91Ó°ÊÓ

Since there are 2 fair 6-sided dice being rolled, there are a total of 6 possible outcomes for each die, so there are a total of \(6^2 = 36\) possible outcomes for the two dice together. This sample space can be represented by the pairs (x,y), where x is the value of the first die and y is the value of the second die.

Each outcome of the dice rolls has an equal probability, and there are 36 total possible outcomes. Hence, the probability of each outcome is given by \[P(x,y)=\frac{1}{36}\]

Now, we need to find the probability values \(P\{X=i\}\) for \(i=1, \ldots, 36\). To do this, we need to count the number of pairs (x,y) whose product equals i and divide this by the total number of possible outcomes (36). For each value of \(i\), do the following: 1. Iterate through possible values of x (1 to 6) and then find the possible values of y, such that the product x*y equals i. Count the number of valid pairs (x,y) for each i. 2. Divide the count of valid pairs by 36 to find the probability of the product being equal to i.

Iterating through each value of i (from 1 to 36) and following the steps above, we find the probability values P{X=i} as: \(P\{X=1\} = \frac{1}{36}\) \(P\{X=2\} = \frac{2}{36}\) \(P\{X=3\} = \frac{2}{36}\) \(P\{X=4\} = \frac{3}{36}\) \(P\{X=5\} = \frac{1}{36}\) \(P\{X=6\} = \frac{4}{36}\) \(P\{X=7\} = \frac{0}{36}\) \(P\{X=8\} = \frac{1}{36}\) \(P\{X=9\} = \frac{4}{36}\) \(P\{X=10\} = \frac{2}{36}\) \(P\{X=12\} = \frac{4}{36}\) \(P\{X=15\} = \frac{1}{36}\) \(P\{X=16\} = \frac{1}{36}\) \(P\{X=18\} = \frac{2}{36}\) \(P\{X=20\} = \frac{1}{36}\) \(P\{X=24\} = \frac{2}{36}\) \(P\{X=25\} = \frac{1}{36}\) \(P\{X=30\} = \frac{2}{36}\) \(P\{X=36\} = \frac{1}{36}\) The other probabilities (P\{X=11\}, P\{X=13\}, P\{X=14\}, P\{X=17\}, P\{X=19\}, P\{X=21\}, P\{X=22\}, P\{X=23\}, P\{X=26\}, P\{X=27\}, P\{X=28\}, P\{X=29\}, P\{X=31\}, P\{X=32\}, P\{X=33\}, P\{X=34\}, P\{X=35\}) are all equal to 0, since there are no valid (x,y) pairs in the sample space with a product equal to these values.