a die, and wanted to know the probability of getting a head on the coin <em>or<\/em> a 6 on the die.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q443646\">Show Solution<\/span><\/p>\n<div id=\"q443646\" class=\"hidden-answer\" style=\"display: none\">\n<p>Here, there are still 12 possible outcomes: {H1, H2, H3, H4, H5, H6, T1, T2, T3, T4, T5, T6}<\/p>\n<p>By simply counting, we can see that 7 of the outcomes have a head on the coin <em>or<\/em> a 6 on the die <em>or<\/em> both \u2013 we use <em>or<\/em> inclusively here (these 7 outcomes are H1, H2, H3, H4, H5, H6, T6), so the probability is [latex]\\frac{7}{12}[\/latex]. How could we have found this from the individual probabilities?<\/p>\n<p>As we would expect, [latex]\\frac{1}{2}[\/latex] of these outcomes have a head, and [latex]\\frac{1}{6}[\/latex] of these outcomes have a 6 on the die. If we add these, [latex]\\frac{1}{2}+\\frac{1}{6}=\\frac{6}{12}+\\frac{2}{12}=\\frac{8}{12}[\/latex], which is not the correct probability. Looking at the outcomes we can see why: the outcome H6 would have been counted twice, since it contains both a head and a 6; the probability of both a head <em>and<\/em> rolling a 6 is [latex]\\frac{1}{12}[\/latex].<\/p>\n<p>If we subtract out this double count, we have the correct probability: [latex]\\frac{8}{12}-\\frac{1}{12}=\\frac{7}{12}[\/latex].<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox exercises\">\n<h3>example<\/h3>\n<p><a href=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/5548\/2016\/10\/04191241\/Deck-of-Cards.png\"><img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-medium wp-image-5985\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/5548\/2016\/10\/04191241\/Deck-of-Cards-223x300.png\" alt=\"\" width=\"223\" height=\"300\" \/><\/a><\/p>\n<p>Suppose we draw one card from a standard deck. What is the probability that we get a Queen or a King?<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q657503\">Show Solution<\/span><\/p>\n<div id=\"q657503\" class=\"hidden-answer\" style=\"display: none\">\n<p>There are 4 Queens and 4 Kings in the deck, hence 8 outcomes corresponding to a Queen or King out of 52 possible outcomes. Thus the probability of drawing a Queen or a King is:<\/p>\n<p style=\"text-align: center;\">[latex]P(\\text{King or Queen})=\\frac{8}{52}[\/latex]<\/p>\n<p>Note that in this case, there are no cards that are both a Queen and a King, so [latex]P(\\text{King and Queen})=0[\/latex]. Using our probability rule, we could have said:<\/p>\n<p style=\"text-align: center;\">[latex]P(\\text{King or Queen})=P(\\text{King})+P(\\text{Queen})-P(\\text{King and Queen})=\\frac{4}{52}+\\frac{4}{52}-0=\\frac{8}{52}=\\frac{2}{13}[\/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-1\" 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<p>When the events are <strong>not mutually exclusive<\/strong>, it is useful to draw a picture or use a diagram.\u00a0 For instance, let&#8217;s use a picture of a deck of cards to calculate probability.<\/p>\n<p><a href=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/5548\/2016\/10\/04191241\/Deck-of-Cards.png\"><img loading=\"lazy\" decoding=\"async\" class=\"size-medium wp-image-5985 aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/5548\/2016\/10\/04191241\/Deck-of-Cards-223x300.png\" alt=\"\" width=\"223\" height=\"300\" \/><\/a><\/p>\n<div class=\"textbox examples\">\n<h3>Example<\/h3>\n<p>Using the picture of a standard deck of cards above, calculate the following probabilities.<\/p>\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<p><a href=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/5548\/2016\/10\/04193205\/Red-Card-and-King1.png\"><img loading=\"lazy\" decoding=\"async\" class=\"size-medium wp-image-5989 aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/5548\/2016\/10\/04193205\/Red-Card-and-King1-201x300.png\" alt=\"\" width=\"201\" height=\"300\" \/><\/a><\/p>\n<p>From the picture, we can determine that the number of cards that are red or a King is the 26 red cards plus the King of spades and the King of clubs for a total of 28 cards.\u00a0 Therefore the P(red card if a King) = 28\/52 which simplifies to 7\/13.<\/p>\n<p>Let&#8217;s look at the same example using the formula.<\/p>\n<p>P(A or B)\u00a0 =\u00a0 P(A) + P(B) &#8211; P(A and B)<\/p>\n<p>P(red or King) = P(red) + P(King) &#8211; P(red and King)<\/p>\n<p>P(red or King) = 26\/52\u00a0 +\u00a0 \u00a04\/52\u00a0 \u00a0 &#8211;\u00a0 \u00a0 2\/52<\/p>\n<p>P(red or King) = 28\/52<\/p>\n<p>P(red or King) = 7\/13<\/p>\n<\/div>\n<p><a href=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/5548\/2016\/10\/04220345\/red-and-blue-die.png\"><img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-full wp-image-6037\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/5548\/2016\/10\/04220345\/red-and-blue-die.png\" alt=\"\" width=\"95\" height=\"55\" \/><\/a>Let&#8217;s calculate probability when rolling 2 dice.\u00a0 How many different outcomes are possible when rolling 2 dice, one red and one blue?\u00a0 Start with drawing a picture of the possible outcomes.<\/p>\n<p><a href=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/5548\/2016\/10\/04220607\/outcome-for-2-standard-dice.png\"><img loading=\"lazy\" decoding=\"async\" class=\"size-medium wp-image-6038 aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/5548\/2016\/10\/04220607\/outcome-for-2-standard-dice-300x232.png\" alt=\"\" width=\"300\" height=\"232\" \/><\/a><\/p>\n<div class=\"textbox examples\">\n<h3>Example<\/h3>\n<p>Roll 2 dice.\u00a0 What&#8217;s the probability that the sum of the 2 dice is either a 7 or 11?\u00a0 That is, P(sum is 7 or 11)<\/p>\n<p><a href=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/5548\/2016\/10\/04221246\/sum-of-7-or-11.png\"><img loading=\"lazy\" decoding=\"async\" class=\"size-medium wp-image-6039 aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/5548\/2016\/10\/04221246\/sum-of-7-or-11-300x216.png\" alt=\"\" width=\"300\" height=\"216\" \/><\/a><\/p>\n<p>P(sum is 7) =\u00a0 6\/36<\/p>\n<p>P(sum is 11) = 2\/36<\/p>\n<p>These events are mutually exclusive since the sum cannot be 7 and 11.<\/p>\n<p>P(sum is 7 or 11) = P(sum is 7) + P(sum is 11)<\/p>\n<p>P(sum is 7 or 11) =\u00a0 \u00a0 \u00a0 6\/36\u00a0 \u00a0 \u00a0 \u00a0+\u00a0 \u00a0 \u00a0 \u00a02\/36<\/p>\n<p>P(sum is 7 or 11) =\u00a0 \u00a08\/36 which simplifies to 2\/9<\/p>\n<\/div>\n<div class=\"textbox examples\">\n<h3>Examples<\/h3>\n<p>Roll 2 dice.\u00a0 What&#8217;s the probability that the sum of the 2 dice is greater than or equal to 10 or the red die shows a 5 on its face?\u00a0 That is, P(sum <span style=\"text-decoration: underline;\">&gt;<\/span> 10 or 5 on red die)<\/p>\n<p><a href=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/5548\/2016\/10\/04223433\/dice-problem-2.png\"><img loading=\"lazy\" decoding=\"async\" class=\"size-medium wp-image-6044 aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/5548\/2016\/10\/04223433\/dice-problem-2-300x223.png\" alt=\"\" width=\"300\" height=\"223\" \/><\/a>Looking at the picture and counting the outcomes that are shaded the probability is 10\/36 which simplifies to 5\/18.<\/p>\n<p>Now, let&#8217;s walk through the <span style=\"text-decoration: underline;\">or<\/span> formula.<\/p>\n<p>P(sum <span style=\"text-decoration: underline;\">&gt;<\/span> 10) =\u00a0 6\/36<\/p>\n<p>P(red die is a 5) = 6\/36<\/p>\n<p>These events are not mutually exclusive since the sum can have a 5 on the red die and be a sum greater than or equal to 10.<\/p>\n<p>P(sum <span style=\"text-decoration: underline;\">&gt;<\/span>10 or red 5) = P( <span style=\"text-decoration: underline;\">&gt;<\/span> 10) + P(red 5) &#8211; P( <span style=\"text-decoration: underline;\">&gt;<\/span>10&amp;red 5)<\/p>\n<p>P(sum <span style=\"text-decoration: underline;\">&gt;<\/span>10 or red 5) =\u00a0 \u00a0 \u00a0 6\/36\u00a0 \u00a0+\u00a0 \u00a06\/36\u00a0 \u00a0&#8211;\u00a0 \u00a02\/36<\/p>\n<p>P(sum &gt;10 or red 5) =\u00a0 \u00a010\/36 which simplifies to 5\/18.<\/p>\n<\/div>\n<p style=\"text-align: center;\"><span style=\"color: #ff0000;\"><strong>This is the end of the section. Close this tab and proceed to the corresponding assignment.<\/strong><\/span><\/p>\n","protected":false},"author":383311,"menu_order":5,"template":"","meta":{"_candela_citation":"[]","CANDELA_OUTCOMES_GUID":"","pb_show_title":"on","pb_short_title":"","pb_subtitle":"","pb_authors":[],"pb_section_license":""},"chapter-type":[],"contributor":[],"license":[],"class_list":["post-6050","chapter","type-chapter","status-publish","hentry"],"part":329,"_links":{"self":[{"href":"https:\/\/courses.lumenlearning.com\/frontrange-mathforliberalartscorequisite1\/wp-json\/pressbooks\/v2\/chapters\/6050","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/courses.lumenlearning.com\/frontrange-mathforliberalartscorequisite1\/wp-json\/pressbooks\/v2\/chapters"}],"about":[{"href":"https:\/\/courses.lumenlearning.com\/frontrange-mathforliberalartscorequisite1\/wp-json\/wp\/v2\/types\/chapter"}],"author":[{"embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/frontrange-mathforliberalartscorequisite1\/wp-json\/wp\/v2\/users\/383311"}],"version-history":[{"count":66,"href":"https:\/\/courses.lumenlearning.com\/frontrange-mathforliberalartscorequisite1\/wp-json\/pressbooks\/v2\/chapters\/6050\/revisions"}],"predecessor-version":[{"id":6740,"href":"https:\/\/courses.lumenlearning.com\/frontrange-mathforliberalartscorequisite1\/wp-json\/pressbooks\/v2\/chapters\/6050\/revisions\/6740"}],"part":[{"href":"https:\/\/courses.lumenlearning.com\/frontrange-mathforliberalartscorequisite1\/wp-json\/pressbooks\/v2\/parts\/329"}],"metadata":[{"href":"https:\/\/courses.lumenlearning.com\/frontrange-mathforliberalartscorequisite1\/wp-json\/pressbooks\/v2\/chapters\/6050\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/courses.lumenlearning.com\/frontrange-mathforliberalartscorequisite1\/wp-json\/wp\/v2\/media?parent=6050"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/frontrange-mathforliberalartscorequisite1\/wp-json\/pressbooks\/v2\/chapter-type?post=6050"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/frontrange-mathforliberalartscorequisite1\/wp-json\/wp\/v2\/contributor?post=6050"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/frontrange-mathforliberalartscorequisite1\/wp-json\/wp\/v2\/license?post=6050"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}