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Re: A certain club has 10 members, including Harry. One of the
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11 Oct 2015, 05:38
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.



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Re: A certain club has 10 members, including Harry. One of the
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11 Oct 2015, 18:24
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.



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Re: A certain club has 10 members, including Harry. One of the
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19 Dec 2015, 14:37
[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.
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A certain club has 10 members, including Harry. One of the
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Updated on: 01 Jun 2017, 10:07
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).
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A certain club has 10 members, including Harry. One of the
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17 May 2016, 03:35
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!



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Re: A certain club has 10 members, including Harry. One of the
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14 Jun 2016, 09:57
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!!



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A certain club has 10 members, including Harry. One of the
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22 Jun 2016, 08:47
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.



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Re: A certain club has 10 members, including Harry. One of the
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17 Oct 2016, 19:48
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?



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Re: A certain club has 10 members, including Harry. One of the
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19 Oct 2016, 05:00
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
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Re: A certain club has 10 members, including Harry. One of the
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17 Nov 2016, 16:17
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



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Re: A certain club has 10 members, including Harry. One of the
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31 May 2017, 21:01
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
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Re: A certain club has 10 members, including Harry. One of the
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09 Jul 2017, 09:56
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



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A certain club has 10 members, including Harry. One of the
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26 May 2018, 07:28
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



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A certain club has 10 members, including Harry. One of the
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26 May 2018, 07:42
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!
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Re: A certain club has 10 members, including Harry. One of the
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05 Jul 2018, 10:04
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




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