Hi!sh.nada wrote:6 married couples are present at the party. if 4 people are selected out of these 12, what is the probability that none of these people will be married to each other ?
A) 1/33
B) 2/33
c) 1/3
D) 16/33
E) 11/12
Whenever you make multiple simultaneous selections, you can treat the question as though you're selecting the items one at a time; let's do that on this question.
Our first person can be anyone, so there's a 12/12 chance that the person is "safe".
Once we choose our first person, there are 10 remaining people who aren't a spouse of that person. So the probability that the second person isn't a spouse is 10/11.
Now that we've chosen two non-spouses, 8 of the remaining 10 people aren't a spouse of either person that we've chosen. Accordingly, the probability that the third person isn't a spouse is 8/10.
Finally, when we choose our fourth person, 6 of the remaining 9 people aren't spouses of any of our 3 choices. So, the probability that the fourth person chosen is a non-spouse is 6/9.
We're making multiple selections, so we multiply the individual probabilities:
12/12 * 10/11 * 8/10 * 6/9
= (10*8*6)/(11*10*9)
= 8*6/11*9
= 48/99
= 16/33
choose (D)!
