GMAT Question of the Day - Daily to your Mailbox; hard ones only

 It is currently 18 Nov 2019, 12:37

### 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

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

Author Message
TAGS:

### Hide Tags

Math Expert
Joined: 02 Sep 2009
Posts: 59125
A certain club has 10 members, including Harry. One of the  [#permalink]

### Show Tags

25 Jun 2012, 02:31
16
142
00:00

Difficulty:

65% (hard)

Question Stats:

62% (02:05) correct 38% (02:24) wrong based on 2482 sessions

### HideShow timer Statistics

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 be either the member chosen to 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

Diagnostic Test
Question: 7
Page: 21
Difficulty: 650

_________________
Math Expert
Joined: 02 Sep 2009
Posts: 59125
Re: A certain club has 10 members, including Harry. One of the  [#permalink]

### Show Tags

25 Jun 2012, 02:31
24
47
SOLUTION

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 be either the member chosen to 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.

Similar questions to practice:
a-certain-team-has-12-members-including-joey-a-three-131321.html
a-medical-researcher-must-choose-one-of-14-patients-to-127396.html
a-certain-class-consists-of-8-students-including-kim-127730.html
_________________
Current Student
Joined: 29 Mar 2012
Posts: 295
Location: India
GMAT 1: 640 Q50 V26
GMAT 2: 660 Q50 V28
GMAT 3: 730 Q50 V38
Re: A certain club has 10 members, including Harry. One of the  [#permalink]

### Show Tags

25 Jun 2012, 04:35
43
16
Hi,

Difficulty level = 650

Probability = (Favorable cases)/(total cases)

Case 1: When Harry is chosen as secretary
probability = 9/10 * 1/9 * 8/8 = 1/10
(9 available for post of president out of 10, then Harry has to be chosen out of 9, and finally out of 8 anyone can be treasurer)

Case 2: When Harry is chosen as treasurer
probability = 9/10 * 8/9 * 1/8 = 1/10
(9 available for post of president out of 10, then out of 8 anyone can be secretary, and finally Harry has to be treasurer)

probability that Harry will be either the member chosen to be the secretary or the member chosen to be the treasurer = 1/10 + 1/10
=1/5

Regards,
##### General Discussion
Math Expert
Joined: 02 Sep 2009
Posts: 59125
Re: A certain club has 10 members, including Harry. One of the  [#permalink]

### Show Tags

29 Jun 2012, 02:43
5
6
SOLUTION

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 be either the member chosen to 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.

Similar questions to practice:
a-certain-team-has-12-members-including-joey-a-three-131321.html
a-medical-researcher-must-choose-one-of-14-patients-to-127396.html
a-certain-class-consists-of-8-students-including-kim-127730.html
_________________
Manager
Joined: 12 Feb 2012
Posts: 114
Re: A certain club has 10 members, including Harry. One of the  [#permalink]

### Show Tags

09 Sep 2012, 17:18
2
Hey Bunuel,

This the qay I did it:

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

But I am confused since:

P(secretary OR treasurer)=P(secretary)+P(treasurer)-P(secretary and treasurer)+ P(Neither secretary nor treasurer)

Now since the two events are mutually exclusive and Harry can neither be both secretary and treasurer, P(secretary and treasurer)=0.

P(Neither secretary nor treasurer)=(9/10)(8/9)(7/8).

What am I doing wrong?
Math Expert
Joined: 02 Sep 2009
Posts: 59125
Re: A certain club has 10 members, including Harry. One of the  [#permalink]

### Show Tags

11 Sep 2012, 02:53
1
alphabeta1234 wrote:
Hey Bunuel,

This the qay I did it:

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

But I am confused since:

P(secretary OR treasurer)=P(secretary)+P(treasurer)-P(secretary and treasurer)+ P(Neither secretary nor treasurer)

Now since the two events are mutually exclusive and Harry can neither be both secretary and treasurer, P(secretary and treasurer)=0.

P(Neither secretary nor treasurer)=(9/10)(8/9)(7/8).

What am I doing wrong?

If you want to do this way then: P(secretary or treasurer)=P(not Harry)*P(Harry)*P(any)+P(not Harry)*P(not Harry)*P(Harry)=9/10*1/9*1+9/10*8/9*1/8=2/10. In the red, you are calculating the probability that Harry will be secretary OR treasurer OR neither and we don't need neither.

Hope it's clear.
_________________
Manager
Joined: 28 Dec 2012
Posts: 96
Location: India
Concentration: Strategy, Finance
WE: Engineering (Energy and Utilities)
A certain club has 10 members, including Harry. One of the  [#permalink]

### Show Tags

06 Jan 2013, 13:49
2
1
E.

[ 9C2 * ( 3! - 2!) ] / 10C3
= 1/5
_________________
Impossibility is a relative concept!!
Manager
Joined: 09 Jan 2013
Posts: 68
Concentration: Entrepreneurship, Sustainability
GMAT 1: 650 Q45 V34
GMAT 2: 740 Q51 V39
GRE 1: Q790 V650
GPA: 3.76
WE: Other (Pharmaceuticals and Biotech)
Re: A certain club has 10 members, including Harry. One of the  [#permalink]

### Show Tags

26 Jan 2013, 22:59
sagarsingh wrote:
E.

[ 9C2 * ( 3! - 2!) ] / 10C3
= 1/5

Hi Sagar,
I tried to solve this problem in a similar way, but couldn't get it right. Please explain the significance of (3! - 2!).
Intern
Joined: 08 Feb 2013
Posts: 2
Re: A certain club has 10 members, including Harry. One of the  [#permalink]

### Show Tags

12 Feb 2013, 03:02
Can you do this with the formula Favorables/Total? Tried but cant get the right result.
Manager
Joined: 24 Sep 2012
Posts: 81
Location: United States
GMAT 1: 730 Q50 V39
GPA: 3.2
WE: Education (Education)
Re: A certain club has 10 members, including Harry. One of the  [#permalink]

### Show Tags

16 Feb 2013, 12:58
11
Favorable method:

Total options=10*9*8 // since any of the ten members can become President, any one of the nine can become secretary and any one of the remaining eight can become treasurer

Favorable options are the ones which include Harry as Treasurer or the secretary

Now consider the number of options in which harry is chosen as the secretary. Since we know that there is only one option for secretary, that position already has one person fixed. Hence, the president must be chosen out of the remaining nine people and the secretary from the remaining eight.

Hence, ways in which Harry can become secretary are 9*1*8=72

Similarly, ways in which Harry can become treasurer are 9*8*1=72

Since the probability asks for "or" we must add both the favorable outputs=72+72=144

Now, (favorable results/total results) = (144/720)=1/5

Hence the answer is E. Let me know if any other clarifications are required.

Victorlp89 wrote:
Can you do this with the formula Favorables/Total? Tried but cant get the right result.
Intern
Joined: 05 Jun 2012
Posts: 6
Re: A certain club has 10 members, including Harry. One of the  [#permalink]

### Show Tags

18 Aug 2013, 05:25
2
1
Probability = favourable outcomes / total outcomes

Total outcomes = 10*9*8=720

favourable outcome( secretary) = 9*1*8=72
favourable outcome( treasurer ) = 9*8*1=72

72/720 +72/720 =1/5

We add because of OR in the question.
Senior Manager
Joined: 10 Jul 2013
Posts: 282
Re: A certain club has 10 members, including Harry. One of the  [#permalink]

### Show Tags

18 Aug 2013, 13:59
1
1
Bunuel 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 be the treasurer. What is the probability that Harry will be either the member chosen to 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

Diagnostic Test
Question: 7
Page: 21
Difficulty: 650

President then Secretary or Treasurer

9/10 * 1/9 * 8/8 = 1/10 (when Harry is the secretary)
or
9/10 * 8/9 * 1/8 = 1/10 (when Harry is the treasurer)

So probability = 1/10 + 1/10 = 1/5 (Answer E)
_________________
Asif vai.....
Intern
Joined: 05 Feb 2014
Posts: 40
Re: A certain club has 10 members, including Harry. One of the  [#permalink]

### Show Tags

21 Jun 2014, 01:53
1
2
30 Second approach

Total = 10
President = 1
Secretary = 1
Treasurer = 1

Probability of anything other than president and the left overs = 8/10

Probability of either Secretary or treasurer = $$1-\frac{8}{10}=\frac{1}{5}$$
Intern
Joined: 05 Sep 2014
Posts: 6
Re: A certain club has 10 members, including Harry. One of the  [#permalink]

### Show Tags

31 Oct 2014, 15:47
Hi Bunnel,

How can we just add 1/10+1/10, when the question clearly states "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"

I did it this way

9/10*1/9*8/8 + 9/10*8/9*1/8 = 1/10 + 1/10 = 1/5.

Just trying to understand the logic in your solution.

Cheers,
Anie
Math Expert
Joined: 02 Sep 2009
Posts: 59125
Re: A certain club has 10 members, including Harry. One of the  [#permalink]

### Show Tags

01 Nov 2014, 05:06
annie2014 wrote:
Hi Bunnel,

How can we just add 1/10+1/10, when the question clearly states "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"

I did it this way

9/10*1/9*8/8 + 9/10*8/9*1/8 = 1/10 + 1/10 = 1/5.

Just trying to understand the logic in your solution.

Cheers,
Anie

Yes, we can that simply add 1/10 and 1/10.

What is the probability that Harry will be either the member chosen to be the secretary or the member chosen to be the treasurer? So, what is the probability that Harry will be either 2nd or 3rd in the row? The probability of each is 1/10, so P = 1/10 + 1/10.

Hope it helps.
_________________
Senior Manager
Status: Verbal Forum Moderator
Joined: 17 Apr 2013
Posts: 447
Location: India
GMAT 1: 710 Q50 V36
GMAT 2: 750 Q51 V41
GMAT 3: 790 Q51 V49
GPA: 3.3
Re: A certain club has 10 members, including Harry. One of the  [#permalink]

### Show Tags

02 Nov 2014, 04:09
Bunuel wrote:
SOLUTION

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 be either the member chosen to 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.

Similar questions to practice:
a-certain-team-has-12-members-including-joey-a-three-131321.html
a-medical-researcher-must-choose-one-of-14-patients-to-127396.html
a-certain-class-consists-of-8-students-including-kim-127730.html

So Bunuel you are saying that events are not defined, I mean to say that we can't assume that First president is selected, then Secretary and then Treasurer. Because if the events are imposed then the total number of available person will decrease in each event, because it is nowhere mentioned that one person can hold more than position. Please reply so that I can get rid of my doubts. Thanks!
_________________
Like my post Send me a Kudos It is a Good manner.
My Debrief: http://gmatclub.com/forum/how-to-score-750-and-750-i-moved-from-710-to-189016.html
Math Expert
Joined: 02 Sep 2009
Posts: 59125
Re: A certain club has 10 members, including Harry. One of the  [#permalink]

### Show Tags

02 Nov 2014, 07:30
1
honchos wrote:
Bunuel wrote:
SOLUTION

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 be either the member chosen to 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.

Similar questions to practice:
a-certain-team-has-12-members-including-joey-a-three-131321.html
a-medical-researcher-must-choose-one-of-14-patients-to-127396.html
a-certain-class-consists-of-8-students-including-kim-127730.html

So Bunuel you are saying that events are not defined, I mean to say that we can't assume that First president is selected, then Secretary and then Treasurer. Because if the events are imposed then the total number of available person will decrease in each event, because it is nowhere mentioned that one person can hold more than position. Please reply so that I can get rid of my doubts. Thanks!

No. The sequence is the way it's given. But it does not matter, whether we select president first and secretary second or vise-versa.
_________________
Intern
Joined: 21 Apr 2014
Posts: 39
Re: A certain club has 10 members, including Harry. One of the  [#permalink]

### Show Tags

26 Jan 2015, 04:56
5
1
It wants to know the probability that Harry is Secretary or treasurer, so we should add the probability that he will be chosen secretary to the probability that he will be chosen treasurer.

The Probability that he is chosen secretary is
9/10*1/9*8/8=1/10

The 9/10 represents the probability that anyone but harry is president
The 1/9 represents the probability that harry is secretary
The 8/8 represents the fact that anyone can be treasurer, so it really does not affect the probability at all

The probability that he is chosen treasurer is
9/10*8/9*1/8=1/10

The 9/10 represents the probability that anyone but harry is president
The 1/9 represents the probability anyone but harry is secretary
The 1/8 represents the probability that harry is treasurer

Add the two together and you get 1/5 (E)
_________________
Eliza
GMAT Tutor
bestgmatprepcourse.com
Intern
Joined: 14 Oct 2013
Posts: 43
Re: A certain club has 10 members, including Harry. One of the  [#permalink]

### Show Tags

25 Apr 2015, 20:34
1
2
I didn't see this way posted so figured I would add it here, this way was fastest for me...

P(H is secretary or treasurer)= 1- [P(H is pres)+P(H not chosen for a role at all)]
=1- [1/10 + 7/10]
=1-[8/10] = 2/10=1/5
Intern
Joined: 13 Sep 2015
Posts: 16
A certain club has 10 members, including Harry. One of the  [#permalink]

### Show Tags

26 Sep 2015, 13:12
5
For an easy way to look at it and not be confused with how the probability can be 1/10 in each instance, think of it this way:

The question is asking the probability of either one or the other happening.

You have to find the probability that Harry will either be chosen as Secretary or treasurer.

Case 1 (Prob of Harry being chosen as Secretary)= Probability that anyone except Harry will be Selected as President (9/10) * Probability that Harry will be chosen as Secretary (1/9) * Probability that any of the others will be selected as treasurer (8/8)

Which is = 9/10 * 1/9 * 8/8 = 1/10

Case 2 (Probability that Harry will be selected as Treasurer) = Probability that anyone except Harry will be chosen as president (9/10) * Probability that anyone except Harry will be chosen as Secretary (8/9) * Probability that Harry will be chosen as Treasurer (8/8)

= 9/10 * 8/9 * 1/8 = 1/10

Now to find the answer, add the two probabilities (as we do in either or cases) to find the probability that Harry will either be chosen Secretary or Treasurer = 1/10 + 1/10 = 2/10 = 1/5

A certain club has 10 members, including Harry. One of the   [#permalink] 26 Sep 2015, 13:12

Go to page    1   2    Next  [ 38 posts ]

Display posts from previous: Sort by