Question 18.5: a. Consider the toss of a single evenly weighted die. What a...
a. Consider the toss of a single evenly weighted die. What are the six possible mutually exclusive results that can occur?
b. Now consider the toss of a pair of evenly weighted dice and add their individual results. What is the probability of each of the resulting sums?
c. What is the most probable sum in item b?
a. Since the die is an evenly weighted cube, all six sides have the same probability of landing face up, so we have N = 6.
Consequently the six possible mutually exclusive results are
P_1 = P_2 = P_3 = P_4 = P_5 = P_6 = 1 /6
Also, we can write
\backsim P_1 = \backsim P_2 = \backsim P_3 = \backsim P_4 = \backsim P_5 = \backsim P_6 = 5/6
b. The total number of combinations of results is shown in Table 18.4. From this table, it is seen that N = 6 × 6 = 36. Using this table, we can construct an event vs. frequency table, as shown in Table 18.5. Thus, we can compute the following probabilities for the sum of the results of the individual die:
| Table 18.4 The Total Number of ombinations of Tossing Two Dice | |||
| Die 1 | Die 2 | Die 1 | Die 2 |
| 1 | 1 | 4 | 1 |
| 1 | 2 | 4 | 2 |
| 1 | 3 | 4 | 3 |
| 1 | 4 | 4 | 4 |
| 1 | 5 | 4 | 5 |
| 1 | 6 | 4 | 6 |
| 2 | 1 | 5 | 1 |
| 2 | 2 | 5 | 2 |
| 2 | 3 | 5 | 3 |
| 2 | 4 | 5 | 4 |
| 2 | 5 | 5 | 5 |
| 2 | 6 | 5 | 6 |
| 3 | 1 | 6 | 1 |
| 3 | 2 | 6 | 2 |
| 3 | 3 | 6 | 3 |
| 3 | 4 | 6 | 4 |
| 3 | 5 | 6 | 5 |
| 3 | 6 | 6 | 6 |
| Table 18.5 An Event-Frequency Table for Tossing Two Dice | ||
| Sum M of Die Values | Number of Results Producing Sum M | Ways of Obtaining Sum M |
| 0 | 0 | |
| 1 | 0 | |
| 2 | 1 | 1 + 1 |
| 3 | 2 | 1 + 2, 2 + 1 |
| 4 | 3 | 2 + 2, 1 + 3, 3 + 1 |
| 5 | 4 | 2 + 3, 3 + 2, 1 + 5, 4 + 1 |
| 6 | 5 | 3 + 3, 4 + 2, 2 + 5, 5 + 1, 1 + 5 |
| 7 | 6 | 6 + 1, 1 + 6, 5 + 2, 2 + 5, 4 + 3, 3 + 4 |
| 8 | 5 | 4 + 4, 5 + 3, 3 + 6, 6 + 2, 2 + 6 |
| 9 | 4 | 4 + 5, 5 + 4, 3 + 7, 6 + 3 |
| 10 | 3 | 6 + 4, 4 + 6, 5 + 6 |
| 11 | 2 | 6 + 5, 5 + 6 |
| 12 | 1 | 6 + 6 |
| P_0 = P_1 = 0 | P_5 = P_9 = 4/36 = 1/9 |
| P_2 = P_{12} = 1/36 | P_6 = P_8 = 5/36 |
| P_3 = P_{11} = 2/36 = 1/18 | P_7 = 6/36 = 1/6 |
| P_4 = P_{10} = 3/36 = 1/12 |
c. From Table 18.5, it is clear that in the toss of two evenly weighted dice, the number 7 is the most probable outcome, appearing on the average of once every six tosses. It has a probability given by Eq. (18.30) of
P_A =\frac{M}{N} (18.30)
P_7 = 1/6 = 0.1667 = 16.67\%