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

Redeem

Board of directors

This topic has expert replies
Post new topic Post Reply
Previous Topic Next Topic
Master | Next Rank: 500 Posts
Posts: 218
Joined: Wed Nov 23, 2011 8:05 pm
Thanked: 26 times
Followed by:4 members

Board of directors

by chieftang » Sun Dec 04, 2011 7:52 pm
Thought I'd throw this one up for you guys. Adapted from an OG question:

A board of directors has 10 members, including Gonzalo. Randomly, one of the 10 members is chosen to be the president, one of the remaining 9 members is randomly chosen to be the treasurer, and one of the remaining 8 members is randomly chosen to be the secretary. What is the probability that Gonzalo will be either the member chosen to be the secretary or the member chosen to be the treasurer?

(A) 1/720
(B) 1/20
(C) 1/10
(D) 1/9
(E) 1/5

OA to come.
Top
Source: — Problem Solving |
Master | Next Rank: 500 Posts
Posts: 385
Joined: Fri Sep 23, 2011 9:02 pm
Thanked: 62 times
Followed by:6 members

by user123321 » Sun Dec 04, 2011 9:45 pm
[spoiler]IMO 1/5[/spoiler]

(9*1*8)/(10*9*8) + (9*8*1)/(10*9*8)

user123321
Just started my preparation :D
Want to do it right the first time.
Top
User avatar
GMAT Instructor
Posts: 15539
Joined: Tue May 25, 2010 12:04 pm
Location: New York, NY
Thanked: 13060 times
Followed by:1906 members
GMAT Score:790

by GMATGuruNY » Mon Dec 05, 2011 4:57 am
chieftang wrote:Thought I'd throw this one up for you guys. Adapted from an OG question:

A board of directors has 10 members, including Gonzalo. Randomly, one of the 10 members is chosen to be the president, one of the remaining 9 members is randomly chosen to be the treasurer, and one of the remaining 8 members is randomly chosen to be the secretary. What is the probability that Gonzalo will be either the member chosen to be the secretary or the member chosen to be the treasurer?

(A) 1/720
(B) 1/20
(C) 1/10
(D) 1/9
(E) 1/5

OA to come.
P(A or B) = P(A) + P(B).

Each of the 10 members has a 1/10 chance of being chosen secretary or treasury.
The intervening elections are IRRELEVANT.

P(Gonzales is chosen secretary or treasurer) = 1/10 + 1/10 = 1/5.

The correct answer is E.
Private tutor exclusively for the GMAT and GRE, with over 20 years of experience.
Followed here and elsewhere by over 1900 test-takers.
I have worked with students based in the US, Australia, Taiwan, China, Tajikistan, Kuwait, Saudi Arabia -- a long list of countries.
My students have been admitted to HBS, CBS, Tuck, Yale, Stern, Fuqua -- a long list of top programs.

As a tutor, I don't simply teach you how I would approach problems.
I unlock the best way for YOU to solve problems.

For more information, please email me (Mitch Hunt) at [email protected].
Student Review #1
Student Review #2
Student Review #3
Top
Master | Next Rank: 500 Posts
Posts: 218
Joined: Wed Nov 23, 2011 8:05 pm
Thanked: 26 times
Followed by:4 members

by chieftang » Mon Dec 05, 2011 7:45 am
GMATGuruNY wrote:
chieftang wrote:Thought I'd throw this one up for you guys. Adapted from an OG question:

A board of directors has 10 members, including Gonzalo. Randomly, one of the 10 members is chosen to be the president, one of the remaining 9 members is randomly chosen to be the treasurer, and one of the remaining 8 members is randomly chosen to be the secretary. What is the probability that Gonzalo will be either the member chosen to be the secretary or the member chosen to be the treasurer?

(A) 1/720
(B) 1/20
(C) 1/10
(D) 1/9
(E) 1/5

OA to come.
P(A or B) = P(A) + P(B).

Each of the 10 members has a 1/10 chance of being chosen secretary or treasury.
The intervening elections are IRRELEVANT.

P(Gonzales is chosen secretary or treasurer) = 1/10 + 1/10 = 1/5.

The correct answer is E.
Yes!!

And it bears out when you multiply the odds of NOT being chosen as president by the odds of being chosen as treasurer: (9/10)(1/9) = 1/10, and the odds of not being chosen as president, not being chosen as treasurer, and being chosen as secretary: (9/10)(8/9)(1/8) = 1/10

OA [spoiler]E[/spoiler]
Top
Master | Next Rank: 500 Posts
Posts: 416
Joined: Thu Jul 28, 2011 12:48 am
Thanked: 28 times
Followed by:6 members

by gunjan1208 » Fri Dec 09, 2011 5:43 am
Hi,
I did it exactly the way described by Cheftang and got 1/10...

But the language does say that remaining 9...Remaining 8 etc....Therefore it needs to be done that way. What am I missing?

Guys, help.
Top
User avatar
Legendary Member
Posts: 588
Joined: Sun Oct 16, 2011 9:42 am
Location: New Delhi, India
Thanked: 130 times
Followed by:9 members
GMAT Score:720

by rijul007 » Sat Dec 10, 2011 11:40 am
chieftang wrote:Thought I'd throw this one up for you guys. Adapted from an OG question:

A board of directors has 10 members, including Gonzalo. Randomly, one of the 10 members is chosen to be the president, one of the remaining 9 members is randomly chosen to be the treasurer, and one of the remaining 8 members is randomly chosen to be the secretary. What is the probability that Gonzalo will be either the member chosen to be the secretary or the member chosen to be the treasurer?

(A) 1/720
(B) 1/20
(C) 1/10
(D) 1/9
(E) 1/5

OA to come.
Total no of ways = 10*9*8
No of possible selections
i) when Gonzalo is the treasurer = 9*1*8
ii) when he is the secr = 9*8*1
Total no of selections = 2*9*8

Probability = 2*9*8/10*9*8 = 1/5

Option E
Top
Post new topic Post Reply
• Page 1 of 1