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Post by Deleted on May 16, 2013 20:45:50 GMT -5
You have one man to start, so he is automatically in the 90% of families that have children. The 10% without children are irrelevant / one extra layer of multiplication to throw in at the end.
Families with 1 child have either one boy or one girl. If our starter man is from one of those families, he does not have a brother.
Families with 2 children are either MM, MF, FM or FF.
Families with 3 children are MMM, MMF, MFF or FFF. Of those, only MMM, MMF and MFF (and various reorderings) would be involved in our game.
Families with 4 children are MMMM, MMMF, MMFF, MFFF or FFFF (and various reorderings). Again, the all girl family is off the table.
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spartan7886
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Post by spartan7886 on May 16, 2013 20:51:15 GMT -5
You have to either add in MFM or remember to count MMF as twice as likely for the 3 children family, and similar for the 4 children.
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Post by Deleted on May 16, 2013 21:23:42 GMT -5
You're right... there's
MMM MMF MFM FMM MFF FMF FFM
FFF
9/12 of those dudes have a brother.
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Virgil Showlion
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Post by Virgil Showlion on May 16, 2013 21:36:15 GMT -5
spartan7886 is correct, with 54.8%. We have: F | p(F) | p(M|F)
| p(B|M,F)
| (no children)
| 1/10 | 0 | 0 | M
| 15/200 | 1 | 0 | F | 15/200
| 0 | 0 | MM | 35/400 | 1 | 1 | MF | 35/400
| 1/2 | 0 | FM | 35/400
| 1/2 | 0 | FF | 35/400
| 0 | 0 | MMM | 25/800
| 1 | 1 | MMF | 25/800
| 2/3 | 1 | MFM | 25/800 | 2/3 | 1 | MFF | 25/800 | 1/3 | 0 | FMM | 25/800 | 2/3 | 1 | FMF | 25/800 | 1/3 | 0 | FFM | 25/800 | 1/3 | 0 | FFF | 25/800 | 0 | 0 | MMMM | 15/1600
| 1 | 1 | MMMF | 15/1600 | 3/4
| 1 | MMFM | 15/1600 | 3/4 | 1 | MMFF | 15/1600 | 1/2 | 1 | MFMM | 15/1600 | 3/4 | 1 | MFMF | 15/1600 | 1/2 | 1 | MFFM | 15/1600 | 1/2 | 1 | MFFF | 15/1600 | 1/4 | 0 | FMMM | 15/1600 | 3/4 | 1 | FMMF | 15/1600 | 1/2 | 1 | FMFM | 15/1600 | 1/2 | 1 | FMFF | 15/1600 | 1/4 | 0 | FFMM | 15/1600 | 1/2 | 1 | FFMF | 15/1600 | 1/4 | 0 | FFFM | 15/1600 | 1/4 | 0 | FFFF | 15/1600 | 0 | 0 |
where p(F) is the probability of a given family, p(M|F) is the probability that a random sibling sampled will be male given family F, p(B|M,F) is the probability that a random male sampled from family F will have a brother. We wish to find the probability of a male having a brother, or p(B|M). We can obtain this using p(B|M) = Σ F p(B|M,F)p(F|M) We already have p(B|M,F). To obtain p(F|M), we use a simple application of Bayes' rule (I know you remember it from high school ): p(F|M) = p(M|F)p(F) / Σ F p(M|F)p(F) Plugging in our data, we have p(B|M) = 79/144 ~= 54.86%. Hence a man is slightly more likely than not to have a brother. Spartan is the winner. She will get two months of ad-free and a special dedication on the EE homepage. But... starting Monday, since I'm busy over the weekend.
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Sammy
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Post by Sammy on May 16, 2013 23:27:15 GMT -5
spartan: (17.5+18.75+13.125)/90=.548
Dang! I was just about to post the exact same answer.
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Post by ՏՇԾԵԵʅՏɧ_LԹՏՏʅҼ on May 16, 2013 23:53:22 GMT -5
Suuuurrree you were.
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Post by Deleted on May 17, 2013 6:44:06 GMT -5
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Virgil Showlion
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Post by Virgil Showlion on May 17, 2013 9:33:32 GMT -5
Actually, spartan7886, I realized last night that based on the problem statement, we're both mistaken. My analysis in Reply #642 mistakenly assumes that the a priori probability of running into an individual from family F is p(F), but this neglects the fact that the latter families are larger than the former ones. The probabilities we should be using must reflect the likelihood of the individual, not of a family. For example, in an extreme case where 99 families had 1 child and 1 family had 1 million children (the Duggars, perhaps?), my above analysis would give us the nonsensical result that virtually nobody sampled at random would have a brother, despite the exact opposite being true. We could account for this by adjusting the a priori p(F), but it turns out it's just easier to denormalize p(M|F), scaling it up family size. And of course that leaves us with just M--the number of males in the family. Hence our correct analysis is: p(B|M) = Σ F p(B|M,F)p(F'|M) where p(F'|M) = M p(F) / Σ F M p(F) And plugging in our data yields p(F'|M) = 115/176 ~= 65.34%, which is quite close to our original answer only by coincidence. But you still get the prize.
I made the same mistake in analysis as above with the mini-example, @steve. The correct answer is 33.3%. For the example with the puppies, suppose we have 400 dogs. 90% of those (360 dogs) will have no puppies. Half of 5% (10 dogs) will have one black puppy. Half of 5% (10 dogs) will have one brown puppy. A quarter of 5% (5 dogs) will have two black puppies. A quarter of 5% (5 dogs) will have two brown puppies. Half of 5% (10 dogs) will have one black and one brown puppy. Hence if we look at all of this and add up the total number of brown puppies, we end up with a total of 10 + 5*2 + 10 = 30 brown puppies. How many of those brown pups have a brown sibling? Well, we can see immediately that the only pups who fit the bill are the 10 that come from the two-brown-puppy family. Hence the total number of brown puppies is 30, the number of brown puppies that have brown siblings is 10, and if we see a brown puppy, we know that the probability of him having a brown sibling is 10/30 = 1/3 ~= 33.3%.
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Colleenz
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Post by Colleenz on May 17, 2013 9:43:24 GMT -5
Wow, my new games are much more fun than this.
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Spellbound454
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Post by Spellbound454 on May 20, 2013 13:16:36 GMT -5
10% of families have no children 15% of families have 1 child 35% of families have 2 children 25% of families have 3 children 15% of families have 4 children hmm 10+15+35= 60 Of the rest 3 children....Total possibilities of two or more boys 4 in 8 ie..... half of 25% =12'5% 4 children....Total possibility of two or more boys 10 in 16 ie 5 in 8 = 62.5% of 15%= 9.37% =81.87% Probably got that totally wrong but that's my best guess... Sorry I forgot to read the question 78.13 of total families will not have a male child with a brother..... Crap.... I've done it wrong again, its not 10/16 its 11/16 so 11 divided by 16 x 100 = 68.75 68.75 of 15% =10.3% 15-10.3% = 4.7 60+12.5+4.7 = 77.2% will not have a male child with a brother... and I'm not looking at it again.
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Virgil Showlion
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Post by Virgil Showlion on May 20, 2013 14:44:37 GMT -5
You're close, Spellbound. If you want to do it that way, you can consider: 10% of families have no children : produce 0.1*(0) males and 0 males with brothers 15% of families have 1 child : produce 0.15*(1/2) males and 0 males with brothers 35% of families have 2 children : produce 0.35*(1) males and 0.35*(1/2) males with brothers 25% of families have 3 children : produce 0.25*(1.5) males and 0.25*(9/8) males with brothers 15% of families have 4 children : produce 0.15*(2) males and 0.15*(28/16) males with brothers Sum up the total # of males with brothers, divide by the total # of males, and you get 115/176 = 65.34%. ...whi-i-i-ich is what I should've posted in Reply #647 (I botched one of the calculations. )
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Spellbound454
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Post by Spellbound454 on May 20, 2013 15:31:00 GMT -5
I forgot this bit but .... Possibility of having two boys with two kids is 1/4 1/4 of 35% is 8.75 ie...26.25% without two boys. three kids 4/8.. 1/2.. of 25% = 12.5% without two boys four kids 5/16...of 15% = 4.7% without two boys Therefore 10% 15% 26.25% 12.5% 4.7%= 68.45% without two or more boys. Well its closer anyway
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Post by Opti on May 20, 2013 15:37:26 GMT -5
I don't know how to do it elegantly, but I think my answer is closer to correct.
Universe of male children
.5(15%) = 7.5 Two Children -> MM FF MF FM -> .75(35%) = 26.25 Three Children -> MMM MFF MMF MFM FMM FFF FMF FFM -> 7/8(25%) = 21.875 Four children 15/16(15%)= 14.0625
Universe of male children is 69.6875% of the original 100% of families.
In that universe, the kids born as only children have no sibs -> 0%
In the two children scenario, oddly since we are selecting only those pairings with a male in them the actual odds of another male is .33, i.e. MM, MF, FM -> .33(26.25)
Looking at the three child scenario we have MFF FFM FMF no male sibs and MMM MMF FMM MFM so 4/7(21.875) for male sibs
For the four child scenario only FFFF(already excluded) and FFFM, FMFF, FFMF, and MFFF will be excluded, so 11/15(14.0625)
(.33(26.25) + 4/7(21.875) + 11/15(14.0625))/ 69.6875
By my math, 45.75% chance a male in those families has a male sib.
The issue is your starting universe is different that the one in the problem. Take the two child scenario: there is a 25% chance of any combination, MM, FF, FM, and MF. However, selecting the 75% of that universe containing a male sib, there is only a one in three chance that the other sib is male.
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Spellbound454
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Post by Spellbound454 on May 20, 2013 15:46:51 GMT -5
Oh right..... I'm not excluding FFF or FFFF
Didn't realise we had to
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Post by Opti on May 20, 2013 16:03:41 GMT -5
Spell, PB ate my other reply, but I believe the problem states given a man selected at random... pretty much excludes the families not having at least one male child IMO.
Given this makeup, what is the likelihood that a man selected at random from the population will have at least one brother?
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Virgil Showlion
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Post by Virgil Showlion on May 20, 2013 17:45:30 GMT -5
I don't know how to do it elegantly, but I think my answer is closer to correct. Universe of male children .5(15%) = 7.5 Two Children -> MM FF MF FM -> .75(35%) = 26.25 Three Children -> MMM MFF MMF MFM FMM FFF FMF FFM -> 7/8(25%) = 21.875 Four children 15/16(15%)= 14.0625 Universe of male children is 69.6875% of the original 100% of families. In that universe, the kids born as only children have no sibs -> 0% In the two children scenario, oddly since we are selecting only those pairings with a male in them the actual odds of another male is .33, i.e. MM, MF, FM -> .33(26.25) Looking at the three child scenario we have MFF FFM FMF no male sibs and MMM MMF FMM MFM so 4/7(21.875) for male sibs For the four child scenario only FFFF(already excluded) and FFFM, FMFF, FFMF, and MFFF will be excluded, so 11/15(14.0625) (.33(26.25) + 4/7(21.875) + 11/15(14.0625))/ 69.6875 By my math, 45.75% chance a male in those families has a male sib. The issue is your starting universe is different that the one in the problem. Take the two child scenario: there is a 25% chance of any combination, MM, FF, FM, and MF. However, selecting the 75% of that universe containing a male sib, there is only a one in three chance that the other sib is male. There are a couple things you need to watch out for. The first is that in a universe where every family had two children, the probability of a male having a brother is 1/2. This is because the four families are equally likely: MM MF FM FF We note there are four males, two of which have brothers, giving us a 2/4 = 1/2 likelihood of a male having a brother. For a three-child family, we have equal possibilities for MMM MMF MFM MFF FMM FMF FFM FFF In this case there are 16 males total, 12 of which have brothers, giving us a 3/4 probability of a male having a brother. In general, for an n-child family, we will have a total over all combinations of 2 n-1n males and (2 n-1-1) n males with brothers, giving us a 1 - 2 1-n probability of a male having a brother. The other thing you have to consider is that x% of families having n children does not mean that a random person you meet will be a member of an n-child family with probability x%. For example, in a world where 50% of families have one child and 50% of families have two children, a random person you meet will have a 2/3 (66.7%) chance of being from a 2-child family. To solve the problem you need to use conditional probability (as I did initially), or alternatively you can compute the total number of males and total number of males with brothers contributed by each family size, and take the ratio of the two. You get the same answer either way: 65.34% If you prefer to use actual numbers, suppose we have a world with 8000 families. Then 800 of these families will have no children 1200 of these families will have 1 child; 600 of these children will be male, and none of these males will have a brother 2800 of these families will have 2 children; 2800 of these children will be male, and 1400 of these males will have a brother 2000 of these families will have 3 children; 3000 of these children will be male, and 2250 of these males will have a brother 1200 of these families will have 4 children; 2400 of these children will be male, and 2100 of these males will have a brother Hence we have a total of 600 + 2800 + 3000 + 2400 = 8800 males, and a total of 1400 + 2250 + 2100 = 5750 males with brothers. A male selected at random will therefore have a brother with probability 5750/8800 = 65.34%.
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Post by Opti on May 20, 2013 19:07:57 GMT -5
The first is that in a universe where every family had two children, the probability of a male having a brother is 1/2. This is because the four families are equally likely: MM MF FM FF We note there are four males, two of which have brothers, giving us a 2/4 = 1/2 likelihood of a male having a brother. Oops, forgot about counting from all brothers in a given situation.
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Post by Opti on May 20, 2013 19:15:21 GMT -5
PB didn't like my post. I guess I solved the problem of the percentage of eldest male child in a family having a male sibling. I was hoping perhaps there was a consolation prize but the PB server does not agree.
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Virgil Showlion
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Post by Virgil Showlion on May 20, 2013 19:42:09 GMT -5
PB didn't like my post. I guess I solved the problem of the percentage of eldest male child in a family having a male sibling. I was hoping perhaps there was a consolation prize but the PB server does not agree. I'll buy you the ad-free along with Spartan if you admit snow leopards are wretched, ugly little monsters.
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ՏՇԾԵԵʅՏɧ_LԹՏՏʅҼ
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Post by ՏՇԾԵԵʅՏɧ_LԹՏՏʅҼ on May 20, 2013 19:48:22 GMT -5
We need a new brain-teaser - and one dumbed-down just a little bit - all these calculations are giving me a headache.
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Opti
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Post by Opti on May 20, 2013 20:33:02 GMT -5
PB didn't like my post. I guess I solved the problem of the percentage of eldest male child in a family having a male sibling. I was hoping perhaps there was a consolation prize but the PB server does not agree. I'll buy you the ad-free along with Spartan if you admit snow leopards are wretched, ugly little monsters. I guess its not enough to admit I'd feel wretched if I was physically in reach when they were hungry.
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Spellbound454
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Post by Spellbound454 on May 21, 2013 11:31:45 GMT -5
...and I solved the likelihood of there being two male children in the given percentages.... but sadly...that wasn't the question. Oh well.
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wyouser
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Post by wyouser on May 21, 2013 14:36:26 GMT -5
Oh Virgil...what have we done? I did a Google for "junk" and did a number search for an answer to your question asking the computor to take the numbers and apply pi R Squared. The computor screen went red...sirens sounded....My house was quickly surrounded by FEDs and a group of large white vans showed up filled with guys in white coats carrying these REALLY Really long sleeved white garments that lace up on the back. It would seem the words "my brothers junk" when sent electronically from Canada into the US triggers an immediated terrorist alert to Homeland Security. They arrested my dog, shot the neighbors cat, impounded my ornamental koi from my pond and sent me to my rock in the backyard........and they sent 4 to make sure I stayed on that rock.
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Post by Peace Of Mind on May 21, 2013 16:44:55 GMT -5
Virgil - Why do you keep shuddering? Are you reading Paul's posts or are you chilly today?
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Post by Opti on May 21, 2013 16:52:13 GMT -5
Personally I think the consequences could have been worse. Had Virgil titled it My Sister's Eggs and had us find the percentage of women with one or more sisters there probably would have been snow leopards loose in Wyouser's yard.
The upside of course would be the Koi, the cat, and possibly the dog would feed the snow leopard enough that he could escape to safety ...
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Virgil Showlion
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Post by Virgil Showlion on May 23, 2013 11:16:24 GMT -5
Problem 73 - Stuck on PurposeConsider the following building floor plan, with a set of rooms labeled A-G and an 'Outside': The lines between each pair of blue dots are walls. Ordinarily in problems like this the goal is to draw a single continuous curve that passes through each wall once and only once, ending up back in the room where the curve was started. However, for this particular set of rooms, the real challenge is to get stuck. Getting "stuck" means you draw your curve so that you eventually have no way of exiting the room you're in (because you've passed through all of its walls once) even though you haven't passed through all the walls in the building. You can start your curve in any room, including the outside of the building. So the question is: Can you get "stuck" on purpose?
To present your solution to the problem, give the sequence of rooms your curve passes through, beginning with the room you start in. For example, a curve that starts at the 'x' and that looks like would have a sequence Outside A B Outside G ...
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wyouser
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Post by wyouser on May 23, 2013 12:41:56 GMT -5
Virgil, you are entering the rooms from outside the walls? Do we add doors and windows? Also, how many of you noticed the layout of this floor plans uncanny resemblence to the lowest level of the Fed Repository at Ft Knox? Correct me if I'm wrong , but don't they come and take you away for entering the vaults of the gold repository "through" a wall? If I remember my Bond movies correctly, they may also cancel your birthday (Goldfinger?) over this. So in conclusion, breaking into the Fed Gold Repository at Ft Knox "on purpose" will result in your being "stuck" in an 8x10 cell for most of the rest of your life.......Did I get it right?? Did I ?? Did I Do I get a cookie
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Post by Jaguar on May 23, 2013 12:45:48 GMT -5
Why would you be satisfied with a silly cookie wyouser, wouldn't you rather have world dominance with a side topping of whipped cream instead ?
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Post by Deleted on May 23, 2013 13:04:37 GMT -5
Yes.
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Virgil Showlion
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Post by Virgil Showlion on May 23, 2013 13:46:07 GMT -5
Virgil, you are entering the rooms from outside the walls? Do we add doors and windows? Also, how many of you noticed the layout of this floor plans uncanny resemblence to the lowest level of the Fed Repository at Ft Knox? Correct me if I'm wrong , but don't they come and take you away for entering the vaults of the gold repository "through" a wall? If I remember my Bond movies correctly, they may also cancel your birthday (Goldfinger?) over this. So in conclusion, breaking into the Fed Gold Repository at Ft Knox "on purpose" will result in your being "stuck" in an 8x10 cell for most of the rest of your life.......Did I get it right?? Did I ?? Did I Do I get a cookie If I go down for this, I'm bringing you all with me.
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