Let X and Y equal the outcomes when two fair six-sided dice are rolled. Let
W = X + Y. Assuming independence, find the p.m.f. of W when
(a) The first die has three faces numbered 0 and three faces numbered 2, and the second die has its faces numbered 0, 1, 4, 5, 8, and 9.
(b) The faces on the first die are numbered 0, 1, 2, 3, 4, and 5, and the faces on the second die are numbered 0, 6, 12, 18, 24, and 30.
Outcomes of the first die:
{0,0,0,2,2,2}
Outcomes of the second die:
{0,1,4,5,8,9}
There are 36 outcomes for W=x+y, some of which are identical. Make a table and compile the sums:
X 0 0 0 2 2 2
0 0 0 0 2 2 2
1 1 1 1 3 3 3
4 4 4 4 6 6 6
5 5 5 5 7 7 7
8 8 8 8 10 10 10
9 9 9 9 11 11 11
Now compile the count of the outcomes, and divide by 36 to get the pmf (probability mass function)
0 3 1/12
1 3 1/12
2 3 1/12
3 3 1/12
4 3 1/12
5 3 1/12
6 3 1/12
7 3 1/12
8 3 1/12
9 3 1/12
10 3 1/12
11 3 1/12
Part b is similar to part a.
Post your answer for a check if you wish.
To find the probability mass function (p.m.f.) of the random variable W, we need to determine the probability of each possible value of W when two fair six-sided dice are rolled.
(a) In this case, the first die has three faces numbered 0 and three faces numbered 2, while the second die has faces numbered 0, 1, 4, 5, 8, and 9.
We can find the p.m.f. of W by counting the number of outcomes with the same sum for each possible value of W.
W can take the following values:
- W = 0: The only possible outcome is (0, 0). The probability is the product of the probabilities of getting 0 on both dice, which is (3/6) * (1/6) = 1/12.
- W = 1: There are two possible outcomes: (0, 1) and (1, 0). The probability of each outcome is (3/6) * (1/6) = 1/12. Since the outcomes are independent, we add the probabilities, giving a total probability of 1/12 + 1/12 = 1/6.
- W = 2: There are three possible outcomes: (0, 2), (1, 1), and (2, 0). The probability of each outcome is (3/6) * (1/6) = 1/12. Adding these probabilities, we get 1/12 + 1/12 + 1/12 = 1/4.
- W = 3: There are three possible outcomes: (1, 2), (2, 1), and (3, 0). The probability of each outcome is (3/6) * (1/6) = 1/12. Adding these probabilities, we get 1/12 + 1/12 + 1/12 = 1/4.
- W = 4: There are four possible outcomes: (0, 4), (1, 3), (2, 2), and (3, 1). The probability of each outcome is (3/6) * (2/6) = 1/12. Adding these probabilities, we get 1/12 + 1/12 + 1/12 + 1/12 = 1/3.
- W = 5: There are four possible outcomes: (0, 5), (1, 4), (2, 3), (3, 2), and (4, 1). The probability of each outcome is (3/6) * (1/6) = 1/12. Adding these probabilities, we get 1/12 + 1/12 + 1/12 + 1/12 + 1/12 = 5/12.
- W = 6: There are four possible outcomes: (2, 4), (3, 3), (4, 2), and (5, 1). The probability of each outcome is (3/6) * (2/6) = 1/12. Adding these probabilities, we get 1/12 + 1/12 + 1/12 + 1/12 = 1/3.
Therefore, the p.m.f. of W for this case is:
W | Probability
--------------
0 | 1/12
1 | 1/6
2 | 1/4
3 | 1/4
4 | 1/3
5 | 5/12
6 | 1/3
(b) In this case, the faces on the first die are numbered 0, 1, 2, 3, 4, and 5, while the faces on the second die are numbered 0, 6, 12, 18, 24, and 30.
Following the same process as above, you can determine the p.m.f. of W for this case by counting the number of outcomes with the same sum for each possible value of W and calculating the corresponding probabilities.
By going through the steps detailed in part (a), you should be able to find the p.m.f. of W for this case as well.