I dont know whether this is the right way to approach the problem , however i get the same ans.
Originally seated A B C D
now when after they get up and when they sit back again .
1st- A has option to sit on 3 seats ( apart from his previous seat . thus he now sits on B's seat.)
2nd- Similarly B has option to sit on 3 seats ( because A has already occupied B's previous seat, thus B sits on a's seat.)
3rd- Now C has only 1 option to sit on D's seat . and similarly D also has one option to sit on C's seat.)
hence total favourable outcomes 3*3*1*1=9
and total possible outcomes =4!=24
probability of the favourable outcome= 9/24.
Please correct me if i am wrong.
eladshus wrote:
Man - you played it!
Thanks. I didn't think of all the possibilities.
My initial solution was:
3/4 probability for the first person to NOT sit on his chair
2/3 probability for the second person to NOT sit on his chair (from the remaining 3 chairs)
1/2 probability for the third person to NOT sit on his chair (from the remaining 2 chairs)
So, my final solution was: 3/4 * 2/3 * 1/2 = 1/4
But I missed some arrangements.
Thanks
+1 from me