tag:blogger.com,1999:blog-15628310.post6144228477814289329..comments2022-05-11T09:54:19.451-04:00Comments on Question of the day: Draw ThreeAnonymoushttp://www.blogger.com/profile/18153935609499338685noreply@blogger.comBlogger4125tag:blogger.com,1999:blog-15628310.post-19464042251939458152009-11-16T09:27:42.156-05:002009-11-16T09:27:42.156-05:00I love the variation on how you solved the answers...I love the variation on how you solved the answers. I can't add anything more to what you've written, but the answer is 0.2857 or so. <br /><br />I love coming back from a weekend like this one to find this type of answer.Anonymoushttps://www.blogger.com/profile/18153935609499338685noreply@blogger.comtag:blogger.com,1999:blog-15628310.post-48540922860621712272009-11-13T18:24:36.401-05:002009-11-13T18:24:36.401-05:00To get 6 we either need '123' (in any orde...To get 6 we either need '123' (in any order) or '222'.<br /><br />P(123 in any order) = 6*P(123 in that order) = 6*(6*5*5)/(16*15*14) = 15/56<br />P(222) = (5*4*3)/(16*15*14) = 1/56<br /><br />so P(sum is a 6) = 16/56 = 2/7<br />which is, as calculated in the previous posts, around 0.2857.Jon Ingramhttp://joningram.org/noreply@blogger.comtag:blogger.com,1999:blog-15628310.post-53064865538862854472009-11-13T11:37:47.079-05:002009-11-13T11:37:47.079-05:00Same answer as the pilot, but I got there a differ...Same answer as the pilot, but I got there a different way.<br /><br />Using the combination formula for choosing from a set where order doesn't matter, nCr=<br /><br /> n!<br />--------<br />r!(n-r)!<br /><br />Total combinations = 16C3 = 560<br /><br />I wasn't sure how to figure out the right combinations, so I subtracted the ones that aren't right:<br /><br />All three 1's = 6C3 = 20<br />All three 3's = 5C3 = 10<br />Two 1's and a 2 or 3 = 6C2 * 10 = 150<br />Two 2's and a 1 or 3 = 5C2 * 11 = 110<br />Two 3's and a 1 or 2 = 5C2 * 11 = 110<br /><br />560-20-10-150-110-110 = 160<br /><br />So the odds of drawing a total value of 6 is:<br /><br />160/560 = .285714 = about 29%Andyhttps://www.blogger.com/profile/14117563810484999900noreply@blogger.comtag:blogger.com,1999:blog-15628310.post-37350561053449756852009-11-13T10:55:56.946-05:002009-11-13T10:55:56.946-05:00As previously stated, I'm no stats genius, but...As previously stated, I'm no stats genius, but here's my guess. Probability says:<br /><br />The probability of any one of several distinct events is the sum of their individual probabilities, provided that the events are mutually exclusive (i.e. occurrence of one event precludes the others, e.g. selection without replacement). <br /><br />So I would say that the odds of drawing three tiles that total 6 are the sum of their individual probabilities.<br /><br />Out of the 27 possible combinations of drawing three tiles, we can get 6 by drawing:<br />1. 1-2-3<br />2. 1-3-2<br />3. 2-1-3<br />4. 2-2-2<br />5. 2-3-1<br />6. 3-1-2<br />7. 3-2-1<br /><br />I found the probability of drawing each one of those is:<br />1. .044643 (6/16*5/15*5/14) and so on for each<br />2. .044643 <br />3. .044643<br />4. .017857<br />5. .044643<br />6. .044643<br />7. .044643<br /><br />The sum of all those probabilities is .285714. <br /><br />So that's my guess, about 29% chance of drawing three that add up to 6.USAF Pilothttps://www.blogger.com/profile/01326339691636594880noreply@blogger.com