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21 Jul 2013, 14:11
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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?
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?
21 Jul 2013, 23:01
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alphabeta1234
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?
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.
28 May 2020, 04:30
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Bunuel
alphabeta1234
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?
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.
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 ?
