Find all School-related info fast with the new School-Specific MBA Forum

 It is currently 25 Jul 2016, 21:46

### GMAT Club Daily Prep

#### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History

# Events & Promotions

###### Events & Promotions in June
Open Detailed Calendar

# In a high school debating team consisting of 2 freshmen, 2

 post reply Question banks Downloads My Bookmarks Reviews Important topics
Author Message
Director
Joined: 29 Oct 2004
Posts: 863
Followers: 3

Kudos [?]: 141 [0], given: 0

In a high school debating team consisting of 2 freshmen, 2 [#permalink]

### Show Tags

24 Dec 2004, 06:23
00:00

Difficulty:

(N/A)

Question Stats:

0% (00:00) correct 0% (00:00) wrong based on 0 sessions

### HideShow timer Statistics

This topic is locked. If you want to discuss this question please re-post it in the respective forum.

In a high school debating team consisting of 2 freshmen, 2 sophomores, 2 juniors, and 2 seniors, two students are selected to represent the school at the state debating championship. The rules stipulate that the representatives must be from different grades, but otherwise the 2 representatives are to be chosen by lottery. What is the probability that the students selected will consist of one freshman and one sophomore?

A. 1/16
B. 1/8
C. 1/7
D. 1/6
E. 1/4

HIGHTLIGHT BELOW TO SEE OA:
OA is (D)

I tried to solve this a different way but encounter a different result. Please look
P(p(1st freshman AND 2nd sophomore) OR p(1st sophomore AND 2nd freshman)
=p1 + p2
=(2/8 * 2/7) + (2/8 * 2/7) (7 because one person already choosen, so OUT)
=1/14 + 1/14
=1/7
This is different from OA. Anyone can point out the problem with solution above?
Manager
Joined: 13 Dec 2004
Posts: 51
Followers: 0

Kudos [?]: 4 [0], given: 0

### Show Tags

24 Dec 2004, 14:23
Quote:
tried to solve this a different way but encounter a different result. Please look

P(p(1st freshman AND 2nd sophomore) OR p(1st sophomore AND 2nd freshman)
=p1 + p2
=(2/8 * 2/7) + (2/8 * 2/7) (7 because one person already choosen, so OUT)
=1/14 + 1/14
=1/7
This is different from OA. Anyone can point out the problem with solution above?

I'll give it a try.
I think what you want is P(F1 & So1) + P(F1 & So2) + P(F2 + So1) + P(F2 + So2). There are four possible desireable outcomes. Since there are 24 possible outcomes, I get 4/24, which is 1/6.

Doing it your way, I think you would want to do it like this:
[P(choosing a freshman first) + P(choosing a sophomore second)] + [P(choosing a sophomore first) + P(choosing a freshman second)] =

(1/4 x 2/6) + (1/4 + 2/6) = 4/24 = 1/6

Your denominator was 7, when it should have been 6, because there was a requirement that the two students be from different grades, so once a student is chosen, the other student in his/her grade can't be chosen.
Senior Manager
Joined: 19 Sep 2004
Posts: 369
Followers: 1

Kudos [?]: 4 [0], given: 0

### Show Tags

25 Dec 2004, 14:05
Hi I am kind of lost!

My way was:

(2C1*2C1)
------------= 2*2
8C2 ----- = 1/7
8*7/2

The order dosen't matter rt?I am kind of lost.
Please help. Hil..I was not able to understand your apporach.I gues syou have some "+" "*" prob in your post. Please check

Thanks
Saurabh Malpani
Director
Joined: 21 Sep 2004
Posts: 610
Followers: 1

Kudos [?]: 25 [0], given: 0

### Show Tags

25 Dec 2004, 20:44
hilairity wrote:
Quote:
tried to solve this a different way but encounter a different result. Please look

P(p(1st freshman AND 2nd sophomore) OR p(1st sophomore AND 2nd freshman)
=p1 + p2
=(2/8 * 2/7) + (2/8 * 2/7) (7 because one person already choosen, so OUT)
=1/14 + 1/14
=1/7
This is different from OA. Anyone can point out the problem with solution above?

I'll give it a try.
I think what you want is P(F1 & So1) + P(F1 & So2) + P(F2 + So1) + P(F2 + So2). There are four possible desireable outcomes. Since there are 24 possible outcomes, I get 4/24, which is 1/6.

Doing it your way, I think you would want to do it like this:
[P(choosing a freshman first) + P(choosing a sophomore second)] + [P(choosing a sophomore first) + P(choosing a freshman second)] =

(1/4 x 2/6) + (1/4 + 2/6) = 4/24 = 1/6

Your denominator was 7, when it should have been 6, because there was a requirement that the two students be from different grades, so once a student is chosen, the other student in his/her grade can't be chosen.

24 possible outcomes? didn't get that..
Intern
Joined: 17 Dec 2004
Posts: 24
Followers: 0

Kudos [?]: 1 [0], given: 0

### Show Tags

26 Dec 2004, 03:38
There should be 28 total possible outcome, calculated by 8C2 (or by counting 7+6+5+4+3+2+1=28). However, as the question says "the representative should be from different grade.......so 28 - 4 = 24.

Total outcome for one F + one SO is 4 (i.e. F1+SO1, F1+SO2, F2+SO1 and F2+SO2) - Order is not important....

So....it should be 4/24 = 1/6
Manager
Joined: 29 Jul 2004
Posts: 61
Followers: 1

Kudos [?]: 0 [0], given: 0

### Show Tags

27 Dec 2004, 12:06
Hcgmat wrote:
There should be 28 total possible outcome, calculated by 8C2 (or by counting 7+6+5+4+3+2+1=28). However, as the question says "the representative should be from different grade.......so 28 - 4 = 24.

Total outcome for one F + one SO is 4 (i.e. F1+SO1, F1+SO2, F2+SO1 and F2+SO2) - Order is not important....

So....it should be 4/24 = 1/6

How did you figure out to subtract 4 from the 28. I got 4/28, but didn't know how to eliminate the ways where the representatives were in the same grade.
[#permalink] 27 Dec 2004, 12:06
Display posts from previous: Sort by

# In a high school debating team consisting of 2 freshmen, 2

 post reply Question banks Downloads My Bookmarks Reviews Important topics

 Powered by phpBB © phpBB Group and phpBB SEO Kindly note that the GMAT® test is a registered trademark of the Graduate Management Admission Council®, and this site has neither been reviewed nor endorsed by GMAC®.