Thread: Dice

Hi, I am currently stuck on this extended response question from my book, I really don't know how to do the parts of the questions, could anyone help me? Thanks

You are given two un-biased dice, one painted red and one painted blue. You roll the dice and you record the numbers showing on each die. Let R by the random variable for the red die and B the corresponding random variable for the blue die.

(a) Write down the probability distribution for each of the random variables R and B.
(b) Write down the joint probability distribution, Pr(R,B). That is, the probability of obtaining, in the one roll, a specic value for R and a specic value for B.
(c) Use the joint distribution to deduce the probability distribution for R - B.
(d) Hence calculate the mean and standard deviation for R - B.

Originally Posted by usagi_killer

Hi, I am currently stuck on this extended response question from my book, I really don't know how to do the parts of the questions, could anyone help me? Thanks

You are given two un-biased dice, one painted red and one painted blue. You roll the dice and you record the numbers showing on each die. Let R by the random variable for the red die and B the corresponding random variable for the blue die.

(a) Write down the probability distribution for each of the random variables R and B.
(b) Write down the joint probability distribution, Pr(R,B). That is, the probability of obtaining, in the one roll, a specic value for R and a specic value for B.
(c) Use the joint distribution to deduce the probability distribution for R - B.
(d) Hence calculate the mean and standard deviation for R - B.

So what have you done? You should be able to do (a) and (b) straight off.

When you have done (b) for each point (r,b) in the sample space you have a probability, and you can compute r-b, then ...

CB

a) $\displaystyle P(R) = P(B) = \frac{1}{6}$

b) $\displaystyle P(R,B) = \frac{1}{36}$

c) $\displaystyle Y= R-B$

$\displaystyle ran(Y) = \{-5,-4,-3,-2,-1,0,1,2,3,4,5\}$

$\displaystyle P(-5) = P(5) = \frac{1}{36}$

$\displaystyle P(-4) = P(4) = \frac{2}{36}$

$\displaystyle P(-3) = P(3) = \frac{3}{36}$

$\displaystyle P(-2) = P(2) = \frac{4}{36}$

$\displaystyle P(-1) = P(1) = \frac{5}{36}$

$\displaystyle P(0) = \frac{6}{36}$

d) Positive and negative factors are canceling, so $\displaystyle \mu(Y)=E(Y)=0$

$\displaystyle \sigma = \sqrt{E(Y^2) - \mu^2} = \sqrt{(\frac{25}{36} +\frac{16}{36} +\frac{9}{36} +\frac{4}{36} + \frac{1}{36}) \times 2 - 0} = \sqrt{3.0555} \approx 1.74 $