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

Question

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.

nishi999 · Answer

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

Bunuel · Answer

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.

nishi999 · Answer

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?

nishi999 · Answer

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

Bunuel · Answer

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.

nishi999 · Answer

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?

Bunuel · Answer

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

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