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

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

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25 Jun 2012, 01:31
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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 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
[Reveal] Spoiler: OA

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

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25 Jun 2012, 03:35
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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

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

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25 Jun 2012, 01:31
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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
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Re: A certain club has 10 members, including Harry. One of the [#permalink]

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16 Feb 2013, 11:58
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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.
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Re: A certain club has 10 members, including Harry. One of the [#permalink]

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29 Jun 2012, 01:43
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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
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A certain club has 10 members, including Harry. One of the [#permalink]

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06 Jan 2013, 12:49
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E.

[ 9C2 * ( 3! - 2!) ] / 10C3
= 1/5
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A certain club has 10 members, including Harry. One of the [#permalink]

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28 Mar 2016, 16:53
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Here is a visual that should help. Notice that the question does not indicate whether Harry was chosen as president, thus his chances of becoming secretary are 1/10 and not 1/9.
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Screen Shot 2016-03-28 at 5.51.47 PM.png [ 93.81 KiB | Viewed 618 times ]

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

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11 Sep 2012, 01:53
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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.
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Re: A certain club has 10 members, including Harry. One of the [#permalink]

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21 Jun 2014, 00:53
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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 = $$1-\frac{8}{10}=\frac{1}{5}$$
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Re: A certain club has 10 members, including Harry. One of the [#permalink]

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26 Jan 2015, 03:56
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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)
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Re: A certain club has 10 members, including Harry. One of the [#permalink]

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19 Oct 2016, 05:00
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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

We are given that a club has 10 members, including Harry. When selecting a president, secretary, and treasurer from the 10 members, we must determine the probability that Harry will either be chosen secretary or treasurer.

Since we have 10 total people the probability that Harry is chosen to be the secretary is 1/10 and the probability that he is chosen to be the treasurer is 1/10.

Thus, the probability that he is chosen to be the secretary or treasurer is 1/10 +1/10 = 1/5.

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

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09 Sep 2012, 16:18
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?
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Re: A certain club has 10 members, including Harry. One of the [#permalink]

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26 Jan 2013, 21:59
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!).
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Re: A certain club has 10 members, including Harry. One of the [#permalink]

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12 Feb 2013, 02:02
Can you do this with the formula Favorables/Total? Tried but cant get the right result.
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Re: A certain club has 10 members, including Harry. One of the [#permalink]

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18 Aug 2013, 04:25
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.
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Re: A certain club has 10 members, including Harry. One of the [#permalink]

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18 Aug 2013, 12:59
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)
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Re: A certain club has 10 members, including Harry. One of the [#permalink]

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31 Oct 2014, 14:47
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,
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Re: A certain club has 10 members, including Harry. One of the [#permalink]

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01 Nov 2014, 04:06
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.
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Re: A certain club has 10 members, including Harry. One of the [#permalink]

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02 Nov 2014, 03:09
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!
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Re: A certain club has 10 members, including Harry. One of the [#permalink]

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02 Nov 2014, 06:30
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.
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Re: A certain club has 10 members, including Harry. One of the   [#permalink] 02 Nov 2014, 06:30

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