Thread: Probability

Can anyone help me with the following probability question please:

18 fair dice are thrown and the uppermost faces are investigated. What is the probability of each face appearing 3 times?
(That is the outcome will be {1,1,1,2,2,2,3,3,3,4,4,4,5,5,5,6,6,6}.

Many thanks to anyone who is able to help

Hello,

The probability of each face appearing once is 3/18 = 1/6 (equiprobability)

If you want this face appear 3 times, it's . Because the throws are independent, this probability equals to P(face)^3, that is to say (1/6)^3

Does this mean then that the probability of each face appearing three times is:

(1/6)^18? (same as (1/6^3)

Originally Posted by smiler

18 fair dice are thrown and the uppermost faces are investigated. What is the probability of each face appearing 3 times?
(That is the outcome will be (1,1,1,2,2,2,3,3,3,4,4,4,5,5,5,6,6,6).

The event (1,1,1,2,2,2,3,3,3,4,4,4,5,5,5,6,6,6) is one 18-tuple out of possible outcomes.
But that selection of individual outcomes can occur in ways.
So the overall probability is .

Woops, am really sorry, i misunderstood the "18 fair dice", thought it was a 18-faces dice :s