**Note:** In writing your solutions
to the problems, be sure to write out complete sentences explaining your
steps and logic. Do not just write down numerical answers with no explanations!
When a problem calls for you to make a graph, make a __neat__ graph
__on graph paper__ (or with a computer).

1) There is no problem (1). Just do problem (2) below.

2) [25 points] This problem may seem a bit tedious to work out the requested calculations, but none of the steps should be conceptually that difficult.

a) [3 points] Consider the roll of
a fair die. Any one of the six faces is equally likely to land on top.
Label these faces by the numbers n = 1, 2, 3, ...,6. The probability that
any particular face lands on top is therefore given by the probability
distribution P_{1}(n)=1/6,
for all n. Make a graph of P_{1}(n)
versus n, and compute the mean and standard deviation of P_{1}(n).

b) [5 points] Consider the roll of
two fair dice. Suppose the first die lands with face m and the second die
lands with face k. Their sum is n=m+k, where n can take any integer value
from n=2 to n=12. In HW#2 problem #2 you found the probability distribution
for this sum, call it P_{2}(n).
Make a graph of P_{2}(n)
versus n, and compute the mean and standard deviation of P_{2}(n).

c) [5 points] Now lets construct
the distribution P_{2}(n)
by another method. View (m,k) as the outcome that the first die lands with
m and the second die lands with k (m and k are both in the range 1,2,3,4,5,6).
Since the roll of the two die are independent, the probability for outcome
(m,k) is P_{1}(m)P_{1}(k).
The probability P_{2}(n)
is then just the sum of all probabilities P_{1}(m)P_{1}(k)
for all outcomes (m,k) such that m+k=n. For example,

P_{2}(4)
= P_{1}(1)P_{1}(3)
+ P_{1}(3)P_{1}(1)
+ P_{1}(2)P_{1}(2)
= (1/6)(1/6)+(1/6)(1/6)+(1/6)(1/6) = (3/36)

Using similar equations as above,
reconstruct the probability distribution P_{2}(n)
for all values of n=2 to n=12.

d) [8 points] In the same manner
as above, one can construct the probability distribution P_{3}(n)
for the sum n of the roll of *three* fair dice. Let (m,k) be the outcome
that the sum of the first two dice is m (now m is in the range 2,3,...,12)
and the third die is k (k in the range 1,2,...,6). The probability for
outcome (m,k) is now P_{2}(m)P_{1}(k).
P_{3}(n) is then the
sum of all probabilities P_{2}(m)P1(k)
such that m+k=n. For example,

P_{3}(5)
= P_{2}(2)P_{1}(3)
+ P_{2}(3)P_{1}(2)
+ P_{2}(4)P_{1}(1)

(for n=5 there are no other combinations of m and k that work!)

Using similar equations as above,
and your earlier results for P_{2}(m)
and P_{1}(k), compute
the distribution P_{3}(n)
for all allowed values of n. Make a graph of P_{3}(n)
versus n, and compute the mean and standard deviation of P_{3}(n).

e) [4 points] Comment on the shape,
mean, and standard deviations of the distributions P_{1}(n),
P_{2}(n), and P_{3}(n).
What trends or other interesting observations do you see?