Data Sufficiency

There are n students in a class. Of them, k boys and k 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?

(1) Other than Harvey, 5 boys are selected for the dance.

(2) 8 of the k boys are not selected for the dance.

Source : Veritas
OA=A

Jay@ManhattanReview wrote:

ziyuenlau wrote:There are n students in a class. Of them, k boys and k 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?

(1) Other than Harvey, 5 boys are selected for the dance.

(2) 8 of the k boys are not selected for the dance.

Source : Veritas
OA=A

Hi ziyuenlau,

Statement 1: Other than Harvey, 5 boys are selected for the dance.

=> There are 5+1 = 6 boys (k=6) and 6 girls

The number of boys and the number of girls are known. Nothing is unknown. Sufficient.

Statement 2: 8 of the k boys are not selected for the dance.

With this information, we cannot get the value of k. Insufficient.

The correct answer: A

Hope this helps!

Relevant book: Manhattan Review GMAT Combinatorics and Probability Guide

-Jay
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Could you help to explain statement (2)? I have no idea what it is all about.

Jay@ManhattanReview wrote:

ziyuenlau wrote:There are n students in a class. Of them, k boys and k 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?

(1) Other than Harvey, 5 boys are selected for the dance.

(2) 8 of the k boys are not selected for the dance.

Source : Veritas
OA=A

Hi ziyuenlau,

Statement 1: Other than Harvey, 5 boys are selected for the dance.

=> There are 5+1 = 6 boys (k=6) and 6 girls

The number of boys and the number of girls are known. Nothing is unknown. Sufficient.

Hi Jay,
Can you please show how to calculate the probability after getting value of k?

Thanks

Can you please show how to calculate the probability after getting value of k?

Thanks

Jay did all the heavy lifting here.

Imagine you're Harvey. You're at a dance. There are 6 girls there, including Jessica. You're going to be randomly paired with one. Well, there's a 1 in 6 probability that you're paired with Jessica, right?

DavidG@VeritasPrep wrote:

Can you please show how to calculate the probability after getting value of k?

Thanks

Jay did all the heavy lifting here.

Imagine you're Harvey. You're at a dance. There are 6 girls there, including Jessica. You're going to be randomly paired with one. Well, there's a 1 in 6 probability that you're paired with Jessica, right?

Hi David,
Thanks for you reply. I understood it as choosing a pair (H & J) out of 12 which will give same prob 1/6 but what if there is not equal number of boys and girls. say 6 boys and 5 girls. I think here the probility will as follows:

1 way to choose Harvey of 6 boys =1/6

1 way to choose Jesscia of 5 boys =1/5

Here I'm confused if I should multiple both or sum them up

Total probability of the pair =1/6 * 1/5 =1/30

or

Total probability of the pair =1/6 + 1/5 =11/30

Any thought or support for above thinking

Mo2men wrote:

DavidG@VeritasPrep wrote:

Can you please show how to calculate the probability after getting value of k?

Thanks

Jay did all the heavy lifting here.

Imagine you're Harvey. You're at a dance. There are 6 girls there, including Jessica. You're going to be randomly paired with one. Well, there's a 1 in 6 probability that you're paired with Jessica, right?

Hi David,
Thanks for you reply. I understood it as choosing a pair (H & J) out of 12 which will give same prob 1/6 but what if there is not equal number of boys and girls. say 6 boys and 5 girls. I think here the probility will as follows:

1 way to choose Harvey of 6 boys =1/6

1 way to choose Jesscia of 5 boys =1/5

Here I'm confused if I should multiple both or sum them up

Total probability of the pair =1/6 * 1/5 =1/30

or

Total probability of the pair =1/6 + 1/5 =11/30

Any thought or support for above thinking

Well, first we need to define the parameters of the question you'd like to discuss. In the one we looked at, we had six boys and six girls, all of whom would be in a pair with a member of the opposite GMAT. If we have six boys and five girls, clearly we can't pair them off this way anymore - one of the boys is going to be left out. So there's two ways we can analyze this.

1) We could say that we're planning on pairing off five girls with five of the boys and one of the boys will be left out. Now we could calculate the probability that Harvey and Jessica will be paired together. (In this case, there'd actually still be 6 possibilities for Harvey - getting paired with each of the 5 girls, or being left on his own.) So the probability is still 1/6 that he ends up with Jessica.

2) We could alter the question to be - we're selecting one boy from the group of 6 and one girl from the group of 5, and what is the probability that our pair is Harvey and Jessica? (Notice this is a significant departure from the original question in which everyone gets paired off.) There'd be a (1/6) probability of selecting Harvey from the boys and (1/5) probability of selecting Jessica from the girls, and thus (1/6) * (1/5) = 1/30 probability that we select Harvey and Jessica for the one group.

DavidG@VeritasPrep wrote:

Mo2men wrote:

DavidG@VeritasPrep wrote:

Can you please show how to calculate the probability after getting value of k?

Thanks

Jay did all the heavy lifting here.

Imagine you're Harvey. You're at a dance. There are 6 girls there, including Jessica. You're going to be randomly paired with one. Well, there's a 1 in 6 probability that you're paired with Jessica, right?

Hi David,
Thanks for you reply. I understood it as choosing a pair (H & J) out of 12 which will give same prob 1/6 but what if there is not equal number of boys and girls. say 6 boys and 5 girls. I think here the probility will as follows:

1 way to choose Harvey of 6 boys =1/6

1 way to choose Jesscia of 5 boys =1/5

Here I'm confused if I should multiple both or sum them up

Total probability of the pair =1/6 * 1/5 =1/30

or

Total probability of the pair =1/6 + 1/5 =11/30

Any thought or support for above thinking

Well, first we need to define the parameters of the question you'd like to discuss. In the one we looked at, we had six boys and six girls, all of whom would be in a pair with a member of the opposite GMAT. If we have six boys and five girls, clearly we can't pair them off this way anymore - one of the boys is going to be left out. So there's two ways we can analyze this.

1) We could say that we're planning on pairing off five girls with five of the boys and one of the boys will be left out. Now we could calculate the probability that Harvey and Jessica will be paired together. (In this case, there'd actually still be 6 possibilities for Harvey - getting paired with each of the 5 girls, or being left on his own.) So the probability is still 1/6 that he ends up with Jessica.

2) We could alter the question to be - we're selecting one boy from the group of 6 and one girl from the group of 5, and what is the probability that our pair is Harvey and Jessica? (Notice this is a significant departure from the original question in which everyone gets paired off.) There'd be a (1/6) probability of selecting Harvey from the boys and (1/5) probability of selecting Jessica from the girls, and thus (1/6) * (1/5) = 1/30 probability that we select Harvey and Jessica for the one group.

Thanks David.
I actually was thinking in case 2, when I asked my question.

Mo2men wrote:

DavidG@VeritasPrep wrote:

Mo2men wrote:

DavidG@VeritasPrep wrote:

Can you please show how to calculate the probability after getting value of k?

Thanks

Jay did all the heavy lifting here.

Imagine you're Harvey. You're at a dance. There are 6 girls there, including Jessica. You're going to be randomly paired with one. Well, there's a 1 in 6 probability that you're paired with Jessica, right?

Hi David,
Thanks for you reply. I understood it as choosing a pair (H & J) out of 12 which will give same prob 1/6 but what if there is not equal number of boys and girls. say 6 boys and 5 girls. I think here the probility will as follows:

1 way to choose Harvey of 6 boys =1/6

1 way to choose Jesscia of 5 boys =1/5

Here I'm confused if I should multiple both or sum them up

Total probability of the pair =1/6 * 1/5 =1/30

or

Total probability of the pair =1/6 + 1/5 =11/30

Any thought or support for above thinking

Well, first we need to define the parameters of the question you'd like to discuss. In the one we looked at, we had six boys and six girls, all of whom would be in a pair with a member of the opposite GMAT. If we have six boys and five girls, clearly we can't pair them off this way anymore - one of the boys is going to be left out. So there's two ways we can analyze this.

1) We could say that we're planning on pairing off five girls with five of the boys and one of the boys will be left out. Now we could calculate the probability that Harvey and Jessica will be paired together. (In this case, there'd actually still be 6 possibilities for Harvey - getting paired with each of the 5 girls, or being left on his own.) So the probability is still 1/6 that he ends up with Jessica.

2) We could alter the question to be - we're selecting one boy from the group of 6 and one girl from the group of 5, and what is the probability that our pair is Harvey and Jessica? (Notice this is a significant departure from the original question in which everyone gets paired off.) There'd be a (1/6) probability of selecting Harvey from the boys and (1/5) probability of selecting Jessica from the girls, and thus (1/6) * (1/5) = 1/30 probability that we select Harvey and Jessica for the one group.

Thanks David.
I actually was thinking in case 2, when I asked my question.

If you're looking for more practice, here's an official question that I think of as a prototypical probability problem: https://www.beatthegmat.com/probability- ... 72268.html