Problem Solving

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?


[spoiler]16/33[/spoiler]

Picking 4 people out of 12 people(order does not matter since its the same 4 people no matter when u pick whom)

12C4 (Total possible outcomes)

Lets take 1 married couple and see how many different groupings of 4 can be formed including them

2c2 (no of ways of picking 1 couple) * 10c2 (picking 2 people out of 10 people-> In this 2 we could pick another married couple and this does not matter since in this grouping count we are trying to find where there is atleast 1 married couple)

Since there are 6 marrued couples there will be 6 * [ 2c2*10c2] ways

The no of groupings with no married couples( No of favorable outcomes)
= 12c4 - 6 * [ 2c2*10c2]


So the probability we need = No of favorable outcomes / Total possible outcomes

= 12c4 - 6 * [ 2c2*10c2] / 12c4

= 15/33

Must be missing something.....

Regards,
CR

We can solve this problem this way:
12/12*10/11*8/10*6/9=16/33

The explanation to this kind of problem is well described here:
https://www.beatthegmat.com/deck-of-cards-t28564.html
Look for shulapa's explanation.

There's a simpler way of doing this..

Let the couples be: Aa, Bb, Cc, Dd, Ee, Ff

Based upon the condition given that none of the person selected may be married to the any of the other three that are selected, implies
Number of way of selecting the first person: 12
Number of way of selecting the second person: 10
Number of way of selecting the third person: 8
Number of way of selecting the Fourth person: 6

So total number of ways of selecting these people who are not married to each other = 12x10x8x6 --- (1)

Total number of ways of selecting 4 people out of 12 = 12x11x10x9 ---(2)

so Probability = (1)/(2)
= 12x10x8x6/(12x11x10x9)
= 16/33

that's the answer

Ooops.. I am sorry.. lilu had not replied by the time I was posting my solution.. please do not consider my solution as plagiarizing.
thnx..

Thanks Guys!

Shulapa's method is very good & the way to go. I think my explanations also make it look a lot longer than it is....

Anyway coming back to how I did:

I counted some grouptings twice so the right solution based on what I did would be

Picking 4 people out of 12 people(order does not matter since its the same 4 people no matter when u pick whom)

12C4 (Total possible outcomes) = 495

Lets take 1 married couple and see how many different groupings of 4 can be formed including them

2c2 (no of ways of picking 1 couple) * 10c2

Since there are 6 married couples there will be 6 * [ 2c2*10c2] = 270

I must subtract 15 since when there are 2 couples in a group I would have double counted

270-15 = 255

The no of groupings with no married couples( No of favorable outcomes)
= 495 - 255 = 240


So the probability we need = No of favorable outcomes / Total possible outcomes

= 240/ 495
= 16/33



Regards,
CR

total=12C4
none of them are married to each other =6C4*2C1*2C1*2C1*2C1=6C4*(2C1)^4 (here we have chosen 4 couples out of 6, and then we got 1 person from each of these 4 couples)

P= (6C4*(2C1)^4)/12C4=16/33

Followed this method:
Total number of outcomes = 12C4 =495
No. of favorable outcomes = 6C4 + 6C3 * 3C1 + 6C2 * 4C2 + 6C1 * 5C3 + 6C4(all 4 males + 3 males and 1 female and so on) =240
Thus,
Probability = 240/495 =16/33.