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

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?

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

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.

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!.

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