• Jan 18th 2011, 02:16 AM
abrahamtim
One evening a drunken hotel clerk tried to hang 11 keys on their 11 hooks, but only managed to hang them independently and at random. there was no limit to the number of keys which could be hung on any hook. Find the probability of:
a) At least 1 key was hung on its own hook
b) At least 1 key was hung on the wrong hook
how did you get there ?
• Jan 18th 2011, 02:27 AM
ahaok
Quote:

Originally Posted by abrahamtim
One evening a drunken hotel clerk tried to hang 11 keys on their 11 hooks, but only managed to hang them independently and at random. there was no limit to the number of keys which could be hung on any hook. Find the probability of:
a) At least 1 key was hung on its own hook
b) At least 1 key was hung on the wrong hook
how did you get there ?

At least 1 key was hung on its own hook= 1-(no key was hung on its own hook)=1-(10/11)^11
b) At least 1 key was hung on the wrong hook=1-(no key was hung on the wrong hook)=1-(all keys were hung on the right hook)=1-(1/11)^11