Physics 104 Homework #4 Due Thurs. Oct. 12, 2000

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?