gmatprep09 wrote:
If 4 people are selected from a group of 6 married couples, what is the probability that none of them would be married to each other?
A. 1/33
B. 2/33
C. 1/3
D. 16/33
E. 11/12
Each couple can send only one "representative" to the committee. We can choose 4 couples (as there should be 4 members) to send that one "representatives" to the committee in
C^4_6 # of ways.
But these 4 chosen couples can send two persons (either husband or wife):
2*2*2*2=2^4.
So # of ways to choose 4 people out 6 married couples so that none of them would be married to each other is:
C^4_6*2^4.
Total # of ways to choose 4 people out of 12 is
C^4_{12}.
P=\frac{C^4_6*2^4}{C^4_{12}}=\frac{16}{33}Answer: D.
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