system (02:05:55:892 PM): Mod1 entered the Probabilities chat room. system (02:07:22:289 PM): greg entered the Probabilities chat room. greg (02:07:26:628 PM): anyone here? Mod1 (02:07:32:422 PM): Hey, Greg.. .sorry for the wait Mod1 (02:07:32:700 PM): greg (02:07:38:758 PM): no worries! Mod1 (02:07:38:778 PM): so what's your question? greg (02:07:48:393 PM): ok, i am gonna type it.. greg (02:08:31:032 PM): there are eight people named adam, tom, harry, mark, wendy, sally, cynthia and time. for people are selected at random; what is the prob. of harry and sally are among the four selected? greg (02:08:38:275 PM): four people* Mod1 (02:08:56:441 PM): Nice... I like these kind of problems... Mod1 (02:09:02:663 PM): So what's your thinking? greg (02:09:07:672 PM): alright.. greg (02:09:12:850 PM): this is what i am thinking... greg (02:09:32:907 PM): the other people have noithing to do with it system (02:09:53:423 PM): greg left the Probabilities chat room. system (02:10:04:789 PM): greg entered the Probabilities chat room. greg (02:10:06:970 PM): ahh Mod1 (02:10:10:167 PM): what do you mean? greg (02:10:10:454 PM): sorry i left by accident greg (02:10:26:749 PM): i need to find the probability of picking pairs? greg (02:10:33:987 PM): keep going, with what u are thinking Mod1 (02:10:42:698 PM): Ah! I am glad you are back! you probably don't see the text that you typed before but I have it greg (02:10:51:263 PM): ok great Mod1 (02:11:17:329 PM): What do you mean by "the other people have nothing to do with it"? greg (02:11:33:542 PM): wont the probabilities of getting a 2 or 7 be defivided by the total number of possibilities? greg (02:12:18:266 PM): yeah i dont know what i mean, i think what i meant was that it is just totally random Mod1 (02:12:28:113 PM): Ah, so you mean that it does not really matte which two we are talking about, right? greg (02:12:32:481 PM): go on with your explanation i will listen greg (02:12:37:035 PM): righttt! greg (02:12:48:016 PM): like it is pretty much the same as saying what are the chances of picking mark and adam? Mod1 (02:13:10:185 PM): No, you are in the right path... I was just trying to clarify greg (02:13:15:495 PM): ok great greg (02:13:17:476 PM): go on Mod1 (02:13:30:984 PM): you think that the probability is 2 out of 7, ritht? greg (02:13:40:642 PM): no Mod1 (02:13:46:844 PM): no??? greg (02:14:16:356 PM): i meant not a 2 or 7 of a harry or sally greg (02:14:21:609 PM): i was confusing another problem greg (02:14:24:559 PM): well there are 8 people Mod1 (02:14:58:767 PM): I see greg (02:15:04:509 PM): 8 people greg (02:15:07:533 PM): we are selecting 4 greg (02:15:21:989 PM): we want to know the probability of getting 2 that we went, harry and sally Mod1 (02:16:17:597 PM): Ah... that is a very different problem... ;-) Mod1 (02:17:09:789 PM): And you said that you found an answer? greg (02:17:15:306 PM): no greg (02:17:20:992 PM): i am asking you how to do it greg (02:17:24:146 PM): do u know? greg (02:17:26:993 PM): how to? Mod1 (02:18:11:814 PM): Well... let's think about it... Mod1 (02:18:33:555 PM): What is the probability that if you pick 8 people out of 8, Mark will be there? greg (02:18:42:777 PM): 1 out of 8 greg (02:19:22:916 PM): right? Mod1 (02:20:01:374 PM): really? that sounds like a low probability for something so easy to achieve, doesn' it? greg (02:20:26:233 PM): ohh 8 people out of 8 greg (02:20:31:722 PM): obviously 100% Mod1 (02:20:34:735 PM): right Mod1 (02:20:54:230 PM): and if we picked only 4 people? greg (02:21:16:471 PM): 1 out of 8? Mod1 (02:22:18:688 PM): are you sure? greg (02:22:23:809 PM): noi greg (02:22:43:509 PM): 4!? Mod1 (02:23:00:676 PM): I see ;-) So you knew that getting Mark in a group of 8 picks was a no brainer becuase... greg (02:23:00:789 PM): or 8 x 7 x 6 x 5? Mod1 (02:24:01:171 PM): you could only make one group of 8 people and it would include Mark, right? greg (02:24:14:882 PM): what? Mod1 (02:24:24:986 PM): so how many groups of 4 can you make and in how many of those would you find Mark? greg (02:24:37:020 PM): u tell me greg (02:24:41:874 PM): thats y i am asking greg (02:25:07:012 PM): 56 Mod1 (02:25:43:073 PM): I know.. but I am sure that you can figure it out, too. That way you will be able to solve not just this problem but any problem of this type. Give me some examples of groups of 4... greg (02:26:02:329 PM): adam tom mark cynthia Mod1 (02:26:08:183 PM): good Mod1 (02:26:22:184 PM): anothe with adam in it? greg (02:26:35:984 PM): adam mark tom wendy Mod1 (02:27:03:437 PM): good, would adam, mark, cynthia, tom be another? greg (02:27:14:056 PM): no greg (02:28:24:173 PM): dude, this is taking too long, i got a a midterm in 2 hours Mod1 (02:28:56:012 PM): Don't worry I promise you would understand this thing like noone else greg (02:29:06:844 PM): hahaha ok Mod1 (02:29:07:083 PM): the point is that order does not matter greg (02:29:10:573 PM): right greg (02:29:12:550 PM): combination Mod1 (02:29:35:370 PM): 4 people are 4 people no matter how you list them, right? greg (02:29:40:706 PM): absolutely! Mod1 (02:29:48:130 PM): you got it! the magic word! greg (02:29:55:919 PM): combination? Mod1 (02:29:56:109 PM): and how do we do combinations? greg (02:30:04:281 PM): 8 combo 4? greg (02:30:08:135 PM): 8c4 greg (02:30:10:631 PM): on the calculator greg (02:30:14:776 PM): he allows us to use them Mod1 (02:30:26:064 PM): Ah.. what would we do without calculators! ;-) greg (02:30:32:600 PM): haha right greg (02:30:46:115 PM): 8C4 is 70 greg (02:30:49:251 PM): is that the right way to do it? Mod1 (02:31:20:275 PM): lets see... I don't have your calculator so let me try it by hand in a second greg (02:31:42:826 PM): ALRIGHT, BUT I AM ASKING IS THAT THE RIGHT calculation to do? 8 C4? Mod1 (02:32:21:691 PM): ah, yes! that will give us how many groups of 4 can we make with 8 people if order does not matter! greg (02:32:31:869 PM): 70 my friend! Mod1 (02:32:42:302 PM): now the question is how many of those will have Mark in them? greg (02:33:09:654 PM): 8C1? Mod1 (02:33:59:258 PM): that is a fancy way to think about it, how much is that? greg (02:34:04:223 PM): 8 greg (02:34:16:804 PM): so is that right? Mod1 (02:34:23:764 PM): it makes sense to me...^M greg (02:34:24:226 PM): for the number of groups mark is in? greg (02:34:31:669 PM): so let me guess from here how to do it... Mod1 (02:34:47:909 PM): great! remember, probability is # success divided by # trials right?^M greg (02:35:48:115 PM): hows that look? greg (02:35:53:941 PM): do u see the canvas? Mod1 (02:36:08:295 PM): why do we need the second 8C1? greg (02:36:14:487 PM): because there are 2 people Mod1 (02:36:18:494 PM): *yes I see it greg (02:36:25:879 PM): we are selecting harry and sally! greg (02:36:28:055 PM): so 2 people greg (02:36:32:828 PM): oh wait! greg (02:36:39:288 PM): it's 7C1 greg (02:36:42:309 PM): right? Mod1 (02:36:44:741 PM): Ah! I thought we were still working with Mark, but I know that you have to go! ;-) greg (02:37:01:753 PM): no no it's fine, lets keep going on this problem Mod1 (02:37:25:863 PM): backup for a second... greg (02:37:34:022 PM): ok greg (02:37:48:356 PM): 8 people, 4 chosen, what are teh chances that harry and sally are chosen Mod1 (02:38:38:858 PM): 70 groups, how many of those have Harry in them ?greg (02:38:42:805 PM): 8 Mod1 (02:38:50:143 PM): yes! greg (02:39:02:148 PM): right but we need harry and sally greg (02:39:08:084 PM): so it would be what i wrote wouldnt it? Mod1 (02:39:09:794 PM): really? greg (02:39:21:186 PM): yes, i said that before greg (02:39:29:131 PM): we need to pick 4 people greg (02:39:40:396 PM): we want to know the probability of picking harry and sally greg (02:39:54:200 PM): so 2 of 4 and those 4 from the total 8 greg (02:39:55:180 PM): get it? Mod1 (02:40:36:402 PM): I know... we have our denominator but the numerator does not seem right... the numerator should tell us how many of those 70 groups have Harry and Sally together! greg (02:40:55:460 PM): ok greg (02:41:14:108 PM): 8C2? Mod1 (02:42:10:403 PM): that will give us all the possible pairs but not our "success" pairs greg (02:42:17:329 PM): ok sgreg (02:42:26:433 PM): ok tell me how how to do it greg (02:42:28:834 PM): i am going nuts here Mod1 (02:42:59:767 PM): sorry... that is not the intention greg (02:43:21:221 PM): it is just taking so long Mod1 (02:44:25:159 PM): yes, but if I gave you the answer, do you think you will be able to do another similar problem? if you think hard about this one you will be able to Mod1 (02:44:33:757 PM): We got the denominator fix... Mod1 (02:44:38:838 PM): any ideas for the denominator? Mod1 (02:44:51:298 PM): sorry *numerator? greg (02:45:14:865 PM): yes the prob of picking harry and sally out of those 4 Mod1 (02:46:13:078 PM): 4? greg (02:46:23:059 PM): because we are picking four people!!!!!!! greg (02:46:28:935 PM): we're have 8 to choose from greg (02:46:33:440 PM): we are picking 4! greg (02:46:41:058 PM): we want to pick 2 specific Mod1 (02:47:54:301 PM): I see, I thought you were guessing 4 for the numerator. So what is your guess for the numerator? Mod1 (02:49:14:808 PM): so? greg (02:49:18:897 PM): i'm thinking greg (02:49:23:977 PM): 8c2? Mod1 (02:51:18:981 PM): don't think formulas so much... if we fix two people in a group of 4 the problem is how many combinations of the other 6 people can be fit in the other two slots, isn't it? greg (02:52:01:165 PM): yea Mod1 (02:52:26:142 PM): so, look at those rectangles... Sally and Harry are fixed greg (02:52:31:877 PM): ok Mod1 (02:52:49:849 PM): how do we fill the other two slots? greg (02:53:01:628 PM): with 2 of the other 6 people greg (02:53:40:570 PM): i think the answer is 15/70 greg (02:53:53:089 PM): just please tell me if that is right greg (02:54:14:203 PM): and if it's not, i am gonna figure it out elsewhere because i am growing extremely impacient with the lack of speed\ Mod1 (02:54:20:061 PM): how many of those pairs can we make with 6 people? greg (02:54:21:689 PM): of this chatting method greg (02:54:27:258 PM): 3 greg (02:54:34:390 PM): no Mod1 (02:54:36:549 PM): one more step and we are done... greg (02:54:42:834 PM): 15 Mod1 (02:55:17:084 PM): YES!!!! how did you get that? greg (02:55:18:502 PM): hello???? greg (02:55:21:768 PM): ok greg (02:55:24:029 PM): 6C2 greg (02:55:28:657 PM): which is the same as i said before greg (02:55:43:059 PM): 8C1 + 7C1/ 8C4 Mod1 (02:55:45:801 PM): yes, but do you know why, now? greg (02:55:55:024 PM): yes greg (02:56:23:202 PM): because there are 8 different pairs ahrry can be in Mod1 (02:56:26:114 PM): so it is not just the mechanical application of the formula... you got it! greg (02:56:33:590 PM): and 7 that sally can be in once harry is selected greg (02:56:41:627 PM): so the answer is 15/70 Mod1 (02:57:11:583 PM): brilliant! Mod1 (02:57:18:722 PM): how does that feel? greg (02:57:30:858 PM): great! greg (02:57:32:497 PM): thank you buddy greg (02:57:38:342 PM): are u positive about the answer?Mod1 (02:57:48:365 PM): u r welcome... I hope you do well in that test, greg greg (02:57:57:901 PM): thanks me too Mod1 (02:58:04:656 PM): think it through don't jump on the formulas to quickly! Mod1 (02:58:12:500 PM): ;-) greg (02:58:14:614 PM): so u are positive that this is the answer? Mod1 (02:58:36:205 PM): I am not sure I am just a machine but you are a fully thinking human! greg (02:58:45:789 PM): lol right Mod1 (02:58:57:606 PM): do you visit Dr. Math a lot? greg (02:59:03:657 PM): have been recently for this class greg (02:59:07:671 PM): i got another question for you greg (02:59:09:593 PM): while we are here Mod1 (02:59:12:188 PM): which class is that? greg (02:59:19:112 PM): business stat and prob Mod1 (02:59:29:796 PM): in college, right? greg (02:59:34:719 PM): yep Mod1 (03:00:08:233 PM): good... perhaps one day you will stop by and help others now that you have mastered combinations, permutations and factorials! greg (03:00:15:723 PM): hahaha greg (03:00:18:672 PM): sure thing greg (03:00:23:232 PM): here is another question for you Mod1 (03:00:30:733 PM): you said that you had found an answer but didn't like the explanation, right? greg (03:00:37:091 PM): yeahh Mod1 (03:01:01:487 PM): was that in Dr. Math or somewhere else? greg (03:01:10:517 PM): dr math greg (03:02:29:824 PM): ok greg (03:02:35:399 PM): that is the table we need to refer to greg (03:02:40:341 PM): so let me tlel u the problem now Mod1 (03:02:57:124 PM): o.k. go ahead greg (03:03:04:654 PM): the number of cars X that a sales person sells on any day has the following probability distribution: (the table) greg (03:03:16:317 PM): salesperson earns $80 for every car he sells: greg (03:03:28:917 PM): find the prob. that he earns between $120 and $280 greg (03:03:40:296 PM): so this si what i did Mod1 (03:04:05:682 PM): makes sense... right? difficult to sell a lot of cars, easier to sell 0 cars. The probability of selling 2 cars looks a little fishy greg (03:04:07:225 PM): i multiplied everyone of the number above times 80 to get the amount of money he makes from those number o cars greg (03:04:33:566 PM): right, but this is the table my statistics book provided me with Mod1 (03:04:46:338 PM): o.k. Mod1 (03:04:51:103 PM): go ahead greg (03:04:59:094 PM): so there u have it greg (03:05:23:128 PM): the questions is FIND THE PROB. THAT HE THE SALESMAN EARNS BETWEEN $120 AND $280 greg (03:05:32:952 PM): THIS IS WHAT I DID SO FAR Mod1 (03:05:43:719 PM): aha greg (03:06:29:297 PM): MAKES IT EASIER THAT WAY greg (03:06:33:880 PM): ANY IDEA NOW HOW TO DO IT? Mod1 (03:06:56:139 PM): well, lets think together about this, o.k.? greg (03:07:04:738 PM): SURE! Mod1 (03:08:18:323 PM): from your table, do you think that the probabilities are exclusive? greg (03:08:28:384 PM): NO Mod1 (03:09:24:677 PM): strange... greg (03:09:30:650 PM): YES greg (03:09:38:137 PM): DO U KNOW HOW TO DO THIS ONE? Mod1 (03:09:47:409 PM): I am thinking... Mod1 (03:10:19:451 PM): how many cars do you need to sell to make $120 greg (03:10:46:162 PM): THAT'S WHAT I DONT KNOW Mod1 (03:11:05:406 PM): well, if you sell one car, how much money do you make? greg (03:11:20:145 PM): $80 Mod1 (03:11:34:749 PM): and if you sell two? (sorry for the stupid questions) greg (03:11:48:945 PM): $160 Mod1 (03:11:49:444 PM): $160, you have it there in your table Mod1 (03:12:01:733 PM): but you can't seel one car an a half, right? greg (03:12:06:267 PM): RIGHT] Mod1 (03:12:53:220 PM): I think that this prob. are exclusive, that is why they add up to 1 Mod1 (03:14:57:016 PM): so selling two cars would satisfy the condition of the problem, right? greg (03:15:08:255 PM): YES Mod1 (03:15:24:947 PM): what about selling 3 cars? greg (03:15:47:825 PM): YEP Mod1 (03:16:13:435 PM): but 1 or 4 would not satisfy it, right? greg (03:16:25:084 PM): RIGHT greg (03:16:28:489 PM): AND OBVIOUSLY NOT 5 Mod1 (03:17:16:445 PM): sure... so we know that he needs to sell 2 or 3 cars to make that money.. how would you rephrase that in terms of probabilities: what is the probability that the seller will... greg (03:17:28:216 PM): .4 greg (03:17:39:458 PM): U THINK THAT IS IT? greg (03:17:43:927 PM): I THOUGHT ABOUT THAT greg (03:17:48:015 PM): BUT IT DOESNT SEEM QUITE RIGHT Mod1 (03:18:05:694 PM): how did you get that? greg (03:18:25:419 PM): .3+.4 greg (03:18:28:814 PM): TABLE Mod1 (03:19:07:985 PM): good thinking, but that works when the probabilities are mutually what? greg (03:19:23:019 PM): U THINK THAT IS THE ANSWER? Mod1 (03:20:31:582 PM): nope... adding the two probabilities only work if the events are mutually something. Do you remember what that "somethign is"? greg (03:20:57:593 PM): INDEPENDANT Mod1 (03:21:51:601 PM): or exclusive, right. But selling 3 cars depends on you selling 2, right? greg (03:22:00:960 PM): WAIT I AM RIGHT Mod1 (03:22:08:676 PM): why? greg (03:22:11:630 PM): I THINK THE ANSWER IS JUST .3+.1 Mod1 (03:22:20:811 PM): how come? greg (03:22:21:253 PM): BECAUSE OF ANOTHER EXAMPLE I HAVE HERE greg (03:22:24:687 PM): DO U DISAGREE? greg (03:22:32:344 PM): THE EVENTS ARE MUTALLY EXCLUSIVE Mod1 (03:22:42:693 PM): It depends on what the book says... why do you think u are right? Mod1 (03:22:59:927 PM): does the problem states that? greg (03:23:03:928 PM): NO Mod1 (03:23:17:445 PM): so? system (03:23:59:678 PM): greg left the Probabilities chat room. Mod1 (03:24:30:594 PM): but if the probability of selling two cars is 3 in 10, does it make sense that the probability of selling one car is only 2 in 10 if the probabilities presentes are not exclusive? system (05:50:43:628 PM): Mod1 left the Probabilities chat room.