Author 
Message 
TAGS:

Hide Tags

Manager
Joined: 13 Apr 2015
Posts: 75
Concentration: General Management, Strategy
GPA: 3.25
WE: Project Management (Energy and Utilities)

Re: A certain club has 10 members, including Harry. One of the
[#permalink]
Show Tags
11 Oct 2015, 06:38
Is 1  1/10 + 7/10 = 1/5 a right aproach. 1/10 is his prob of becoming president and 7/10 is his prob of becoming nothing.
Suggestion would be of a great help.



Math Expert
Joined: 02 Sep 2009
Posts: 49303

Re: A certain club has 10 members, including Harry. One of the
[#permalink]
Show Tags
11 Oct 2015, 06:43



Manager
Joined: 13 Apr 2015
Posts: 75
Concentration: General Management, Strategy
GPA: 3.25
WE: Project Management (Energy and Utilities)

Re: A certain club has 10 members, including Harry. One of the
[#permalink]
Show Tags
11 Oct 2015, 19:24
Bunuel wrote: goldfinchmonster wrote: Is 1  1/10 + 7/10 = 1/5 a right aproach. 1/10 is his prob of becoming president and 7/10 is his prob of becoming nothing.
Suggestion would be of a great help. 1  1/10 + 7/10 = 16/10 > 1 not 1/5. So, this approach is not correct. Several correct approaches and links to similar questions are given on page 1. Hey extreamly sorry, i missed out the bracket. It should be 1 [ 1/10 + 7/10 ] = 1/5.



Intern
Joined: 05 Mar 2015
Posts: 32
Location: United States
Concentration: Marketing, General Management

Re: A certain club has 10 members, including Harry. One of the
[#permalink]
Show Tags
19 Dec 2015, 15:37
[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= 11/10=9/10 Secretary= 1/9 Not Secretary= 11/9=8/9 Treasurer= 1/8 Not Treasurer= 11/8= 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.
_________________
"You have to learn the rules of the game. And then you have to play better than anyone else". Albert Einstein



Director
Status: Professional GMAT Tutor
Affiliations: AB, cum laude, Harvard University (Class of '02)
Joined: 10 Jul 2015
Posts: 669
Location: United States (CA)
Age: 38
GMAT 1: 770 Q47 V48 GMAT 2: 730 Q44 V47 GMAT 3: 750 Q50 V42
WE: Education (Education)

A certain club has 10 members, including Harry. One of the
[#permalink]
Show Tags
Updated on: 01 Jun 2017, 11:07
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).
Attachments
Screen Shot 20160328 at 5.51.47 PM.png [ 93.81 KiB  Viewed 1990 times ]
_________________
Harvard grad and 99% GMAT scorer, offering expert, private GMAT tutoring and coaching, both inperson (San Diego, CA, USA) and online worldwide, since 2002.
One of the only known humans to have taken the GMAT 5 times and scored in the 700s every time (700, 710, 730, 750, 770), including verified section scores of Q50 / V47, as well as personal bests of 8/8 IR (2 times), 6/6 AWA (4 times), 50/51Q and 48/51V (1 question wrong).
You can download my official testtaker score report (all scores within the last 5 years) directly from the Pearson Vue website: https://tinyurl.com/y94hlarr Date of Birth: 09 December 1979.
GMAT Action Plan and Free EBook  McElroy Tutoring
Contact: mcelroy@post.harvard.edu



Manager
Status: 2 months to go
Joined: 11 Oct 2015
Posts: 121
GPA: 3.8

A certain club has 10 members, including Harry. One of the
[#permalink]
Show Tags
17 May 2016, 04:35
I admit that probabilities are my Achilles' heel, but I wanted to ask something to clear my mind. I obviously love Bunuel's answer because of its simplicity but I'm still puzzled by it. I don't get why there's no sequence, if we have 10 balls and then extract 1, unless we put it back in the group the probability should vary. Can someone explain why it doesn't vary (maybe with an example ) ? Thanks a lot!



Intern
Joined: 21 Mar 2013
Posts: 12

Re: A certain club has 10 members, including Harry. One of the
[#permalink]
Show Tags
14 Jun 2016, 10:57
Hi All, I began my GMAT prep just two days ago. So forgive me if this looks a bit too obvious. I have been focusing on Combinatorics and Probability since last two days. While this question looks straightforward to many of you here, the way the OG explained it, it took me on a spin. So my question is two folds: P(Harry is Pres) = 1/10. Therefore him not being Pres = (11/10) = 9/10 P(Harry is Sec) = P(Harry NOT being Pres) * P(Harry is Sec) * P(Harry is NOT Tres) = 9/10 * 8/9 * 8/8 = 1/10 > (Equation 1) Thus P(Harry is Sec) = 1/10 > (Equation 2) P(Harry is Tres) = P(Harry NOT Pres) *P (Harry NOT Sec) * P(Harry is Tres) = 9/10/ * 8/9 * 1/8 = 1/10 > (Equation 3) Q:1 How is P(Harry is Sec) calculated with P(Harry is Sec) being part of the equation??? See the highlights in (Equation 1). It is a case of a circular reference?? Q:2 Just like how we calculated P(Harry NOT Pres) = 1  P(Harry is Pres), why aren't we plugging in the value of [1P(Harry is Sec)] = 11/10 = 9/10 in the (Equation 3) where it calls for P(Harry NOT Sec) and instead taking 8/9? I am beyond lost on this topic. The more I read the deeper I dig my own grave. Thanks!!



Intern
Joined: 10 Aug 2015
Posts: 32
Location: India
GPA: 3.5
WE: Consulting (Computer Software)

A certain club has 10 members, including Harry. One of the
[#permalink]
Show Tags
22 Jun 2016, 09:47
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
Okay let me give a very simple solution to the question. Counting chapter is one of the part of our exam so lets just use counting to solve the problem rather than listing out all the probabilities. Step1 We know here that order matters as Harry as president is very different from Harry as Secretary. So total no of ways we can arrange 3 people from a group of 10 = 10P3 Step2  Here we have two cases Harry as Sec. or Harry as Tres.  Mark the word "OR" so we have to add the cases. Now fix Harry as Sec. then we can arrange rest 2 persons from a group of 9 in 9P2 ways and similarly keeping Harry as Tres. we can get 9P2 ways. So P(H as Sec or Tres)= 2*9P2/10P3 = 1/5.



Intern
Joined: 21 Jan 2015
Posts: 39
Location: United States
GPA: 3.2

Re: A certain club has 10 members, including Harry. One of the
[#permalink]
Show Tags
17 Oct 2016, 20:48
I understood how the answer is E, but I am also getting confused now. The probability of harry not being president is (9/10). The probability of harry being secretary is (1/9). Wouldn't the probability of harry not being treasurer be (7/8) then? I know if I did it this way, there is no answer choice that would match this, but I still want to know why the last probability is not 7/8?



Target Test Prep Representative
Status: Head GMAT Instructor
Affiliations: Target Test Prep
Joined: 04 Mar 2011
Posts: 2835

Re: A certain club has 10 members, including Harry. One of the
[#permalink]
Show Tags
19 Oct 2016, 06:00
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. Answer: E
_________________
Jeffery Miller
Head of GMAT Instruction
GMAT Quant SelfStudy Course
500+ lessons 3000+ practice problems 800+ HD solutions



Current Student
Status: DONE!
Joined: 05 Sep 2016
Posts: 390

Re: A certain club has 10 members, including Harry. One of the
[#permalink]
Show Tags
17 Nov 2016, 17:17
Probability Secretary > Prob(Not President)xProb(Secretary) > (9/10)x(1/9) = 9/90 = 1/10
Probability Treasurer > Prob(Not President)x(Prob(Not Secretary)xProb(Treasurer)=(9/10)x(8/9)x(1/8) = 72/720 = 1/10
Prob (Secretary or Treasurer) = 1/10 + 1/10 = 2/10 = 1/5



Director
Joined: 17 Dec 2012
Posts: 636
Location: India

Re: A certain club has 10 members, including Harry. One of the
[#permalink]
Show Tags
31 May 2017, 22:01
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 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
_________________
Srinivasan Vaidyaraman Sravna Holistic Solutions http://www.sravnatestprep.com
Holistic and Systematic Approach



Intern
Joined: 26 Nov 2014
Posts: 7

Re: A certain club has 10 members, including Harry. One of the
[#permalink]
Show Tags
09 Jul 2017, 10:56
For those looking for formula:
Total number of ways of selecting 3 people out of 10=10C3*3!(multiplied by 3! since these 3 positions are different)
Harry's chances that he would be chosen either as Sec or treas= Don't chose harry as prez(101C1)*Chose Harry as Sec(1C1)*Any one to be chosen Treas(8c1)/Total number of ways of selecting 3 people out of 10(10C3*3!) OR (+) Don't chose harry as prez(101C1)*Chose Harry as Treasurer(1C1)*Any one to be chosen Secretary(8c1)/Total number of ways of selecting 3 people out of 10(10C3*3!)
=(9C1*1C1*8C1+9C1*1C1*8C1)/10C3*3!=144/720=1/5



Director
Joined: 09 Mar 2016
Posts: 878

A certain club has 10 members, including Harry. One of the
[#permalink]
Show Tags
26 May 2018, 08:28
elizaanne wrote: 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) Hey pushpitkc the above solution is nicely explained, just one thing i dont understand in this solution as we know the probability formula is # of possible outcomes / total number of outcomes but i dont see the pattern of formula in the above solution why ? Also it says "we should add the probability" but arent these dependent events ? here is my approach below, can yu please explain why it is wrong ? Since the first probability out of 10 is probability of choosing president The second probability is choosing harry as secretary 1/9 The third probability is choosing harry as treasurer 1/8 1/9*1/8 = 1/72 thank you and have a great weekend
_________________
In English I speak with a dictionary, and with people I am shy.



BSchool Forum Moderator
Joined: 26 Feb 2016
Posts: 3136
Location: India
GPA: 3.12

A certain club has 10 members, including Harry. One of the
[#permalink]
Show Tags
26 May 2018, 08:42
dave13 wrote: elizaanne wrote: 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) Hey pushpitkc the above solution is nicely explained, just one thing i dont understand in this solution as we know the probability formula is # of possible outcomes / total number of outcomes but i dont see the pattern of formula in the above solution why ? Also it says "we should add the probability" but arent these dependent events ? here is my approach below, can yu please explain why it is wrong ? Since the first probability out of 10 is probability of choosing president The second probability is choosing harry as secretary 1/9 The third probability is choosing harry as treasurer 1/8 1/9*1/8 = 1/72 thank you and have a great weekend Hey dave13The reason we need to add the probabilities is either of two things is possible 1. Harry is chosen as the Secretary 2. Harry is chosen as the Treasurer Total probability that Harry is chosen Secretary or Treasurer is the sum of the individual probabilities. We multiply when we are asked to find about both the events. Case 1: P(Harry is chosen Secretary) = \(\frac{9}{10}*\frac{1}{9}*\frac{8}{8} = \frac{1}{10}\) Case 2: P(Harry is chosen Treasurer) = \(\frac{9}{10}*\frac{8}{9}*\frac{1}{8} = \frac{1}{10}\) In either cases, you will observe that there are 9 possibilities of choosing President as Harry can't be chosen and 8 possibilities of choosing either Secretary/Treasurer as Harry can't be chosen as Treasurer is the first case and can't be chosen Secretary in the second case. P(Harry is chosen Secretary OR Treasurer) = \(\frac{1}{10} + \frac{1}{10} =\frac{1}{5}\) Hope this helps you!
_________________
You've got what it takes, but it will take everything you've got



Intern
Joined: 07 Feb 2017
Posts: 13

Re: A certain club has 10 members, including Harry. One of the
[#permalink]
Show Tags
05 Jul 2018, 11:04
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 Approach using Permutations: No. of ways to be chosen as secretary : 9P1*1*8P1 = 9*8 No. of ways to be chosen as treasurer : 9P1*8P1*1 = 9*8 No. of ways of choosing three people out of 10 : 10P3 = 10*9*8 Probability = [9*8+9*8]/[10*9*8] = 1/5




Re: A certain club has 10 members, including Harry. One of the &nbs
[#permalink]
05 Jul 2018, 11:04



Go to page
Previous
1 2
[ 36 posts ]



