Answer:
When two dice are thrown, the possible values for the sum X range from 2 (when both dice show a 1) to 12 (when both dice show a 6).
Now, we want to calculate the possible values of Y, where Y = 2X + 3:
1. When X = 2, Y = 2(2) + 3 = 7.
2. When X = 3, Y = 2(3) + 3 = 9.
3. When X = 4, Y = 2(4) + 3 = 11.
4. When X = 5, Y = 2(5) + 3 = 13.
5. When X = 6, Y = 2(6) + 3 = 15.
6. When X = 7, Y = 2(7) + 3 = 17.
7. When X = 8, Y = 2(8) + 3 = 19.
8. When X = 9, Y = 2(9) + 3 = 21.
9. When X = 10, Y = 2(10) + 3 = 23.
10. When X = 11, Y = 2(11) + 3 = 25.
11. When X = 12, Y = 2(12) + 3 = 27.
So, the possible values of Y are 7, 9, 11, 13, 15, 17, 19, 21, 23, 25, and 27.
Now, let's calculate E(Y), the expected value of Y:
E(Y) = Σ(Y * P(Y))
To find E(Y), we need to calculate P(Y) for each possible value of Y:
- P(Y = 7) = P(X = 2) = 1/36
- P(Y = 9) = P(X = 3) = 2/36
- P(Y = 11) = P(X = 4) = 3/36
- P(Y = 13) = P(X = 5) = 4/36
- P(Y = 15) = P(X = 6) = 5/36
- P(Y = 17) = P(X = 7) = 6/36
- P(Y = 19) = P(X = 8) = 5/36
- P(Y = 21) = P(X = 9) = 4/36
- P(Y = 23) = P(X = 10) = 3/36
- P(Y = 25) = P(X = 11) = 2/36
- P(Y = 27) = P(X = 12) = 1/36
Now, calculate E(Y):
E(Y) = (7 * 1/36) + (9 * 2/36) + (11 * 3/36) + (13 * 4/36) + (15 * 5/36) + (17 * 6/36) + (19 * 5/36) + (21 * 4/36) + (23 * 3/36) + (25 * 2/36) + (27 * 1/36)
E(Y) ≈ 16.3333
Now, let's calculate 2E(X) + 3:
E(X) is the expected value of X, which is the sum of two dice rolls. On average, E(X) is 7.
2E(X) + 3 = 2 * 7 + 3 = 14 + 3 = 17
E(Y) is indeed equal to 2E(X) + 3:
E(Y) ≈ 16.3333
2E(X) + 3 = 17
They are very close, but the small difference can be attributed to rounding when calculating probabilities.