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# A certain club has 10 members, including Harry. One of the

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Director
Joined: 26 Feb 2006
Posts: 899
A certain club has 10 members, including Harry. One of the [#permalink]

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09 Jan 2007, 00:22
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Difficulty:

55% (hard)

Question Stats:

62% (02:16) correct 38% (01:32) wrong based on 280 sessions

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A certain club has 10 members, including Harry. One of the 10 members is to be chosen at random to be the president, one of the remaining 9 members is to be chosen at random to be the secretary, and one of the remaining 8 members is to be chosen at random to be the treasurer. What is the probability that Harry will either be the secretary or the member chosen to be the treasurer?

A) 1/720
B) 1/80
C) 1/10
D) 1/9
E) 1/5

OPEN DISCUSSION OF THIS QUESTION IS HERE: a-certain-club-has-10-members-including-harry-one-of-the-134891.html
[Reveal] Spoiler: OA
Director
Joined: 28 Dec 2005
Posts: 919

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09 Jan 2007, 04:14
Here is my try.

The probability will be:

(Probability of not becoming president * prob. of becoming secretary) + (prob of not becoming president*prob. of not becoming secretary*probability of becoming treasurer)

= 9/10*1/9 + 9/10*8/9*1/8 = 1/5
Intern
Joined: 21 Nov 2006
Posts: 44

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09 Jan 2007, 14:56
I came to the same answer a different way and may be lucky it worked. I took prob. of sec. plus tres. 1/9+1/8 =17/72 *9/10 which is prob of not president. This equals 153/720 or .21 or 1/5.

Anyone else have a cleaner method.

hsampath- like yours better, not sure if mine is right way.

Eric
Intern
Joined: 02 Jan 2007
Posts: 41

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09 Jan 2007, 17:56
Hi guys,

Silly question: why is it that you multiply the different probablilities instead of adding them...?

Jeff
Senior Manager
Joined: 12 Mar 2006
Posts: 365
Schools: Kellogg School of Management

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09 Jan 2007, 18:27
number of ways harry can be treasurer

choose harry then among the 9 select 2, so 9C2*2 (*2 since order matters as designation is based on order)

number of ways harry can be secretary is the same

Total number of ways of choosing 3 ppl is 10C3*3!

(9C2*2)/(10C3*3) = (9*8*2)/(10*9*8) = 1/5

so E ?
Senior Manager
Joined: 24 Nov 2006
Posts: 349

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09 Jan 2007, 20:13
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There are two scenarios:
1) Harry is NOT elected President AND is elected Secretary.
2) Harry is NOT elected President AND is not elected Secretary AND is selected Treasurer.

Prob1 = 9/10 * 1/9 = 1/10
Prob2 = 9/10 * 8/9 * 1/8 = 1/10

As these 2 scenarios are mutually exclusive, we can find the total probability simply by adding up prob1 and prob2 = 1/5.

Director
Joined: 26 Feb 2006
Posts: 899

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10 Jan 2007, 23:22
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Himalayan wrote:
A certain club has 10 members, including Harry. One of the 10 members is to be chosed at random to be the president, one of the remaining 9 members is to be chosed at random to be the secretary, and one of the remaining 8 members is to be chosen at random to be the treasurer. What is the probability that Harry will either be the secretary or the member chosen to be the treasurer?

A 1/720
B 1/80
C 1/10
D 1/9
E 1/5

Not sure the OA I have is correct.

its correct.
total possibilities = 10x9x8
no of possibilities for sec. = 1x9x8
no of possibilities for tres. = 1x9x8
prob = 2(9x8)/(10x9x8) = 1/5
GMAT Club Legend
Joined: 07 Jul 2004
Posts: 5043
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11 Jan 2007, 01:07
# of ways to pick 3 people to fill 3 spots = 10P3 = 720

# of ways to pick harry as secretary = 9*1*8 = 72
# of ways to pick harry as treasurer = 9*8*1 = 72

P = 144/720=1/5
Manager
Joined: 28 Feb 2006
Posts: 83

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11 Jan 2007, 01:33
I dont think my reasoning is correct but this is what I did:

Prob of being secratary or treasurer = 1 - (Prob of not being president + Prob of not being secratary or treasurer)

Therefore,

Prob of being secratary or treasurer = 1 - (1/10 + 7/10) = 1/5

So E for me.

Could someone please tell me if this is a logical reasoning?
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Manager
Joined: 09 Nov 2006
Posts: 128

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11 Jan 2007, 02:10
DreamMBA wrote:
I dont think my reasoning is correct but this is what I did:

Prob of being secratary or treasurer = 1 - (Prob of not being president + Prob of not being secratary or treasurer)

Therefore,

Prob of being secratary or treasurer = 1 - (1/10 + 7/10) = 1/5

So E for me.

Could someone please tell me if this is a logical reasoning?

I think it is wrong because Prob of not being president is not 1/10, but 9/10
Manager
Joined: 28 Feb 2006
Posts: 83

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11 Jan 2007, 03:33
DreamMBA wrote:
I dont think my reasoning is correct but this is what I did:

Prob of being secratary or treasurer = 1 - (Prob of not being president + Prob of not being secratary or treasurer)

Therefore,

Prob of being secratary or treasurer = 1 - (1/10 + 7/10) = 1/5

So E for me.

Could someone please tell me if this is a logical reasoning?

I think it is wrong because Prob of not being president is not 1/10, but 9/10

Sorry I meant:

Prob of being secratary or treasurer = 1 - (Prob of 'being' president + Prob of not being secratary or treasurer)

because if Harry is president he cannot be secratary or treasurer be default.
_________________

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Manager
Joined: 09 Nov 2006
Posts: 128

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11 Jan 2007, 05:29
DreamMBA wrote:
DreamMBA wrote:
I dont think my reasoning is correct but this is what I did:

Prob of being secratary or treasurer = 1 - (Prob of not being president + Prob of not being secratary or treasurer)

Therefore,

Prob of being secratary or treasurer = 1 - (1/10 + 7/10) = 1/5

So E for me.

Could someone please tell me if this is a logical reasoning?

I think it is wrong because Prob of not being president is not 1/10, but 9/10

Sorry I meant:

Prob of being secratary or treasurer = 1 - (Prob of 'being' president + Prob of not being secratary or treasurer)

because if Harry is president he cannot be secratary or treasurer be default.

Correct me if I am wrong, but probability of being neither secretary nor treasurer is 8/9 * 7/8 = 7/9
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Joined: 17 Jan 2008
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28 Apr 2008, 18:46
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sondenso wrote:
a certain club has 10 members, including Harry. One of the 10 members is to be chosen at random to be the president, one of the remaining 9 members is to be chosen at random to be the secretary, and one of the remaining 8 members is to be chosen at random to b the treasurer. What is the probability that Harry will be either the member chosen to be the scretary of the member chosen to be the treasurer?
A. 1/720
B. 1/80
C. 1/10
D. 1/9
E. 1/5

P(secretary)=P(not Harry)*P(Harry)*P(not Harry)=(9/10)(1/9)(8/8)=1/10
P(treasurer)=P(not Harry)*P(not Harry)*P(Harry)=(9/10)(8/9)(1/8)=1/10
P(secretary OR treasurer)=P(secretary)+P(treasurer)=(1/10)+(1/10)=1/5

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Manager
Joined: 04 Sep 2010
Posts: 74
A certain club has 10 members, including Harry. One of the 1 [#permalink]

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16 Jun 2011, 21:28
Can this be solved using symmetry rule that - each and every has equal chance of choosing specific position ... so taking Choosing Sec or Tres position would be
1/10 or 1/10 => 1/10+1/10 = 1/5
Intern
Joined: 20 Aug 2010
Posts: 38
Re: A certain club has 10 members, including Harry. One of the [#permalink]

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26 Nov 2011, 10:37
Can somebody explain why we use in the first part of the calculation P{that he is not secretary} =1 and in the second part P{he is not treasurer} = 1/8. Should not it be 7/8 in the first part instead of 1. because we cannot be sure that he is not chosen as secretary ?????
9/10 * 1/9 *1 + 9/10 *8/9 *1/8
Manager
Joined: 25 May 2011
Posts: 152
Re: A certain club has 10 members, including Harry. One of the [#permalink]

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13 Dec 2011, 03:51
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The probability of being secretory: $$\frac{9}{10}*\frac{1}{9}=\frac{1}{10}$$

The probability of being treasurer: $$\frac{9}{10}*\frac{8}{9}*\frac{1}{8}=\frac{1}{10}$$
Total probability: $$\frac{1}{10}+\frac{1}{10}=\frac{1}{5}$$
Manager
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WE: Accounting (Insurance)
Re: A certain club has 10 members, including Harry. One of the [#permalink]

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13 Dec 2011, 13:39
The probability calcs are killing me. The situation is random. What is the probability that anyone is the president. 1/10.

The probability of Harry becoming trasurer is 1/10, and secretary is 1/10. 1/10 + 1/10 os 2/10. .1 +.1 = .2
Senior Manager
Joined: 12 Oct 2011
Posts: 262
Re: A certain club has 10 members, including Harry. One of the [#permalink]

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14 Dec 2011, 00:25
Just solved a similar question, so found this one very easy. The answer is E. The reasons have already been mentioned by most people here.

Probability that Harry is chosen as Secretary = 9/10*1/9 = 1/10
Probability that Harry is chosen as Treasurer = 9/10*8/9*1/8 = 1/10

Thus, the total probability that harry is either the Secretary or the Treasurer = 1/10 + 1/10 = 2/10 = 1/5 (E).
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Re: A certain club has 10 members, including Harry. One of the [#permalink]

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19 May 2014, 10:07
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Re: A certain club has 10 members, including Harry. One of the [#permalink]

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19 May 2014, 23:48
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A certain club has 10 members, including Harry. One of the 10 members is to be chosen at random to be the president, one of the remaining 9 members is to be chosen at random to be the secretary, and one of the remaining 8 members is to be chosen at random to be the treasurer. What is the probability that Harry will either be the secretary or the member chosen to be the treasurer?

A) 1/720
B) 1/80
C) 1/10
D) 1/9
E) 1/5

This question is much easier than it appears.

Each member out of 10, including Harry, has equal chances to be selected for any of the positions (the sequence of the selection is given just to confuse us). The probability that Harry will be selected to be the secretary is 1/10 and the probability that Harry will be selected to be the treasurer is also 1/10. So, the probability that Harry will be selected to be either the secretary or the the treasurer is 1/10+1/10=2/10.

OPEN DISCUSSION OF THIS QUESTION IS HERE: a-certain-club-has-10-members-including-harry-one-of-the-134891.html

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Re: A certain club has 10 members, including Harry. One of the   [#permalink] 19 May 2014, 23:48
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