The full quiz is here. The answers appear below. Comments, which are not part of the answers, are italicized.
1. Explicitly describe the tickets in a box representing a Binom(2, 1/3) variable. ... Use the least possible number of tickets.
First solution: A Binom(2, 1/3) variable is a sum of two independent Binom(1, 1/3) variables. (See the text discussion at page 179 and following.) A Binom(1, 1/3) variable can be represented by three tickets in a box because the success probability is 1/3. On one is written the value 1 (for one success) and on the other two tickets are written the value 0 (no successes). The two independent draws are summarized by the following table showing all 3 * 3 = 9 possible combinations of tickets. The right column provides the sum. This is what must be written on each of the tickets put into the Binom(2, 1/3) box.
| Box 1 | Box 2 | |||
| Ticket | Value | Ticket | Value | Sum |
| 1 | 1 | 1 | 1 | 2 |
| 1 | 1 | 2 | 0 | 1 |
| 1 | 1 | 3 | 0 | 1 |
| 2 | 0 | 1 | 1 | 1 |
| 2 | 0 | 2 | 0 | 0 |
| 2 | 0 | 3 | 0 | 0 |
| 3 | 0 | 1 | 1 | 1 |
| 3 | 0 | 2 | 0 | 0 |
| 3 | 0 | 3 | 0 | 0 |
Therefore we may place nine (9) tickets into a box. One has the value 2, four have the value 1, and four have the value 0. Evidently we cannot use fewer tickets, because it is impossible to split up the single ticket with the value 2.
Second solution: The probabilities for a Binom(2, 1/3) variable are given by Comb(2,K)*(1/3)K*(2/3)2-K for K=0, 1, and 2 (text, formulas 4.37 and 4.38; or see the class notes web page, Probability paradoxes and simulation). These probabilities are
Comb(2,0)*(1/3)0*(2/3)2-0 = 1 * 1 * 4/9 = 4/9
Comb(2,1)*(1/3)1*(2/3)2-1 = 2/1 * 1/3 * 2/3 = 4/9
Comb(2,2)*(1/3)2*(2/3)2-2 = 2*1/(2*1) * 1/9 * 1 = 1/9
Since all probabilities are multiples of 1/9, we use nine tickets, writing K=0 on four of them, K=1 on four of them, and K=2 on one of them.
Third solution: This problem is identical to that of characterizing two flips of a fair coin (a Binom(2, 1/2) variable), except now the coin is weighted to come up heads ("success") 1/3 of the time. From our class notes on distributions, or from our in-class discussion of this situation, recall that there are three possibilities: two heads with probability 1/3 * 1/3 = 1/9, two tails with probability 2/3 * 2/3 = 4/9, and one head, which therefore must have probability 1 - 1/9 - 4/9 = 4/9. To represent these probabilities will require nine tickets in a box, with one of them labeled "2" (for two heads or successes), four of them labeled "0", and the remaining four labeled "1".
2. Provide an Excel expression that will simulate a single draw from a box containing exactly three tickets with the values 1, 2, 3, and 4.
There are many solutions. You will need RAND() to generate uniformly distributed random values between 0 and 1 (not including 1). Multiplying these by four will give uniformly distributed values between 0 and 4 (not including 4). Truncating them with INT() will therefore produce values of 0, 1, 2, and 3 with equal probability. Adding 1 to those results will give the desired simulation. The formula just described is 1 + INT(4*RAND()). The functions INT() and RAND() are documented on the class notes page, Probability paradoxes and simulation
Scoring: The passing score is 90.
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This page was created 18 February 2001 and modified 21 February to include additional comments and the third solution to #1.