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.
"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?
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.
[(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