2. Let us define a new variable Y = the sum of two dice. Complete both the probability and cumulative probability tables for Y. (Hint: First think about what values Y can take on, and then notice rolling a die produces 6 outcomes but rolling two dice produces 36 outcomes. The following table can help you enumerate all 36 outcomes and find out what is the associated Y value for each outcome.) 1 2 3 4 5 6 1 Y=2 . . . . . 2 . . . . . 3 . . . . . . 4 . . . . . . 5 . . . . . . 6 . . . . . . Can someone help show how to calculate the cumulative probability for this question?
Solution:-) The total possible outcomes of two random dice.
1 | 2 | 3 | 4 | 5 | 6 | |
1 | (1, 1) | (1, 2) | (1, 3) | (1, 4) | (1, 5) | (1, 6) |
2 | (2, 1) | (2, 2) | (2, 3) | (2, 4) | (2, 5) | (2, 6) |
3 | (3, 1) | (3, 2) | (3, 3) | (3, 4) | (3, 5) | (3, 6) |
4 | (4, 1) | (4, 2) | (4, 3) | (4, 4) | (4, 5) | (4, 6) |
5 | (5, 1) | (5, 2) | (5, 3) | (5, 4) | (5, 5) | (5, 6) |
6 | (6, 1) | (6, 2) | (6, 3) | (6, 4) | (6, 5) | (6, 6) |
So, Y can take value from 2,3,4,5,..................12.
y | P(Y=y) | Cumulative prob. |
2 | 1/36 | 1/36 |
3 | 2/.36 | 1/36+2/36 = 3/36 |
4 | 3/36 | 6/36 |
5 | 4/36 | 10/36 |
6 | 5/36 | 15/36 |
7 | 6/36 | 21/36 |
8 | 5/36 | 26/36 |
9 | 4/36 | 30/36 |
10 | 3/36 | 33/36 |
11 | 2/36 | 35/36 |
12 | 1/36 | 36/36 =1 |
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