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Help with an easy probability question

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Help with an easy probability question [#permalink] New post 18 Mar 2010, 22:56
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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: Help with an easy probability question [#permalink] New post 19 Mar 2010, 04:20
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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{49-1}{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: Help with an easy probability question [#permalink] New post 19 Mar 2010, 08:12
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: Help with an easy probability question [#permalink] New post 19 Mar 2010, 08:19
Thank you for the wonderful explanation.
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Re: Help with an easy probability question [#permalink] New post 19 Mar 2010, 11:51
Good explanation. And +1 for E 48/49
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Re: Help with an easy probability question   [#permalink] 19 Mar 2010, 11:51
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