# Thread: Simple Porobability question

1. ## 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?

2. 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.

3. Originally Posted by mosespreciado
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?
No the answer is $\displaystyle \frac{1}{6}$
There are only $\displaystyle 3!=6$ ways to serve the three dishes.