Answer : ( 1/2 ),
Solution : First we should calculate the total no. of possibilities,
The followings are the possibilities and no. of possible cases :
# "1st person grabs his own coat", No. of ways = 1;
# "1st person grabs r'th person's coat and r'th person grabs 1'st person's coat",
// where ( 2 <= r <= n);
No. of ways to select 'r' = (n-1).
# "1st grabs r1's, r1 grabs r2's and r2 grabs 1st's coat",
// where ( 2 <= r1 < r2 <= n ),
No. of ways to select r1 and r2 = C(n-1,2)
So total no. of ways = 1+C(n-1,1)+C(n-1,2)+....+C(n-1,n-1) = 2^(n-1);
If the last person gets his own coat :
Then r's (r1, r2....) are selected from 2,3,4...(n-1),
So no. of ways the last person gets his coat = 1+C(n-2,1)+C(n-2,2)+...+C(n-2,n-2) = 2^(n-2);
So the required probability = (2^(n-2)) / (2^(n-1)) = (1/2)...