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A committee of 3 persons is to be formed from 3 company secretaries,

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A committee of 3 persons is to be formed from 3 company secretaries,  [#permalink]

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New post 23 Jun 2016, 01:50
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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.
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Re: A committee of 3 persons is to be formed from 3 company secretaries,  [#permalink]

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New post 23 Jun 2016, 01:57
1
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
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Joined: 02 Sep 2009
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Re: A committee of 3 persons is to be formed from 3 company secretaries,  [#permalink]

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New post 23 Jun 2016, 01:58
1
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.
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Re: A committee of 3 persons is to be formed from 3 company secretaries,  [#permalink]

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New post 23 Jun 2016, 02:11
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?
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Re: A committee of 3 persons is to be formed from 3 company secretaries,  [#permalink]

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New post 23 Jun 2016, 02:25
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
User avatar
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Joined: 02 Sep 2009
Posts: 50570
Re: A committee of 3 persons is to be formed from 3 company secretaries,  [#permalink]

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New post 23 Jun 2016, 02:29
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.
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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.


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Re: A committee of 3 persons is to be formed from 3 company secretaries,  [#permalink]

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New post 23 Jun 2016, 02:44
Bunuel

Hi,

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
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Re: A committee of 3 persons is to be formed from 3 company secretaries,  [#permalink]

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New post 23 Jun 2016, 02:49
1
nishi999 wrote:
Bunuel

Hi,

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?
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Re: A committee of 3 persons is to be formed from 3 company secretaries,  [#permalink]

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