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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.
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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)

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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.
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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]
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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.
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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
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