There are 40 students in a class. Of them, 10 boys and 10 girls (including Harvey and Jessica) are selected for a dance performance in which students will dance in pairs of one boy and one girl. What is the probability that Harvey will be paired with Jessica?
A) 1/10
B) 1/100
C) 1/400
D ) 1/1000
E) 1/50
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Hi nahid078,
This question should really posted in the PS Forum (this is the SC Forum), but I'll be happy to answer it.
This question is worded in a "quirky" way. Based on the second sentence, the information in the first sentence seems to be irrelevant. We're essentially told that there are 10 boys (one is named Harvey) and 10 girls (one is named Jessica) and pairs of 1 boy/1 girl will be formed. We're asked for the probability that Harvey will be paired with Jessica.
Since Harvey could be paired with any of the 10 girls, the probability that he'll be paired with Jessica is 1/10.
Final Answer: A
GMAT assassins aren't born, they're made,
Rich
This question should really posted in the PS Forum (this is the SC Forum), but I'll be happy to answer it.
This question is worded in a "quirky" way. Based on the second sentence, the information in the first sentence seems to be irrelevant. We're essentially told that there are 10 boys (one is named Harvey) and 10 girls (one is named Jessica) and pairs of 1 boy/1 girl will be formed. We're asked for the probability that Harvey will be paired with Jessica.
Since Harvey could be paired with any of the 10 girls, the probability that he'll be paired with Jessica is 1/10.
Final Answer: A
GMAT assassins aren't born, they're made,
Rich
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MartyMurray
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This is a cool question, with a nice twist that makes it a little different from many GMAT probability questions.nahid078 wrote:There are 40 students in a class. Of them, 10 boys and 10 girls (including Harvey and Jessica) are selected for a dance performance in which students will dance in pairs of one boy and one girl. What is the probability that Harvey will be paired with Jessica?
A) 1/10
B) 1/100
C) 1/400
D ) 1/1000
E) 1/50
The twist is that rather than look for a relationship between one instance and the total instances, you can look at just one part of what's going on.
We know there are ten boys and ten girls. So we could use combinatorics to figure out how many possible ways there are to match them up and how many of those ways include Harvey being partnered with Jessica. The thing is getting it right, or at least efficiently getting it right, takes seeing that the total number of match ups is not really what matters.
The point is simply that Harvey will be matched up with one of ten girls and the probability that Jessica will be the one matched up with Harvey is 1 in 10. So the answer is 1/10.
Choose A.
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Brent@GMATPrepNow
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Rich and Marty have shown some nice, fast approaches.nahid078 wrote:There are 40 students in a class. Of them, 10 boys and 10 girls (including Harvey and Jessica) are selected for a dance performance in which students will dance in pairs of one boy and one girl. What is the probability that Harvey will be paired with Jessica?
A) 1/10
B) 1/100
C) 1/400
D ) 1/1000
E) 1/50
However, the great thing about probability questions is that they can typically be solved using a variety of approaches.
Here's one that involves counting.
P(Harvey paired with Jessica) = (number of arrangements in which Harvey is paired with Jessica)/(TOTAL number of pairings)
TOTAL number of pairings
Let A,B,C,D,..,J represent the 10 girls
Now we'll find a boy to pair with each girl
Stage 1: Select a boy for girl A. There are 10 boys, so we can complete this stage in 10 ways.
Stage 2: Select a boy for girl B. There are 9 boys remaining, so we can complete this stage in 9 ways.
Stage 3: Select a boy for girl C. There are 8 boys remaining, so we can complete this stage in 8 ways.
.
.
.
Stage 10: Select a boy for girl J. There is 1 boy remaining, so we can complete this stage in 1 way.
By the Fundamental Counting Principle (FCP), we can complete all 10 stages (and thus pair up all of the children) in (10)(9)(8)(7)(6)(5)(4)(3)(2)(1) ways
Number of arrangements in which Harvey is paired with Jessica
Stage 1: Pair Harvey with Jessica. This can be accomplished in 1 way.
Let A,B,C,D,..,I represent the REMAINING 9 girls
Stage 2: Select a boy for girl A. There are 9 boys remaining, so we can complete this stage in 9 ways.
Stage 3: Select a boy for girl B. There are 8 boys remaining, so we can complete this stage in 8 ways.
Stage 4: Select a boy for girl C. There are 7 boys remaining, so we can complete this stage in 7 ways.
.
.
.
Stage 9: Select a boy for girl I. There is 1 boy remaining, so we can complete this stage in 1 way.
By the Fundamental Counting Principle (FCP), we can complete all 10 stages (and thus pair up all of the children) in (1)(9)(8)(7)(6)(5)(4)(3)(2)(1) ways
So, P(Harvey paired with Jessica) = (1)(9)(8)(7)(6)(5)(4)(3)(2)(1) /(10)(9)(8)(7)(6)(5)(4)(3)(2)(1) ways
= [spoiler]1/10 [/spoiler][since most of the other values cancel out]
Answer: A
--------------------------
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Brent
Last edited by Brent@GMATPrepNow on Sun Feb 22, 2015 10:38 am, edited 1 time in total.
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