BREAKING: Target Test Prep releases Brand New 2026 On Demand GMAT prep course

Redeem

probability!

This topic has expert replies
Post new topic Post Reply
Previous Topic Next Topic
User avatar
Master | Next Rank: 500 Posts
Posts: 407
Joined: Tue Jan 25, 2011 9:19 am
Thanked: 25 times
Followed by:7 members

probability!

by Ozlemg » Thu Aug 04, 2011 8:36 am
In a group of 8 semifinalists, all but 2 will advance to the final round. If in the final round only the top 3 will be awarded medals, then how many groups of medal winners are possible?
(A) 20
(B) 56
(C) 120
(D) 560
(E) 720
The more you suffer before the test, the less you will do so in the test! :)
Top
Source: — Data Sufficiency |
Legendary Member
Posts: 1448
Joined: Tue May 17, 2011 9:55 am
Location: India
Thanked: 375 times
Followed by:53 members

by Frankenstein » Thu Aug 04, 2011 8:40 am
Hi,
Of the 8 people, we need to pick a group of 3.
It will be done in 8C3 = 56 ways

Hence, B

Btw, this is not probability :)
Cheers!

Things are not what they appear to be... nor are they otherwise
Top
User avatar
Legendary Member
Posts: 1101
Joined: Fri Jan 28, 2011 7:26 am
Thanked: 47 times
Followed by:13 members
GMAT Score:640

by HSPA » Thu Aug 04, 2011 9:08 am
Why havent you chose 3 out of 6.. and make it 20
Frankenstein wrote:Hi,
Of the 8 people, we need to pick a group of 3.
It will be done in 8C3 = 56 ways

Hence, B

Btw, this is not probability :)
First take: 640 (50M, 27V) - RC needs 300% improvement
Second take: coming soon..
Regards,
HSPA.
Top
Legendary Member
Posts: 1448
Joined: Tue May 17, 2011 9:55 am
Location: India
Thanked: 375 times
Followed by:53 members

by Frankenstein » Thu Aug 04, 2011 9:16 am
HSPA wrote:Why havent you chose 3 out of 6.. and make it 20
Frankenstein wrote:Hi,
Of the 8 people, we need to pick a group of 3.
It will be done in 8C3 = 56 ways

Hence, B

Btw, this is not probability :)
Hi,
Each of the 8 semifinalists is equally likely to win a medal. So, we have to consider all 8.
Cheers!

Things are not what they appear to be... nor are they otherwise
Top
Master | Next Rank: 500 Posts
Posts: 298
Joined: Tue Feb 16, 2010 1:09 am
Thanked: 2 times
Followed by:1 members

by Deepthi Subbu » Fri Aug 05, 2011 4:02 am
Please help me understand this

1. Only 6 will be entering the final round so I am still confused the reason behind considering all 8.
Top
Legendary Member
Posts: 1448
Joined: Tue May 17, 2011 9:55 am
Location: India
Thanked: 375 times
Followed by:53 members

by Frankenstein » Fri Aug 05, 2011 5:31 am
Deepthi Subbu wrote:Please help me understand this

1. Only 6 will be entering the final round so I am still confused the reason behind considering all 8.
Hi,
We do not know which two specific persons will get knocked out. If we know the two persons who are definitely knocked out, then we need to pick 3 from the remaining 6.
Lets say : Semi Finalists are: 1,2,3,4,5,6,7,8 and we know that 7,8 are knocked out. We are sure that 1,2,3,4,5,6 are the only ones making final. So, we can use 6C3.

But, in the given question everyone is equally likely to get knocked out. So, we have to select 3 from the given 8. Please, think about the difference between the two questions.
Cheers!

Things are not what they appear to be... nor are they otherwise
Top
Newbie | Next Rank: 10 Posts
Posts: 1
Joined: Tue Aug 16, 2011 1:00 am

by wellsronald18 » Tue Aug 16, 2011 1:13 am
It's got to be D. the equation would be 8C6 * 6C3= (8! /6! 2!)*(6! /3! 3!) = 28*20 = 560
Top
Post new topic Post Reply
• Page 1 of 1