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.

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