Given that there are 6 married couples. If we select only 4 people out of the 12, what is the probability that none of them are married to each other? Produce your own response, links

I'd say 1*10/11*8/10*6/9 = 16/33, but probabilities is not really my strong point.....

Can someone please explain this ? Thanks, Deepak

sanju09 wrote:Given that there are 6 married couples. If we select only 4 people out of the 12, what is the probability that none of them are married to each other? Produce your own response, links One way is to treat this as a counting problem. P(no couples) = (# ways to select 4 people with no couples)/(total number of ways to select 4 people) Denominator: 12 people, choose 4. Can be accomplished in 12C4 ways Numerator. First, select 4 couples from the 6 couples. This can be accomplished 6C4 ways. From each of the 4 selected couples, choose 1 member. There are - 2 ways to select a member from the first couple - 2 ways to select a member from the second couple - 2 ways to select a member from the third couple - 2 ways to select a member from the fourth couple Total number of ways to select 4 people with no couples is 6C4x2x2x2x2 Probability = 6C4x2x2x2x2/12C4 = 16/33

sanju09 wrote:Given that there are 6 married couples. If we select only 4 people out of the 12, what is the probability that none of them are married to each other? Produce your own response, links (12C1*10C1*8C1*6C1/4!)/12C4 = 12*10*8*6/12*11*10*9 = 48/99 = 16/33