2 employees are to be randomly chosen to form a committee

Question

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?

Let B-Bob and R-Rachael
Using a reverse probability approach:

1) P(other than B&R)*P(other than B&R)=(6/8)(5/7)
2) P(B)*P(other than B&R)=(1/8)(6/7)
3) P(R)*P(other than B&R)=(1/8)(6/7)

(Let O-other) Situation 2) can happen in two ways BO or OB. Same thing with 3) RO or OR so we have to multiply 2) and 3) by 2!.

Hence the answer should be 1- . But this isn't the answer. What am I doing wrong?

Bunuel · Answer

Nothing is wrong.

The correct answer is  \frac{1}{C^2_8}=\frac{1}{28}  and  1-(\frac{6}{8}*\frac{5}{7}+2*\frac{1}{8}*\frac{6}{7}+2*\frac{1}{8}*\frac{6}{7})  is also equals to 1/28.

kunalc20 · Answer

Bunuel,

using probability approach, why we taking the probability of choosing Bob and Rachel as a fist person
Its a committee formation a selection not an arrangement 
i was taking it as 1/8*1/7
Please let me know what am i missing here ?

SpiceRunner · Answer

I would encourage you to look at this problem as a counting problem instead.

First, let's count the total number of 2 people committee from a 8 people pool. The answer is 8C2 i.e. 28. So there are 28 possible sets of 2 people committees.

Now, let's count the total number of ways Bob and Rachel are chosen together. That will be 1. (This is the key step. If this step is not evident, make up letters for remaining 6 employees, and list out all the possible 2 employee groups for the sake of understanding how we landed with 1)

Probability of Rachel and Bob being chosen = Number of ways Bob and Rachel are chosen/Total number of 2 people committees.

Therefore, Answer = 1/28.

bumpbot · Answer

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2 employees are to be randomly chosen to form a committee

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2 employees are to be randomly chosen to form a committee

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