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 P₁(n)=1/6, for all n. Make a graph of P₁(n) versus n, and compute the mean and standard deviation of P₁(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 P₂(n). Make a graph of P₂(n) versus n, and compute the mean and standard deviation of P₂(n).

c) [5 points] Now lets construct the distribution P₂(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 P₁(m)P₁(k). The probability P₂(n) is then just the sum of all probabilities P₁(m)P₁(k) for all outcomes (m,k) such that m+k=n. For example,

P₂(4) = P₁(1)P₁(3) + P₁(3)P₁(1) + P₁(2)P₁(2) = (1/6)(1/6)+(1/6)(1/6)+(1/6)(1/6) = (3/36)

Using similar equations as above, reconstruct the probability distribution P₂(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 P₃(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 P₂(m)P₁(k). P₃(n) is then the sum of all probabilities P₂(m)P1(k) such that m+k=n. For example,

P₃(5) = P₂(2)P₁(3) + P₂(3)P₁(2) + P₂(4)P₁(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 P₂(m) and P₁(k), compute the distribution P₃(n) for all allowed values of n. Make a graph of P₃(n) versus n, and compute the mean and standard deviation of P₃(n).

e) [4 points] Comment on the shape, mean, and standard deviations of the distributions P₁(n), P₂(n), and P₃(n). What trends or other interesting observations do you see?