Assignments - Winter 2011

- Section 3.1, numbers 2, 6 (and also find E[XY] and Var[XY])
- Section 6.4, numbers 1d, 5 (and also find the marginal distribution of Y)
- Choose a number X uniformly at random from the numbers 1, 2, 3. Given the value of X, choose a number uniformly at random from the numbers 1, ... , X; Find the joint distribution of (X,Y). Then find the marginal distributions of X and Y. Then find the correlation and covariance of X and Y.

- Section 2.4, number 4
- Section 3.4, number 4
- Section 3.5, numbers 2, 8, 10ab, 11a
- Section 4.2, numbers 3bc, 4abc

70 marks

- Section 3.3, numbers 2, 3, 8bc, 12, 14, 20
- Section 4.1, numbers 2c, 3e (variance only), 9

Total: 30 marks.

- Section 3.2, numbers 5, 8, 13ad, 14
- Section 4.1, numbers 2b, 3e (mean only)

All 6 questions were graded, for a total of 60 marks.

- Section 2.1, numbers 4, 7
- Section 2.2, numbers 4, 8, 9

All 5 problems were graded, for a total of 10 marks.

(originally due Friday, February 18, 2011)

- Section 3.1, number 9
- Section 4.1, numbers 2a, 3abcd, 12a
- Section 4.4, number 10c (using cdf's: write the cdf of 1/Z as an integral using the density of Z, and then differentiate using the fundamental theorem of calculus)
- Review problems to Chapter 4, number 4c (find the cdf and then also the density)
- Section 4.5, numbers 2a, 5, 6ab

The 5 problems graded were 4.1 numbers 3, 12; 4.5 numbers 5, 6; Chapter 4 review problem 4, for a total of 25 marks.

- Section 1.5, numbers 3, 6ac
- Section 1.6, numbers 1, 6, 7, 8
- The Monty Hall problem: You are playing a game show, in which there are prizes behind three doors. One prize is good (eg a car), and the others aren't (eg. a goat and a rabbit). Once you pick a door, the host (named Monty Hall) will open one of the other doors, showing you a bad prize. And he will ask you if you want to switch your choice. Should you?
The argument against is that since he always opens a door, he hasn't really given you any information that would favor one door over the other. So both remaining doors are equally likely to be correct (conditional prob 1/2 each), and there is no point in switching. The argument for switching is that you have no new information about your original choice. So the probability you picked correctly in the first place is unchanged at 1/3, and the (conditional) probability the other door is correct is now 2/3. Which argument is right?

Six problems were graded, for a total out of 60 marks. The only ungraded one was the Monty Hall problem.

- Section 1.4, numbers 4, 5, 6, 7, 8

All 5 problems were marked, giving a score out of 50.

- Section 1.1, numbers 2, 7. In number 2, write down explicitly what Omega you are using, and what the event is (as a set) in each case.
- Section 1.3, numbers 2, 4, 5, 6, 9
- Appendix I, numbers vii, viii, x, xii, xiii

The 5 problems graded were 1.1 number 2, 1.3 numbers 4 and 6, Appendix numbers vii and xii, giving a score out of 10