# Simple Porobability question

• Apr 21st 2011, 08:57 PM
Simple Porobability question
Ms X orders cod, Mr.Y orders trout, and Ms.Z orders tuna. The server forgets who ordered which and instead of asking, simply places them down randomly on the table. What is the probability that

Ms.X gets the cod
Mr.Y gets the trout
and
Ms.Z gets the tuna?

my initial thoughts are 1/9 probability for each. is this correct?
• Apr 21st 2011, 09:38 PM
NathanElder
i think i would be 1/27 because of the and i know you multiple and there are 3 different kind of fish so 1/3 multip;lied by 1/3 multiplied by 1/3

someone please correct me if I am wrong.
• Apr 22nd 2011, 02:55 AM
Plato
Quote:

No the answer is $\frac{1}{6}$
There are only $3!=6$ ways to serve the three dishes.