Note that we cannot count an outcome several times. There are, for example, four combinations that result in an outcome of 5. With that said, we do not have four outcomes of 5. If we get a 5, that is one possible outcome of rolling the two dices. With this, we can list the sample space.
{2,3,4,5,6,7,8,9,10,11,12}
b The more combinations that results in a certain outcome, the more likely it is. Let's complete the sample space and mark combinations resulting in an outcome of 7 and 3
There are more combinations that result in an outcome of 7 than 3. Therefore, we know that all outcomes are not equally likely.
c First of all, P(15) is not possible since the sum of the dice is at most 12. Therefore, this outcome has a 0 % chance of occurring. To determine the probability of P(even) and P(10), we will mark all combinations that result in these outcomes in two separate sample spaces.
From the sample spaces, we see that 18 combinations produce an even result, and 3 combinations produce a result of 10. Since the sample space is 6 by 6, the total number of possible outcomes is 6* 6= 36. With this, we can calculate the probability of these events occurring.
P(even)&=18/36= 1/2 [0.8em]
P(10)&=3/36=1/12
d From Part B, we know that 7 has the most combinations of the sample space, there are 6 of them. Additionally, the number of possible outcomes is 36. With this, we can calculate the probability of rolling a 7.
