bibha 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
Total # of ways to select 4 people out of 12 people = 12C4 = (12*11*10*9)/1*2*3*4) = 11 * 5 * 9
Favorable outcomes = # of ways to select 4 people out of 6 couples such that none of them are married to each other
The First person can be chosen in 12 ways.
The Second person can be chosen in 10 ways, since we will remove the spouse of first person selected & then choose.
The Third person can be chosen in 8 ways, since we will remove the spouse of second person selected, as well & then choose.
The Fourth person can be chosen in 6 ways, since we will remove the spouse of third person selected, as well & then choose.
Now we need the # of ways of selection & not the arrangement of the persons, hence order does not matter. To account for this we divide by 4!, which is the # of possible arrangements of 4 persons.
Hence Favorable outcomes = (12 * 10 * 8 * 6)/4! = 240
Required Probability = 240/(11 * 5 *9) = 16/33
Answer D.
Thanks,
GyM
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