Author 
Message 
TAGS:

Hide Tags

Intern
Joined: 06 Mar 2015
Posts: 27

A committee of 3 persons is to be formed from 3 company secretaries,
[#permalink]
Show Tags
23 Jun 2016, 02:50
Question Stats:
70% (01:19) correct 30% (00:57) wrong based on 73 sessions
HideShow timer Statistics
A committee of 3 persons is to be formed from 3 company secretaries, 4 economists and 1 chartered accountant. What is the probability that each of the three professions is represented in the committee? A. 1/28 B. 3/28 C. 1/14 D. 4/28 E. 3/14 Can anyone solve it using probability method. I am only able to crack it by using combinatorial method.
Official Answer and Stats are available only to registered users. Register/ Login.





Intern
Joined: 06 Mar 2015
Posts: 27

Re: A committee of 3 persons is to be formed from 3 company secretaries,
[#permalink]
Show Tags
23 Jun 2016, 02:57
nishi999 wrote: A committee of 3 persons is to be formed from 3 company secretaries, 4 economists and 1 chartered accountant. What is the probability that each of the three professions is represented in the committee?
A. 1/28 B. 3/28 C. 1/14 D. 4/28 E. 3/14
Can anyone solve it using probability method. I am only able to crack it by using combinatorial method. A. Combinatorial Method  No of ways of choosing 1 person out of 3, 1 out of 4 and 1 out of 1, divided by ways of choosing 3 people out of 10. 3C1*4C1*1C1/8C3 = 3/14 B. Probability Method  I can't understand as to what am i doing wrong. I just can't get the answer. The probability of choosing 1 company secretary is 3/8. The probability of choosing 1 economist out of is 4/7. The probability of choosing 1 chartered accountant is 1/6. The probability that each of the 3 professions are represented in the committee are  3/8*4/7*1/6 = 1/28 Don't know where am i going wrong in this approach?? Help needed Thanks



Math Expert
Joined: 02 Sep 2009
Posts: 47898

Re: A committee of 3 persons is to be formed from 3 company secretaries,
[#permalink]
Show Tags
23 Jun 2016, 02:58
nishi999 wrote: A committee of 3 persons is to be formed from 3 company secretaries, 4 economists and 1 chartered accountant. What is the probability that each of the three professions is represented in the committee?
A. 1/28 B. 3/28 C. 1/14 D. 4/28 E. 3/14
Can anyone solve it using probability method. I am only able to crack it by using combinatorial method. 3/8*4/7*1/6*3! = 3/14. We are multiplying by 3! because we can choose secretary, economist, and accountant is several way: SEA, SAE, ESA, EAS, ASE, AES. Answer: E.
_________________
New to the Math Forum? Please read this: Ultimate GMAT Quantitative Megathread  All You Need for Quant  PLEASE READ AND FOLLOW: 12 Rules for Posting!!! Resources: GMAT Math Book  Triangles  Polygons  Coordinate Geometry  Factorials  Circles  Number Theory  Remainders; 8. Overlapping Sets  PDF of Math Book; 10. Remainders  GMAT Prep Software Analysis  SEVEN SAMURAI OF 2012 (BEST DISCUSSIONS)  Tricky questions from previous years.
Collection of Questions: PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.
What are GMAT Club Tests? Extrahard Quant Tests with Brilliant Analytics



Intern
Joined: 06 Mar 2015
Posts: 27

Re: A committee of 3 persons is to be formed from 3 company secretaries,
[#permalink]
Show Tags
23 Jun 2016, 03:11
Bunuel wrote: nishi999 wrote: A committee of 3 persons is to be formed from 3 company secretaries, 4 economists and 1 chartered accountant. What is the probability that each of the three professions is represented in the committee?
A. 1/28 B. 3/28 C. 1/14 D. 4/28 E. 3/14
Can anyone solve it using probability method. I am only able to crack it by using combinatorial method. 3/8*4/7*1/6*3! = 3/14. We are multiplying by 3! because we can choose secretary, economist, and accountant is several way: SEA, SAE, ESA, EAS, ASE, AES. Answer: E. Hi Bunuel, The chapter on probability in the GMAT MATH BOOK states as follows  Example #1 Q: There are 8 employees including Bob and Rachel. If 2 employees are to be randomly chosen to form a committee, what is the probability that the committee includes both Bob and Rachel? Solution: probability approach: The probability of choosing Bob or Rachel as a first person in committee is 2/8. The probability of choosing Rachel or Bob as a second person when first person is already chosen is 1/7. The probability that the committee includes both Bob and Rachel is. P = 2/8*1/7 = 1/28 Shouldn't the above answer be multiplied by 2! as well?



Intern
Joined: 06 Mar 2015
Posts: 27

Re: A committee of 3 persons is to be formed from 3 company secretaries,
[#permalink]
Show Tags
23 Jun 2016, 03:25
Bunuel wrote: nishi999 wrote: A committee of 3 persons is to be formed from 3 company secretaries, 4 economists and 1 chartered accountant. What is the probability that each of the three professions is represented in the committee?
A. 1/28 B. 3/28 C. 1/14 D. 4/28 E. 3/14
Can anyone solve it using probability method. I am only able to crack it by using combinatorial method. 3/8*4/7*1/6*3! = 3/14. We are multiplying by 3! because we can choose secretary, economist, and accountant is several way: SEA, SAE, ESA, EAS, ASE, AES. Answer: E. Hi Bunuel, As per my rudimentary knowledge of probability, i understand that in the above question we are trying to find the probability by considering  Favourable Outcomes/Total Outcomes. Then in such a case, why do we need to consider how the secretary, economist, and accountant are being chosen i.e. No of ways. If you could shed some light on that? Thanks



Math Expert
Joined: 02 Sep 2009
Posts: 47898

Re: A committee of 3 persons is to be formed from 3 company secretaries,
[#permalink]
Show Tags
23 Jun 2016, 03:29
nishi999 wrote: Bunuel wrote: nishi999 wrote: A committee of 3 persons is to be formed from 3 company secretaries, 4 economists and 1 chartered accountant. What is the probability that each of the three professions is represented in the committee?
A. 1/28 B. 3/28 C. 1/14 D. 4/28 E. 3/14
Can anyone solve it using probability method. I am only able to crack it by using combinatorial method. 3/8*4/7*1/6*3! = 3/14. We are multiplying by 3! because we can choose secretary, economist, and accountant is several way: SEA, SAE, ESA, EAS, ASE, AES. Answer: E. Hi Bunuel, The chapter on probability in the GMAT MATH BOOK states as follows  Example #1 Q: There are 8 employees including Bob and Rachel. If 2 employees are to be randomly chosen to form a committee, what is the probability that the committee includes both Bob and Rachel? Solution: probability approach: The probability of choosing Bob or Rachel as a first person in committee is 2/8. The probability of choosing Rachel or Bob as a second person when first person is already chosen is 1/7. The probability that the committee includes both Bob and Rachel is. P = 2/8*1/7 = 1/28 Shouldn't the above answer be multiplied by 2! as well? No, in this solution we already have multiplication by 2. It's 1/8*1/7*2. The difference is that the solution says that "the probability of choosing Bob or Rachel as a first person in committee is 2/8", so this takes care of x2. Hope it's clear.
_________________
New to the Math Forum? Please read this: Ultimate GMAT Quantitative Megathread  All You Need for Quant  PLEASE READ AND FOLLOW: 12 Rules for Posting!!! Resources: GMAT Math Book  Triangles  Polygons  Coordinate Geometry  Factorials  Circles  Number Theory  Remainders; 8. Overlapping Sets  PDF of Math Book; 10. Remainders  GMAT Prep Software Analysis  SEVEN SAMURAI OF 2012 (BEST DISCUSSIONS)  Tricky questions from previous years.
Collection of Questions: PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.
What are GMAT Club Tests? Extrahard Quant Tests with Brilliant Analytics



Intern
Joined: 06 Mar 2015
Posts: 27

Re: A committee of 3 persons is to be formed from 3 company secretaries,
[#permalink]
Show Tags
23 Jun 2016, 03:44
BunuelHi, I am still a bit confused. In the above solution, I assumed that 2/8 is considered, because there are 2 favourable outcomes i.e. Bob or Rachel can be chosen from 8. After that is done, then only 1 favourable outcome remains i.e. choosing 1 out of remaining seven via 1/7. Hence total probability is 2/8*1/7 = 1/28. But as per what i understand from you, it should be 1/8*1/7*2!. (2! as they can both be arranged in 2 ways) In such a case, when taking 1/8 we are only considering 1 favourable outcome out of eight outcomes. But then it should be 2/8 for the reasons mentioned above i.e. Both Rachel and bob are favourable and only after 1 of them is chosen, are we left with only 1 favourable option. Could you kindly confirm where am i going wrong? Secondly, can i take it as a thumb rule, that after calculating probability, i will need to always consider ways of arrangement, if applicable?



Math Expert
Joined: 02 Sep 2009
Posts: 47898

Re: A committee of 3 persons is to be formed from 3 company secretaries,
[#permalink]
Show Tags
23 Jun 2016, 03:49
nishi999 wrote: BunuelHi, I am still a bit confused. In the above solution, I assumed that 2/8 is considered, because there are 2 favourable outcomes i.e. Bob or Rachel can be chosen from 8. After that is done, then only 1 favourable outcome remains i.e. choosing 1 out of remaining seven via 1/7. Hence total probability is 2/8*1/7 = 1/28. But as per what i understand from you, it should be 1/8*1/7*2!. (2! as they can both be arranged in 2 ways) In such a case, when taking 1/8 we are only considering 1 favourable outcome out of eight outcomes. But then it should be 2/8 for the reasons mentioned above i.e. Both Rachel and bob are favourable and only after 1 of them is chosen, are we left with only 1 favourable option. Could you kindly confirm where am i going wrong? Secondly, can i take it as a thumb rule, that after calculating probability, i will need to always consider ways of arrangement, if applicable? Let me try again. The logic is 100% the same as with the original question. We want both B and R. There is one B and one R. Choosing B first and then R: 1/8*1/7. Choosing R first and then B: 1/8*1/7. The sum = 2*1/8*1/7. Or the way we did in original question: 1/8*1/7*2!.
_________________
New to the Math Forum? Please read this: Ultimate GMAT Quantitative Megathread  All You Need for Quant  PLEASE READ AND FOLLOW: 12 Rules for Posting!!! Resources: GMAT Math Book  Triangles  Polygons  Coordinate Geometry  Factorials  Circles  Number Theory  Remainders; 8. Overlapping Sets  PDF of Math Book; 10. Remainders  GMAT Prep Software Analysis  SEVEN SAMURAI OF 2012 (BEST DISCUSSIONS)  Tricky questions from previous years.
Collection of Questions: PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.
What are GMAT Club Tests? Extrahard Quant Tests with Brilliant Analytics



NonHuman User
Joined: 09 Sep 2013
Posts: 7696

Re: A committee of 3 persons is to be formed from 3 company secretaries,
[#permalink]
Show Tags
06 Dec 2017, 07:58
Hello from the GMAT Club BumpBot! Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up  doing my job. I think you may find it valuable (esp those replies with Kudos). Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.
_________________
GMAT Books  GMAT Club Tests  Best Prices on GMAT Courses  GMAT Mobile App  Math Resources  Verbal Resources




Re: A committee of 3 persons is to be formed from 3 company secretaries, &nbs
[#permalink]
06 Dec 2017, 07:58






