{"id":2481,"date":"2017-03-31T22:56:41","date_gmt":"2017-03-31T22:56:41","guid":{"rendered":"https:\/\/courses.lumenlearning.com\/waymakermath4libarts\/?post_type=chapter&#038;p=2481"},"modified":"2019-05-30T17:01:15","modified_gmt":"2019-05-30T17:01:15","slug":"probability-using-permutations-and-combinations","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/waymakermath4libarts\/chapter\/probability-using-permutations-and-combinations\/","title":{"raw":"Probability Using Permutations and Combinations","rendered":"Probability Using Permutations and Combinations"},"content":{"raw":"<div class=\"textbox learning-objectives\">\r\n<h3>Learning Outcomes<\/h3>\r\n<ul>\r\n \t<li>Compute a conditional probability for an event<\/li>\r\n \t<li>Use Baye\u2019s theorem to compute a conditional probability<\/li>\r\n \t<li>Calculate the expected value of an event<\/li>\r\n<\/ul>\r\n<\/div>\r\nWe can use permutations and combinations to help us answer more complex probability questions.\r\n<div class=\"textbox exercises\">\r\n<h3>examples<\/h3>\r\nA 4 digit PIN number is selected. What is the probability that there are no repeated digits?\r\n[reveal-answer q=\"108643\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"108643\"]\r\n\r\nThere are 10 possible values for each digit of the PIN (namely: 0, 1, 2, 3, 4, 5, 6, 7, 8, 9), so there are 10 \u00b7 10 \u00b7 10 \u00b7 10 = 10<sup>4<\/sup> = 10000 total possible PIN numbers.\r\n\r\nTo have no repeated digits, all four digits would have to be different, which is selecting without replacement. We could either compute 10 \u00b7 9 \u00b7 8 \u00b7 7, or notice that this is the same as the permutation 10<em>P<\/em>4 = 5040.\r\n\r\nThe probability of no repeated digits is the number of 4 digit PIN numbers with no repeated digits divided by the total number of 4 digit PIN numbers. This probability is [latex]\\frac{{}_{10}{{P}_{4}}}{{{10}^{4}}}=\\frac{5040}{10000}=0.504[\/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<iframe id=\"mom10\" class=\"resizable\" src=\"https:\/\/www.myopenmath.com\/multiembedq.php?id=7158&amp;theme=oea&amp;iframe_resize_id=mom10\" width=\"100%\" height=\"400\"><\/iframe>\r\n\r\n<\/div>\r\n<div class=\"textbox exercises\">\r\n<h3>Example<\/h3>\r\nIn a certain state's lottery, 48 balls numbered 1 through 48 are placed in a machine and six of them are drawn at random. If the six numbers drawn match the numbers that a player had chosen, the player wins $1,000,000. \u00a0\u00a0\u00a0In this lottery, the order the numbers are drawn in doesn\u2019t matter. Compute the probability that you win the million-dollar prize if you purchase a single lottery ticket.\r\n[reveal-answer q=\"767485\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"767485\"]\r\n\r\nIn order to compute the probability, we need to count the total number of ways six numbers can be drawn, and the number of ways the six numbers on the player\u2019s ticket could match the six numbers drawn from the machine. Since there is no stipulation that the numbers be in any particular order, the number of possible outcomes of the lottery drawing is 48<em>C<\/em>6 = 12,271,512. Of these possible outcomes, only one would match all six numbers on the player\u2019s ticket, so the probability of winning the grand prize is:\r\n\r\n[latex]\\frac{{}_{6}{{C}_{6}}}{{}_{48}{{C}_{6}}}=\\frac{1}{12271512}\\approx0.0000000815[\/latex]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n&nbsp;\r\n<div class=\"textbox exercises\">\r\n<h3>Example<\/h3>\r\nIn the state lottery from the previous example, if five of the six numbers drawn match the numbers that a player has chosen, the player wins a second prize of $1,000. Compute the probability that you win the second prize if you purchase a single lottery ticket.\r\n[reveal-answer q=\"651173\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"651173\"]\r\n\r\nAs above, the number of possible outcomes of the lottery drawing is 48<em>C<\/em>6 = 12,271,512. In order to win the second prize, five of the six numbers on the ticket must match five of the six winning numbers; in other words, we must have chosen five of the six winning numbers and one of the 42 losing numbers. The number of ways to choose 5 out of the 6 winning numbers is given by 6<em>C<\/em>5 = 6 and the number of ways to choose 1 out of the 42 losing numbers is given by 42<em>C<\/em>1 = 42. Thus the number of favorable outcomes is then given by the Basic Counting Rule: 6<em>C<\/em>5 \u00b7 42<em>C<\/em>1 = 6 \u00b7 42 = 252. So the probability of winning the second prize is.\r\n\r\n[latex]\\frac{\\left({}_{6}{{C}_{5}}\\right)\\left({}_{42}{{C}_{1}}\\right)}{{}_{48}{{C}_{6}}}=\\frac{252}{12271512}\\approx0.0000205[\/latex]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\nThe previous examples are worked in the following video.\r\n\r\nhttps:\/\/youtu.be\/b9LFbB_aNAo\r\n<div class=\"textbox exercises\">\r\n<h3>examples<\/h3>\r\nCompute the probability of randomly drawing five cards from a deck and getting exactly one Ace.\r\n\r\n[reveal-answer q=\"542442\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"542442\"]\r\n\r\nIn many card games (such as poker) the order in which the cards are drawn is not important (since the player may rearrange the cards in his hand any way he chooses); in the problems that follow, we will assume that this is the case unless otherwise stated. Thus we use combinations to compute the possible number of 5-card hands, 52<em>C<\/em>5. This number will go in the denominator of\u00a0our probability formula, since it is the number of possible outcomes.\r\n\r\nFor the numerator, we need the number of ways to draw one Ace and four other cards (none of them Aces) from the deck.\u00a0 Since there are four Aces and we want exactly one of them, there will be 4<em>C<\/em>1 ways to select one Ace; since there are 48 non-Aces and we want 4 of them, there will be 48<em>C<\/em>4 ways to select the four non-Aces.\u00a0 Now we use the Basic Counting Rule to calculate that there will be 4<em>C<\/em>1 \u00b7 48<em>C<\/em>4 ways to choose one ace and four non-Aces.\r\n\r\nPutting this all together, we have\r\n<p style=\"text-align: center;\">[latex]P(\\text{oneAce})=\\frac{\\left({}_{4}{{C}_{1}}\\right)\\left({}_{48}{{C}_{4}}\\right)}{{}_{52}{{C}_{5}}}=\\frac{778320}{2598960}\\approx0.299[\/latex]<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox exercises\">\r\n<h3>Example<\/h3>\r\nCompute the probability of randomly drawing five cards from a deck and getting exactly two Aces.\r\n[reveal-answer q=\"49360\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"49360\"]\r\n\r\nThe solution is similar to the previous example, except now we are choosing 2 Aces out of 4 and 3 non-Aces out of 48; the denominator remains the same:\r\n\r\n[latex]P(\\text{twoAces})=\\frac{\\left({}_{4}{{C}_{2}}\\right)\\left({}_{48}{{C}_{3}}\\right)}{{}_{52}{{C}_{5}}}=\\frac{103776}{2598960}\\approx0.0399[\/latex]\r\n\r\nIt is useful to note that these card problems are remarkably similar to the lottery problems discussed earlier.\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\nView the following for further demonstration of these examples.\r\n\r\nhttps:\/\/youtu.be\/RU3e3KTkjoA\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Try It<\/h3>\r\n<iframe id=\"mom1\" class=\"resizable\" src=\"https:\/\/www.myopenmath.com\/multiembedq.php?id=7157&amp;theme=oea&amp;iframe_resize_id=mom1\" width=\"100%\" height=\"400\"><\/iframe>\r\n\r\n<\/div>\r\n<h2>Birthday Problem<\/h2>\r\nLet's take a pause to consider a famous problem in probability theory:\r\n<div class=\"textbox shaded\">Suppose you have a room full of 30 people. What is the probability that there is at least one shared birthday?<\/div>\r\nTake a guess at the answer to the above problem. Was your guess fairly low, like around 10%? That seems to be the intuitive answer (30\/365, perhaps?). Let's see if we should listen to our intuition. Let's start with a simpler problem, however.\r\n<div class=\"textbox exercises\">\r\n<h3>example<\/h3>\r\nSuppose three people are in a room.\u00a0 What is the probability that there is at least one shared birthday among these three people?\r\n[reveal-answer q=\"226841\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"226841\"]\r\n\r\nThere are a lot of ways there could be at least one shared birthday. Fortunately there is an easier way. We ask ourselves \u201cWhat is the alternative to having at least one shared birthday?\u201d In this case, the alternative is that there are <strong>no<\/strong> shared birthdays. In other words, the alternative to \u201cat least one\u201d is having <strong>none<\/strong>. In other words, since this is a complementary event,\r\n<p style=\"text-align: center;\">P(at least one) = 1 \u2013 P(none)<\/p>\r\nWe will start, then, by computing the probability that there is no shared birthday.\u00a0 Let's imagine that you are one of these three people.\u00a0 Your birthday can be anything without conflict, so there are 365 choices out of 365 for your birthday. What is the probability that the second person does not share your birthday?\u00a0 There are 365 days in the year (let's ignore leap years) and removing your birthday from contention, there are 364 choices that will guarantee that you do not share a birthday with this person, so the probability that the second person does not share your birthday is 364\/365.\u00a0 Now we move to the third person.\u00a0 What is the probability that this third person does not have the same birthday as either you or the second person?\u00a0 There are 363 days that will not duplicate your birthday or the second person's, so the probability that the third person does not share a birthday with the first two is 363\/365.\r\n\r\nWe want the second person not to share a birthday with you <em>and<\/em> the third person not to share a birthday with the first two people, so we use the multiplication rule:\r\n<p style=\"text-align: center;\">[latex]P(\\text{nosharedbirthday})=\\frac{365}{365}\\cdot\\frac{364}{365}\\cdot\\frac{363}{365}\\approx0.9918[\/latex]<\/p>\r\nand then subtract from 1 to get\r\n<p style=\"text-align: center;\">P(shared birthday) = 1 \u2013 P(no shared birthday) = 1 \u2013 0.9918 = 0.0082.<\/p>\r\nThis is a pretty small number, so maybe it makes sense that the answer to our original problem will be small.\u00a0 Let's make our group a bit bigger.\r\n\r\n[\/hidden-answer]\r\n\r\n<hr \/>\r\n\r\nSuppose five people are in a room.\u00a0 What is the probability that there is at least one shared birthday among these five people?\r\n[reveal-answer q=\"6410\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"6410\"]\r\n\r\nContinuing the pattern of the previous example, the answer should be\r\n<p style=\"text-align: center;\">[latex]P(\\text{sharedbirthday})=1-\\frac{365}{365}\\cdot\\frac{364}{365}\\cdot\\frac{363}{365}\\cdot\\frac{362}{365}\\cdot\\frac{361}{365}\\approx0.0271[\/latex]<\/p>\r\nNote that we could rewrite this more compactly as\r\n<p style=\"text-align: center;\">[latex]P(\\text{sharedbirthday})=1-\\frac{{}_{365}{{P}_{5}}}{{{365}^{5}}}\\approx0.0271[\/latex]<\/p>\r\nwhich makes it a bit easier to type into a calculator or computer, and which suggests a nice formula as we continue to expand the population of our group.\r\n\r\n[\/hidden-answer]\r\n\r\n<hr \/>\r\n\r\nSuppose 30 people are in a room.\u00a0 What is the probability that there is at least one shared birthday among these 30 people?\r\n[reveal-answer q=\"741907\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"741907\"]\r\n\r\nHere we can calculate\r\n<p style=\"text-align: center;\">[latex]P(\\text{sharedbirthday})=1-\\frac{{}_{365}{{P}_{30}}}{{{365}^{30}}}\\approx0.706[\/latex]<\/p>\r\nwhich gives us the surprising result that when you are in a room with 30 people there is a 70% chance that there will be at least one shared birthday!\r\n\r\n[\/hidden-answer]\r\n\r\nThe birthday problem\u00a0is examined in detail in the following.\r\n\r\nhttps:\/\/youtu.be\/UUmTfiJ_0k4\r\n\r\n<\/div>\r\nIf you like to bet, and if you can convince 30 people to reveal their birthdays, you might be able to win some money by betting a friend that there will be at least two people with the same birthday in the room anytime you are in a room of 30 or more people. (Of course, you would need to make sure your friend hasn't studied probability!) You wouldn't be guaranteed to win, but you should win more than half the time.\r\n\r\nThis is one of many results in probability theory that is counterintuitive; that is, it goes against our gut instincts.\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Try It<\/h3>\r\nSuppose 10 people are in a room. What is the probability that there is at least one shared birthday among these 10 people?\r\n\r\n<\/div>","rendered":"<div class=\"textbox learning-objectives\">\n<h3>Learning Outcomes<\/h3>\n<ul>\n<li>Compute a conditional probability for an event<\/li>\n<li>Use Baye\u2019s theorem to compute a conditional probability<\/li>\n<li>Calculate the expected value of an event<\/li>\n<\/ul>\n<\/div>\n<p>We can use permutations and combinations to help us answer more complex probability questions.<\/p>\n<div class=\"textbox exercises\">\n<h3>examples<\/h3>\n<p>A 4 digit PIN number is selected. What is the probability that there are no repeated digits?<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q108643\">Show Solution<\/span><\/p>\n<div id=\"q108643\" class=\"hidden-answer\" style=\"display: none\">\n<p>There are 10 possible values for each digit of the PIN (namely: 0, 1, 2, 3, 4, 5, 6, 7, 8, 9), so there are 10 \u00b7 10 \u00b7 10 \u00b7 10 = 10<sup>4<\/sup> = 10000 total possible PIN numbers.<\/p>\n<p>To have no repeated digits, all four digits would have to be different, which is selecting without replacement. We could either compute 10 \u00b7 9 \u00b7 8 \u00b7 7, or notice that this is the same as the permutation 10<em>P<\/em>4 = 5040.<\/p>\n<p>The probability of no repeated digits is the number of 4 digit PIN numbers with no repeated digits divided by the total number of 4 digit PIN numbers. This probability is [latex]\\frac{{}_{10}{{P}_{4}}}{{{10}^{4}}}=\\frac{5040}{10000}=0.504[\/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=\"mom10\" class=\"resizable\" src=\"https:\/\/www.myopenmath.com\/multiembedq.php?id=7158&amp;theme=oea&amp;iframe_resize_id=mom10\" width=\"100%\" height=\"400\"><\/iframe><\/p>\n<\/div>\n<div class=\"textbox exercises\">\n<h3>Example<\/h3>\n<p>In a certain state&#8217;s lottery, 48 balls numbered 1 through 48 are placed in a machine and six of them are drawn at random. If the six numbers drawn match the numbers that a player had chosen, the player wins $1,000,000. \u00a0\u00a0\u00a0In this lottery, the order the numbers are drawn in doesn\u2019t matter. Compute the probability that you win the million-dollar prize if you purchase a single lottery ticket.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q767485\">Show Solution<\/span><\/p>\n<div id=\"q767485\" class=\"hidden-answer\" style=\"display: none\">\n<p>In order to compute the probability, we need to count the total number of ways six numbers can be drawn, and the number of ways the six numbers on the player\u2019s ticket could match the six numbers drawn from the machine. Since there is no stipulation that the numbers be in any particular order, the number of possible outcomes of the lottery drawing is 48<em>C<\/em>6 = 12,271,512. Of these possible outcomes, only one would match all six numbers on the player\u2019s ticket, so the probability of winning the grand prize is:<\/p>\n<p>[latex]\\frac{{}_{6}{{C}_{6}}}{{}_{48}{{C}_{6}}}=\\frac{1}{12271512}\\approx0.0000000815[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<p>&nbsp;<\/p>\n<div class=\"textbox exercises\">\n<h3>Example<\/h3>\n<p>In the state lottery from the previous example, if five of the six numbers drawn match the numbers that a player has chosen, the player wins a second prize of $1,000. Compute the probability that you win the second prize if you purchase a single lottery ticket.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q651173\">Show Solution<\/span><\/p>\n<div id=\"q651173\" class=\"hidden-answer\" style=\"display: none\">\n<p>As above, the number of possible outcomes of the lottery drawing is 48<em>C<\/em>6 = 12,271,512. In order to win the second prize, five of the six numbers on the ticket must match five of the six winning numbers; in other words, we must have chosen five of the six winning numbers and one of the 42 losing numbers. The number of ways to choose 5 out of the 6 winning numbers is given by 6<em>C<\/em>5 = 6 and the number of ways to choose 1 out of the 42 losing numbers is given by 42<em>C<\/em>1 = 42. Thus the number of favorable outcomes is then given by the Basic Counting Rule: 6<em>C<\/em>5 \u00b7 42<em>C<\/em>1 = 6 \u00b7 42 = 252. So the probability of winning the second prize is.<\/p>\n<p>[latex]\\frac{\\left({}_{6}{{C}_{5}}\\right)\\left({}_{42}{{C}_{1}}\\right)}{{}_{48}{{C}_{6}}}=\\frac{252}{12271512}\\approx0.0000205[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<p>The previous examples are worked in the following video.<\/p>\n<p><iframe loading=\"lazy\" id=\"oembed-1\" title=\"Probabilities using combinations\" width=\"500\" height=\"281\" src=\"https:\/\/www.youtube.com\/embed\/b9LFbB_aNAo?feature=oembed&#38;rel=0\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/p>\n<div class=\"textbox exercises\">\n<h3>examples<\/h3>\n<p>Compute the probability of randomly drawing five cards from a deck and getting exactly one Ace.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q542442\">Show Solution<\/span><\/p>\n<div id=\"q542442\" class=\"hidden-answer\" style=\"display: none\">\n<p>In many card games (such as poker) the order in which the cards are drawn is not important (since the player may rearrange the cards in his hand any way he chooses); in the problems that follow, we will assume that this is the case unless otherwise stated. Thus we use combinations to compute the possible number of 5-card hands, 52<em>C<\/em>5. This number will go in the denominator of\u00a0our probability formula, since it is the number of possible outcomes.<\/p>\n<p>For the numerator, we need the number of ways to draw one Ace and four other cards (none of them Aces) from the deck.\u00a0 Since there are four Aces and we want exactly one of them, there will be 4<em>C<\/em>1 ways to select one Ace; since there are 48 non-Aces and we want 4 of them, there will be 48<em>C<\/em>4 ways to select the four non-Aces.\u00a0 Now we use the Basic Counting Rule to calculate that there will be 4<em>C<\/em>1 \u00b7 48<em>C<\/em>4 ways to choose one ace and four non-Aces.<\/p>\n<p>Putting this all together, we have<\/p>\n<p style=\"text-align: center;\">[latex]P(\\text{oneAce})=\\frac{\\left({}_{4}{{C}_{1}}\\right)\\left({}_{48}{{C}_{4}}\\right)}{{}_{52}{{C}_{5}}}=\\frac{778320}{2598960}\\approx0.299[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox exercises\">\n<h3>Example<\/h3>\n<p>Compute the probability of randomly drawing five cards from a deck and getting exactly two Aces.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q49360\">Show Solution<\/span><\/p>\n<div id=\"q49360\" class=\"hidden-answer\" style=\"display: none\">\n<p>The solution is similar to the previous example, except now we are choosing 2 Aces out of 4 and 3 non-Aces out of 48; the denominator remains the same:<\/p>\n<p>[latex]P(\\text{twoAces})=\\frac{\\left({}_{4}{{C}_{2}}\\right)\\left({}_{48}{{C}_{3}}\\right)}{{}_{52}{{C}_{5}}}=\\frac{103776}{2598960}\\approx0.0399[\/latex]<\/p>\n<p>It is useful to note that these card problems are remarkably similar to the lottery problems discussed earlier.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<p>View the following for further demonstration of these examples.<\/p>\n<p><iframe loading=\"lazy\" id=\"oembed-2\" title=\"Probabilities using combinations: cards\" width=\"500\" height=\"281\" src=\"https:\/\/www.youtube.com\/embed\/RU3e3KTkjoA?feature=oembed&#38;rel=0\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/p>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p><iframe loading=\"lazy\" id=\"mom1\" class=\"resizable\" src=\"https:\/\/www.myopenmath.com\/multiembedq.php?id=7157&amp;theme=oea&amp;iframe_resize_id=mom1\" width=\"100%\" height=\"400\"><\/iframe><\/p>\n<\/div>\n<h2>Birthday Problem<\/h2>\n<p>Let&#8217;s take a pause to consider a famous problem in probability theory:<\/p>\n<div class=\"textbox shaded\">Suppose you have a room full of 30 people. What is the probability that there is at least one shared birthday?<\/div>\n<p>Take a guess at the answer to the above problem. Was your guess fairly low, like around 10%? That seems to be the intuitive answer (30\/365, perhaps?). Let&#8217;s see if we should listen to our intuition. Let&#8217;s start with a simpler problem, however.<\/p>\n<div class=\"textbox exercises\">\n<h3>example<\/h3>\n<p>Suppose three people are in a room.\u00a0 What is the probability that there is at least one shared birthday among these three people?<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q226841\">Show Solution<\/span><\/p>\n<div id=\"q226841\" class=\"hidden-answer\" style=\"display: none\">\n<p>There are a lot of ways there could be at least one shared birthday. Fortunately there is an easier way. We ask ourselves \u201cWhat is the alternative to having at least one shared birthday?\u201d In this case, the alternative is that there are <strong>no<\/strong> shared birthdays. In other words, the alternative to \u201cat least one\u201d is having <strong>none<\/strong>. In other words, since this is a complementary event,<\/p>\n<p style=\"text-align: center;\">P(at least one) = 1 \u2013 P(none)<\/p>\n<p>We will start, then, by computing the probability that there is no shared birthday.\u00a0 Let&#8217;s imagine that you are one of these three people.\u00a0 Your birthday can be anything without conflict, so there are 365 choices out of 365 for your birthday. What is the probability that the second person does not share your birthday?\u00a0 There are 365 days in the year (let&#8217;s ignore leap years) and removing your birthday from contention, there are 364 choices that will guarantee that you do not share a birthday with this person, so the probability that the second person does not share your birthday is 364\/365.\u00a0 Now we move to the third person.\u00a0 What is the probability that this third person does not have the same birthday as either you or the second person?\u00a0 There are 363 days that will not duplicate your birthday or the second person&#8217;s, so the probability that the third person does not share a birthday with the first two is 363\/365.<\/p>\n<p>We want the second person not to share a birthday with you <em>and<\/em> the third person not to share a birthday with the first two people, so we use the multiplication rule:<\/p>\n<p style=\"text-align: center;\">[latex]P(\\text{nosharedbirthday})=\\frac{365}{365}\\cdot\\frac{364}{365}\\cdot\\frac{363}{365}\\approx0.9918[\/latex]<\/p>\n<p>and then subtract from 1 to get<\/p>\n<p style=\"text-align: center;\">P(shared birthday) = 1 \u2013 P(no shared birthday) = 1 \u2013 0.9918 = 0.0082.<\/p>\n<p>This is a pretty small number, so maybe it makes sense that the answer to our original problem will be small.\u00a0 Let&#8217;s make our group a bit bigger.<\/p>\n<\/div>\n<\/div>\n<hr \/>\n<p>Suppose five people are in a room.\u00a0 What is the probability that there is at least one shared birthday among these five people?<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q6410\">Show Solution<\/span><\/p>\n<div id=\"q6410\" class=\"hidden-answer\" style=\"display: none\">\n<p>Continuing the pattern of the previous example, the answer should be<\/p>\n<p style=\"text-align: center;\">[latex]P(\\text{sharedbirthday})=1-\\frac{365}{365}\\cdot\\frac{364}{365}\\cdot\\frac{363}{365}\\cdot\\frac{362}{365}\\cdot\\frac{361}{365}\\approx0.0271[\/latex]<\/p>\n<p>Note that we could rewrite this more compactly as<\/p>\n<p style=\"text-align: center;\">[latex]P(\\text{sharedbirthday})=1-\\frac{{}_{365}{{P}_{5}}}{{{365}^{5}}}\\approx0.0271[\/latex]<\/p>\n<p>which makes it a bit easier to type into a calculator or computer, and which suggests a nice formula as we continue to expand the population of our group.<\/p>\n<\/div>\n<\/div>\n<hr \/>\n<p>Suppose 30 people are in a room.\u00a0 What is the probability that there is at least one shared birthday among these 30 people?<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q741907\">Show Solution<\/span><\/p>\n<div id=\"q741907\" class=\"hidden-answer\" style=\"display: none\">\n<p>Here we can calculate<\/p>\n<p style=\"text-align: center;\">[latex]P(\\text{sharedbirthday})=1-\\frac{{}_{365}{{P}_{30}}}{{{365}^{30}}}\\approx0.706[\/latex]<\/p>\n<p>which gives us the surprising result that when you are in a room with 30 people there is a 70% chance that there will be at least one shared birthday!<\/p>\n<\/div>\n<\/div>\n<p>The birthday problem\u00a0is examined in detail in the following.<\/p>\n<p><iframe loading=\"lazy\" id=\"oembed-3\" title=\"Probability: the birthday problem\" width=\"500\" height=\"281\" src=\"https:\/\/www.youtube.com\/embed\/UUmTfiJ_0k4?feature=oembed&#38;rel=0\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/p>\n<\/div>\n<p>If you like to bet, and if you can convince 30 people to reveal their birthdays, you might be able to win some money by betting a friend that there will be at least two people with the same birthday in the room anytime you are in a room of 30 or more people. (Of course, you would need to make sure your friend hasn&#8217;t studied probability!) You wouldn&#8217;t be guaranteed to win, but you should win more than half the time.<\/p>\n<p>This is one of many results in probability theory that is counterintuitive; that is, it goes against our gut instincts.<\/p>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p>Suppose 10 people are in a room. What is the probability that there is at least one shared birthday among these 10 people?<\/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-2481\">\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\/4.0\/\">CC BY: Attribution<\/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":9,"template":"","meta":{"_candela_citation":"[{\"type\":\"cc\",\"description\":\"Math in Society\",\"author\":\"Lippman, David\",\"organization\":\"\",\"url\":\"http:\/\/www.opentextbookstore.com\/mathinsociety\/\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"\"},{\"type\":\"original\",\"description\":\"Revision and Adaptation\",\"author\":\"\",\"organization\":\"Lumen Learning\",\"url\":\"\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"\"}]","CANDELA_OUTCOMES_GUID":"2d60a362-3c4b-48e0-adac-3af70abbd2be","pb_show_title":"on","pb_short_title":"","pb_subtitle":"","pb_authors":[],"pb_section_license":""},"chapter-type":[],"contributor":[],"license":[],"class_list":["post-2481","chapter","type-chapter","status-publish","hentry"],"part":329,"_links":{"self":[{"href":"https:\/\/courses.lumenlearning.com\/waymakermath4libarts\/wp-json\/pressbooks\/v2\/chapters\/2481","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/courses.lumenlearning.com\/waymakermath4libarts\/wp-json\/pressbooks\/v2\/chapters"}],"about":[{"href":"https:\/\/courses.lumenlearning.com\/waymakermath4libarts\/wp-json\/wp\/v2\/types\/chapter"}],"author":[{"embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/waymakermath4libarts\/wp-json\/wp\/v2\/users\/21"}],"version-history":[{"count":7,"href":"https:\/\/courses.lumenlearning.com\/waymakermath4libarts\/wp-json\/pressbooks\/v2\/chapters\/2481\/revisions"}],"predecessor-version":[{"id":3043,"href":"https:\/\/courses.lumenlearning.com\/waymakermath4libarts\/wp-json\/pressbooks\/v2\/chapters\/2481\/revisions\/3043"}],"part":[{"href":"https:\/\/courses.lumenlearning.com\/waymakermath4libarts\/wp-json\/pressbooks\/v2\/parts\/329"}],"metadata":[{"href":"https:\/\/courses.lumenlearning.com\/waymakermath4libarts\/wp-json\/pressbooks\/v2\/chapters\/2481\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/courses.lumenlearning.com\/waymakermath4libarts\/wp-json\/wp\/v2\/media?parent=2481"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/waymakermath4libarts\/wp-json\/pressbooks\/v2\/chapter-type?post=2481"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/waymakermath4libarts\/wp-json\/wp\/v2\/contributor?post=2481"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/waymakermath4libarts\/wp-json\/wp\/v2\/license?post=2481"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}