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605-655 (Medium)|   Probability|                        
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Bunuel
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A certain club has 10 members, including Harry. One of the 25 Jun 2012, 02:31
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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
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E
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A certain club has 10 members, including Harry. One of the Updated on: 03 Jul 2026, 10:07
Originally posted by Bunuel on 25 Jun 2012, 02:31.
Last edited by Bunuel on 03 Jul 2026, 10:07, edited 1 time in total.
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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 an equal chance of being selected for any of the positions. The sequence of 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 treasurer is 1/10 + 1/10 = 2/10 = 1/5.

Answer: E.

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A certain club has 10 members, including Harry. One of the 25 Jun 2012, 04:35
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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

Answer (E),

Regards,
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A certain club has 10 members, including Harry. One of the 16 Feb 2013, 12:58
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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.
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A certain club has 10 members, including Harry. One of the 09 Sep 2012, 17:18
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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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A certain club has 10 members, including Harry. One of the 11 Sep 2012, 02:53
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alphabeta1234
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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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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A certain club has 10 members, including Harry. One of the 18 Aug 2013, 05:25
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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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A certain club has 10 members, including Harry. One of the 21 Jun 2014, 01:53
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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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A certain club has 10 members, including Harry. One of the 31 Oct 2014, 15:47
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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
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A certain club has 10 members, including Harry. One of the 01 Nov 2014, 05:06
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annie2014
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
Show more

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.

If you still have some doubts, please follow the links to the similar questions given above.

Hope it helps.
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A certain club has 10 members, including Harry. One of the 26 Jan 2015, 04:56
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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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A certain club has 10 members, including Harry. One of the 25 Apr 2015, 20:34
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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
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A certain club has 10 members, including Harry. One of the 26 Sep 2015, 13:12
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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

Answer: E
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A certain club has 10 members, including Harry. One of the 19 Dec 2015, 15:37
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[quote="Bunuel"]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

There are a lot of explanations as to how solve this problem, however many answers are overly complicated. Here is the simple way:

President= 1/10
Not President= 1-1/10=9/10

Secretary= 1/9
Not Secretary= 1-1/9=8/9

Treasurer= 1/8
Not Treasurer= 1-1/8= 7/8

Thus,
1. NOT President but Secretary = 9/10*1/9=1/10
2. NOT President and NOT Secretary but Treasurer= 9/10*8/9*1/8=1/10
3. Either Secretary or Treasurer= 1/10+1/10=2/10=1/5


A.
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A certain club has 10 members, including Harry. One of the Updated on: 01 Jun 2017, 11:07
Originally posted by mcelroytutoring on 28 Mar 2016, 17:53.
Last edited by mcelroytutoring on 01 Jun 2017, 11:07, edited 2 times in total.
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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 also 1/10 (and not 1/9).
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A certain club has 10 members, including Harry. One of the 19 Oct 2016, 06:00
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Bunuel
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
Show more

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.

Answer: E
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A certain club has 10 members, including Harry. One of the 31 May 2017, 22:01
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Bunuel
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
Show more
1.The final probability is probability of Harry as secretary + probability of Harry as treasurer
2.One is selected as a president. Harry could be the president , the probability being 9/10. So the probability of Harry not as a president is 9/10.
3.The probability of Harry as a Secretary is 9/10*1/9= 1/10
4. Similarly the probability of Harry as a treasurer is 8/10*1/8=1/10
5. Final Probability is 1/10+1/10=1/5­
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A certain club has 10 members, including Harry. One of the 14 Jun 2020, 03:54
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Dear IanStewart GMATGuruNY VeritasKarishma,

Combinatorics can't be used to solve this question because we deal with RANKS/ORDER (i.e. secretary and treasurer), right?

I hesitate because I saw similar question in which Bunuel used Combinatorics to solve the problem.(https://gmatclub.com/forum/a-certain-cl ... 27730.html)
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A certain club has 10 members, including Harry. One of the 14 Jun 2020, 04:42
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varotkorn

Combinatorics can't be used to solve this question because we deal with RANKS/ORDER (i.e. secretary and treasurer), right?

I hesitate because I saw similar question in which Bunuel used Combinatorics to solve the problem.(https://gmatclub.com/forum/a-certain-cl ... 27730.html)
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I'm not sure what you mean when you say "use Combinatorics" to solve a problem. Combinatorics is just the branch of math that deals with counting things. Every solution to problems of this type are using combinatorics in one way or another.

Most math problems can be solved in several ways, this one included, and the problem in this thread and the problem in your link are conceptually identical, so any method that works for one of the two problems will work for the other.
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A certain club has 10 members, including Harry. One of the 14 Jun 2020, 04:46
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Dear IanStewart,

I meant using Combination (choosing a subgroup from a larger group) formula as quoted below:
Bunuel

One can also do: \(P=\frac{C^1_1*C^2_7}{C^3_8}=\frac{3}{8}\), where \(C^1_1\) is # of ways to select Kim, \(C^2_7\) is # of ways to select any 2 students out of 7 left and \(C^3_8\) is total # of ways to select 3 students from 8;

Or: you can find the probability that among 3 students selected to complete the tasks there won't be Kim and subtract it from 1:
\(P=1-\frac{C^3_7}{C^3_8}=\frac{3}{8}\);
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I tried and failed to apply the above method to this question :(

BTW, I know that permutation works for this problem. But not sure about the combination one.
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