need help with an expectation question

• Nov 10th 2009, 11:21 AM
helloworld101
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
• Nov 10th 2009, 12:47 PM
Soroban
Hello, helloworld101!

Quote:

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: .$\displaystyle 6^3 = 216$ possible outcomes.

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

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

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

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

• Nov 10th 2009, 01:13 PM
helloworld101
much thanks soroban :) really appriciate it
• Nov 10th 2009, 05:11 PM
mr fantastic
Quote:

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 = ....