Probability of the Expectation Sum of Die

(1 pt) Three dice are rolled. Let the random variable X represent the sum of the 3 dice. By assuming that each of the possible outcomes is equally likely, find the probability that X equals 11.
P(X=11)=________

I know that the expected value is equal to 3.5n where n=the number of dice=3 which is 10.5 but I do not know how to find the probability of this. We are working using random variables, expectation, and the poison distribution.

Hello, Jessica!

They are certainly trying to scare us with the wording . . .

Three dice are rolled.
Let the random variable X represent the sum of the 3 dice.
Assuming that each of the possible outcomes is equally likely, find the probability that X equals 11.
. . \(\displaystyle P(X=11)\:=\:\_\_\)

I know that the expected value is equal to 3.5n, where n = the number of dice (3),
which is 10.5, but I do not know how to find the probability of this.
We are working using random variables, expectation, and the poisson distribution.
These topics have little to do with the problem.

Click to expand...

The problem is a simple probability problem:
. . "Three dice are rolled. .Find the probability that the sum is 11."


\(\displaystyle \text{There are: }\:6^3 \,=\,216\text{ possible outcomes.}\)

\(\displaystyle \text{How many of them have a sum of eleven?}\)


\(\displaystyle \text{Let's make a list:}\)

. . \(\displaystyle \begin{array}{ccc}(1,4,6) & \text{6 orders} \\ (1,5,5) & \text{3 orders} \\ (2,3,6) & \text{6 orders} \\ (2,4,5) & \text{6 orders} \\ (3,3,5) & \text{3 orders} \\ (3,4,4) & \text{3 orders} \\ \hline \text{Total:} & \text{27 ways} \end{array}\)

\(\displaystyle \text{Therefore: }\;P(\text{sum of 11}) \;=\;\frac{27}{216} \;=\;\frac{1}{8}\)