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Sam and Jessica are invited to a dance. If there are 7 men and 7 women [#permalink]
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18 Mar 2010, 23:56
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Sam and Jessica are invited to a dance. If there are 7 men and 7 women in total at the dance, and one woman and one man are chosen to lead the dance, what is the probability that Sam and Jessica will NOT be the pair chosen to lead the dance? A) 1/49 B) 1/7 C) 6/7 D) 47/49 E) 48/49 I came up with the answer but I tried to solve it in a different way originally and failed miserably.
Why can't i come up with the right fraction if I say, there are 6/7 chance that Sam will NOT be picked and 6/7 chance that Jessican will NOT be picked. So the probably is 36/49 that Sam and Jessica will NOT be the pair choosen. Why wouldn't this method work?
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Re: Sam and Jessica are invited to a dance. If there are 7 men and 7 women [#permalink]
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19 Mar 2010, 05:20
changhiskhan wrote: Sam and Jessica are invited to a dance. If there are 7 men and 7 women in total at the dance, and one woman and one man are chosen to lead the dance, what is the probability that Sam and Jessica will NOT be the pair chosen to lead the dance?
A) 1/49 B) 1/7 C) 6/7 D) 47/49 E) 48/49
I came up with the answer but I tried to solve it in a different way originally and failed miserably.
Why can't i come up with the right fraction if I say, there are 6/7 chance that Sam will NOT be picked and 6/7 chance that Jessican will NOT be picked. So the probably is 36/49 that Sam and Jessica will NOT be the pair choosen. Why wouldn't this method work? Welcome to the Gmat Club. The easiest way to solve this problem is to calculate opposite probability and subtract this value from 1: The opposite probability is that Sam and Jessica WILL be chosen to lead \(=\frac{1}{7}*\frac{1}{7}=\frac{1}{49}\). So, the probability that they will NOT be chosen to lead \(= 1\frac{1}{49}=\frac{48}{49}\). Answer: E. The way you are doing is wrong because \(\frac{36}{49}\) is the probability that in the leading pair won't be neither Sam nor Jessica. But in the leading pair can be Sam with any woman but Jessica and can be Jessica with any man but Sam. So, if you are doing this way you should count \(\frac{7}{7}*\frac{6}{7}\) probability of pairing any man (\(\frac{7}{7}\)) with any woman but Jessica (\(\frac{6}{7}\)) PLUS \(\frac{1}{7}*\frac{6}{7}\) probability of pairing Jessica (\(\frac{1}{7}\)) with any man but Sam (\(\frac{6}{7}\)) \(=\frac{7}{7}*\frac{6}{7}+\frac{1}{7}*\frac{6}{7}=\frac{48}{49}\). OR you can think about it this way: there are total \(7*7=49\) pairs possible, and only one pair is when Jessica and Sam are together, hence probability \(=\frac{491}{49}=\frac{48}{49}\) OR: pairs without Jessica \(6*7=42\) and pairs with Jessica but without Sam \(1*6=6\), hence probability \(=\frac{42+6}{49}=\frac{48}{49}\). Hope it helps.
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Re: Sam and Jessica are invited to a dance. If there are 7 men and 7 women [#permalink]
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19 Mar 2010, 09:12
I agree with the previous poster. For probability, the basic equation is:
# of ways event can happen / total # of possible ways
The numbers of pairs that have Sam and Jessica is just 1, out of a total possibility of 49 pairs. So 1/49 chance they will be chosen.
The chance they will NOT be chosen is 1 minus this value, or 1 1/49 = 48/49.



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Re: Sam and Jessica are invited to a dance. If there are 7 men and 7 women [#permalink]
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19 Mar 2010, 09:19
Thank you for the wonderful explanation.



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Re: Sam and Jessica are invited to a dance. If there are 7 men and 7 women [#permalink]
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19 Mar 2010, 12:51
Good explanation. And +1 for E 48/49
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probabilty problem "Sam & Jessica" [#permalink]
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26 Jul 2010, 23:27
Hi! Could one of the moderators please help me with it? "Sam and Jessica are invited to a dance. If there are 7 men and 7 women in total at the dance, and one woman and one man ar chosen to lead the dance, what is the probability that Sam and Jessica will NOT be chosen to lead the dance?" I know, the thread was posted a looooooonnnnnnng time ago, but I am so confused with which method I should take and WHY? My question is: why can't I solve the problem in this way? 1  C 1 7/C2 14= 12/13 C1 7 : I would choose one group out of 7 (since there are 14 people) C2 14: altogether I would have chosen 2 people out of 14 The "funny" point is, I do also understand, how the original answer has been calculated. Can someone please help me out? Many thanks!!!! OA:



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Re: probabilty problem "Sam & Jessica" [#permalink]
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29 Jul 2010, 10:35
1. You use p = 1  q formula in which q means "Sam and Jessica" 2. \(C^{14}_2\) as the total number of all possibilities are incorrect as it counts manman and womanwoman pairs. So, you have to use \(C^7_1 * C^7_1\) = 49 or any man and any woman combinations. 3. \(C^7_1\). I don't get the reasoning behind the formula. There is only one way to choose "Sam and Jessica" and why do you think that there is 7 groups? There is no special assignment for each woman and man. Let me know if you need further help.
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Re: probabilty problem "Sam & Jessica" [#permalink]
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29 Jul 2010, 10:47
By the way, welcome to GMAT Club!
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Re: probabilty problem "Sam & Jessica" [#permalink]
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31 Jul 2010, 10:10
walker wrote: 1. You use p = 1  q formula in which q means "Sam and Jessica"
2. \(C^{14}_2\) as the total number of all possibilities are incorrect as it counts manman and womanwoman pairs. So, you have to use \(C^7_1 * C^7_1\) = 49 or any man and any woman combinations.
3. \(C^7_1\). I don't get the reasoning behind the formula. There is only one way to choose "Sam and Jessica" and why do you think that there is 7 groups? There is no special assignment for each woman and man.
Let me know if you need further help. Hi, "walker"! Thank you for your explanation! Could you be so kind and explain to me, when to use what kind of formula? My problem is, I often mix combinatoricsprobability formula with simple probability formula. Is there a helpful strategy? Again, thank you so much!



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Re: probabilty problem "Sam & Jessica" [#permalink]
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01 Aug 2010, 04:36
Hi fledgling2010, I think I understand why GMAC use "counting" rather than "prob/comb" term. It seems you are trying to remember where which formula to use rather than understand how to "count". So, my advice is to read this [http://gmatclub.com/forum/mathprobability87244.html][http://gmatclub.com/forum/mathcombinatorics87345.html] and each time you see prob/comb problem try to get idea how to count and then use an appropriate formula.
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Re: probabilty problem "Sam & Jessica" [#permalink]
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02 Aug 2010, 17:44
I personally think this problem is solved so much easier using simple probability rules The trick here is to estimate the individual probabilities of selecting him and her and multiply them because the events are independent from each other. the probability of selecting 1 individual out of 7 is 1/7 and the probability of selecting him and her is 1/7*1/7=1/49. P of not selecting them both is what is asked for. 11/49 = 48/49. I hope u find this method much more clear and concise.
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Re: Sam and Jessica are invited to a dance. If there are 7 men and 7 women [#permalink]
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