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# 2 employees are to be randomly chosen to form a committee

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Manager
Joined: 12 Feb 2012
Posts: 109
Followers: 1

Kudos [?]: 14 [0], given: 28

2 employees are to be randomly chosen to form a committee [#permalink]  21 Jul 2013, 13:11
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-[(6/8)(5/7)+2(1/8)(6/7)+2(1/8)(6/7)]. But this isn't the answer. What am I doing wrong?
Math Expert
Joined: 02 Sep 2009
Posts: 27468
Followers: 4311

Kudos [?]: 42201 [0], given: 5957

Re: 2 employees are to be randomly chosen to form a committee [#permalink]  21 Jul 2013, 22:01
Expert's post
alphabeta1234 wrote:
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-[(6/8)(5/7)+2(1/8)(6/7)+2(1/8)(6/7)]. But this isn't the answer. What am I doing wrong?

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.
_________________
Re: 2 employees are to be randomly chosen to form a committee   [#permalink] 21 Jul 2013, 22:01
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