Cinderella Dance
Rules of the Cinderella dance: All the ladies will throw one of there shoes in a heap. The gentlemen in the party will jump on the heap and select one of the shoe and can dance with the respective owner.
Ten couples were present in the Cinderella's party. What are the chances that exactly nine of them dance with their own spouse/partner ?
Ten couples were present in the Cinderella's party. What are the chances that exactly nine of them dance with their own spouse/partner ?
posted by Rajesh Lal at
10:44 AM
28 Comments:
Impossible, there would have to be 10 actually couples.
Pretty good if the "gentlemen" are in any way perceptive. I know I can recognize my girlfriend's shoes in a second...
that is inpossible! all the gentlemen and ladies would have to have a shoe. or else they break the rules
Impossible, are they all wearing the same shoes, different, or some are wearing the same, what if there is one girl there with all of the 10 guys???!??!?!?!
2 possibilities: 1)its impossible because if 9 had the right shoe, than all 10 would have the right shoe. 2)1 couple could be a gay couple, which would mean they wouldn't throw a shoe in. making the chances 1/9! or 1/362880
impossible as then all of them would have their spouces' shoe. even if all the ladies had the same shoes on they would still have a matching pair.
:0)
ps - is the plural of spouces spoui?!
it says there are ten COUPLES, so there could be any number of single men or women there, so the chances of exactly 9 of the couples dancing together, is 18/>21, or something like that
Would a Gentleman actually jump onto a pile of shoes??
I would think that high heels would hurt
The chances that exactly nine of the gentlemen dance with their respective spouse would be great, since any self-respecting man would make sure he only grabbed the shoe that belonged to his spouse--especially since only nine of the couples are married, and the gentlemen do not have a death wish.
Then again, it would truly depend on how many of the couples were married couples and how many of the couples were just boyfriend and girlfriend.
THere is no chance that exactly nine or only nine would get their spouse unless somebody got into a bad fight during the dance and broke up then the odds are good
if you ever wanted to dance with your classmates girl friend you know now what to do :)
First, assume the problem is solvable.
For the problem to be solvable, we must know exactly how many people there are, not only how many couples there are.
Since we only know the number of couples, we must assume that the couples are the only attendees, if we assume the problem is solvable.
Bearing that in mind, the chance of exactly 9 being correct is zero, because if 9 are correct, the final tenth must also be correct due to elimination of all incorrect choices, rgardlesss of how well each recognize ANYONEs shoe. The case of 9 is physically impossible.
It is impossible unless a woman didn`t take her shoe off.
The question is asking for chances or the PROBABILITY in mathematics. As there are 10 couples and we are asked to find the probability of 9 couples having exactly their same partners then it's the same (or equal to) as asking to find the probability of 10 couples having exactly their same partners because if 9 couples had exactly their same partners automatically the 2 people left would be an exact couple. 10 couples were present in the Cinderella's party. Now let's begin the analysis. If only one couple attended the party then it would be [(1/1) x 100%] or 100% probable that they would find each other. If two couples only attended the party then it would be [(1/2) x 100%] or 50% probable that they would find each other and so on until if 10 couples attended the party then it would be [(1/10) x 100%] or 10% that they would find each other. So the answer is a 10% chance.
This post has been removed by the author.
An addendum to what I just said above. The question is asking for exactly 9 couples dancing with their own spouse/partner. I'm assuming the 10th couple didn't dance. :-) Without this assumption it's impossible to solve.
impossible. There is 8 couples that would be able to dance with their own spouses, but 9 will be impossible unless the couples were odd numbers.
only 1: if 9 couples dance with their respective other, than so is the 10th. therefore thats there is only once chance in how ever many that 9 out of 10 (there fore 10/10) dance with their spouse!
Pretty obvious, no chance.
If 9 couples are together, the remaining man and woman would be the 10th couple. So EXACTLY 9 can never occur.
Chances for exactly nine shoes taken are none. Why?
Chances for one are 1/10. Chances for 2 are 1/10 x 1/10 and so on. By this logic it might look that for 9 right choices you will have one in billion chances, but the reminder shoe is still match therefore it can't happen exactly nine matches. It can be either exactly 8 or exactly 10 (as two highest possibilities)
10 to 1
ANSWER - 0 percent chance
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If Nine of them dance with their spouse/partner, the last gentleman is automatically left with his own spouse/partner for dance.
GOT RIGHT
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Jacquelyn, Clay, louey, bera
and
a few guys/gals who were not confident enough about their answer and prefered to remain anonymous.
well if it's a math question..would this be close to the answer?
there are 10 pairs of shoes. 10 * 2= the # of shoes. 2 out of 20 shoes = the chance of 1 couple getting to dance with each other.
then i took 2/20=.1
.1*.1*.1*.1*.1.*.1*.1*.1*.1=
.000000001 *10 (for percentage)=.00000001% chance that 9 out of 10 dancers will dance with each other
(they don't care about the 10 couples they just want to know about 9 couples)
i mean 9 couples not dancers
ynaomiy... tell me where you went to school so that I know not to send my kids there. There is a 0 percent chance, as if 9 had the correct partner, the 10th would as well.
ynaomiy... tell me where you went to school so that I know not to send my kids there. There is a 0 percent chance, as if 9 had the correct partner, the 10th would as well.
i think it is 9/10000000000 because there is a 1/10 shot for one guy to get his spouse, and there are ten guys so you multiply 1/10x1/10x1/10x1/10x1/10x1/10x1/10x1/10x1/10x1/10 which is 1/10000000000. Then you have to add it 9 times because you want 9 of the couples to be right. so 1+1+1+1+1+1+1+1+1/10000000000 or 9/10000000000, we did these in biology for genetics.
"What are the chances that exactally nine of them dance with their own spouse/partner."
This could mean any number of those couples could have both been male or female with having one less or one more shoe thrown in pile. Therefore we cannot assume that it is impossible.
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