# Thread: need help with an expectation question

1. ## need help with an expectation question

hi hi
am not sure where to begin so all i need is a push in the right direction not the answer. so here's the question

find the expected number of ones when three ordinary fair dice are thrown.

i know that E(X) = xP(X=x) but the only thing throwing me off is the three dice so if u can please point me in the right direction
Thank You so veryyyy much

2. Hello, helloworld101!

Find the expected number of 1's when three fair dice are thrown.

We need the probability of getting: one 1, two 1's, three 1's.

There are: . $6^3 = 216$ possible outcomes.

To get three 1's:
There is one way to get three 1's.
. . $P(\text{three 1's}) \:=\:\frac{1}{216}$

To get two 1's:
There are three cases: .$$33x, \;3x3,\;x33$$
. . And there are 5 choices for the third die.
There are: . $3\cdot5 \:=\:15$ ways to get two 1's.
. . $P(\text{two 1's}) \:=\:\frac{15}{215}$

To get one 1:
There are three cases: . $3xx,\;x3x,\;xx3$
. . And there are $5^2$ choices for the other two dice.
There are: . $3\cdot25 \:=\:75$ ways to get one 1.
. . $P(\text{one 1}) \:=\:\frac{75}{216}$

$\text{Expected number of 1's} \;\;=\;\;(3)\left(\frac{1}{216}\right) + (2)\left(\frac{15}{216}\right) + (1)\left(\frac{75}{215}\right) \;\;=\;\;\frac{108}{216} \;\;=\;\;\frac{1}{2}$

3. much thanks soroban really appriciate it

4. Originally Posted by helloworld101
hi hi
am not sure where to begin so all i need is a push in the right direction not the answer. so here's the question

find the expected number of ones when three ordinary fair dice are thrown.

i know that E(X) = xP(X=x) but the only thing throwing me off is the three dice so if u can please point me in the right direction
Thank You so veryyyy much
If X is the random variable 'number of ones' then X ~ Binomial(n = 3, p = 1/6) and so E(X) = np = ....