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A committee of 3 persons is to be formed from 3 company secretaries,
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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,
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23 Jun 2016, 01:57
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



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Re: A committee of 3 persons is to be formed from 3 company secretaries,
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23 Jun 2016, 01:58
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.
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Re: A committee of 3 persons is to be formed from 3 company secretaries,
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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,
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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



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Re: A committee of 3 persons is to be formed from 3 company secretaries,
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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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Re: A committee of 3 persons is to be formed from 3 company secretaries,
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23 Jun 2016, 02:44
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?



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Re: A committee of 3 persons is to be formed from 3 company secretaries,
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23 Jun 2016, 02:49
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!.
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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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