Thread: totally stuck

"If sixteen married couples are in an accident where only six people survive,
what is the probability that there is at least one married couple among the survivors?"

I know this should be simple, but I don't know how to approach this problem.

1 - P(no survivors are a couple)? Can somebody start me off?

Is it:

P(X1 spouse dead)*P(X2 spouse dead)*P(X3 spouse dead)*P(X4 spouse dead)* P(X5 spouse dead)*P(X6 spouse dead) = no couples survive. So...

( 26 / 31 )^6 = 0.348

P(one or more couples survived) = 1 - 0.348 = 0.65 ?

Doesn't seem right.

or is it:

[(30*29*28*27*26)/(31*30*29*28*27)]
*[(29*28*27*26)/(30*29*28*27)]
*[(28*27*26)/(29*28*27)]
*[(27*26)/(28*27)]
*(26/27)

= 0.583

So P(at least one couple survived) = 1 - 0.583 = 0.417

The total number of possibilities, almost like you said is
32*31*30*29*28*27

For P(no survivors are a couple), let us use
__ __ __ __ __ __ as 6 possible "spaces" for the six survivors out of the whole group. The number in the space is the number of possible entries.

32 __ __ __ __ __ (trivial, we can choose out of 32 people)
32 30 __ __ __ __ (this space cannot be the previous survivor OR that one person who is the mate of the previous survivor)
32 30 28 __ __ __ (same as above)
...
32 30 28 26 24 22

so 32*30*28*26*24*22 is the total number of possibilities that no whole couple survived

so P(no survivors are a couple) = (32*30*28*26*24*22)/(32*31*30*29*28*27)

Correct me if I made a mistake. I'm a bit rusty.

In my way, you line them up and then the first person goes through the other people checking if they are their spouse. And then the next one does the same thing and so on. Your method gives a difference of .013 or something...

[(30*29*28*27*26)/(31*30*29*28*27)]
*[(28*27*26*25)/(29*28*27*26)]
*[(26*25*24)/(27*26*25)]
*[(24*23)/(25*24)]
*(22/23)

I know my way is more tedious but I wanted to see what was wrong with the method. I revised it and it gives the same result as your method so I guess it's 0.434