Let X equal the outcome (1, 2
Let X equal the outcome (1, 2 , 3 or 4) when a fair four-sideddie is rolled; let Y equal the outcome (1, 2, 3, 4, 5 or 6) when afair six-sided die is rolled. Let W=X+Y.
a. What is the pdf of W?
b What is E(W)?
Answer:
(a)
Since each of the four outcomes are equally likely soprobability of getting any outcome when a four sided die is rolledis 1/4.
Since each of the six outcomes are equally likely so probabilityof getting any outcome when a six sided die is rolled is 1/6.
Possible number of outcomes when a four sided and a 6 sided dieis rolled is 4*6 = 24.
Probability of each outcome is 1 / 24 = 0.0417
Following table shows all the possible outcomes and probability:
| X | Y | W=X+Y | P(W=w) |
| 1 | 1 | 2 | 0.04167 |
| 1 | 2 | 3 | 0.04167 |
| 1 | 3 | 4 | 0.04167 |
| 1 | 4 | 5 | 0.04167 |
| 1 | 5 | 6 | 0.04167 |
| 1 | 6 | 7 | 0.04167 |
| 2 | 1 | 3 | 0.04167 |
| 2 | 2 | 4 | 0.04167 |
| 2 | 3 | 5 | 0.04167 |
| 2 | 4 | 6 | 0.04167 |
| 2 | 5 | 7 | 0.04167 |
| 2 | 6 | 8 | 0.04167 |
| 3 | 1 | 4 | 0.04167 |
| 3 | 2 | 5 | 0.04167 |
| 3 | 3 | 6 | 0.04167 |
| 3 | 4 | 7 | 0.04167 |
| 3 | 5 | 8 | 0.04167 |
| 3 | 6 | 9 | 0.04167 |
| 4 | 1 | 5 | 0.04167 |
| 4 | 2 | 6 | 0.04167 |
| 4 | 3 | 7 | 0.04167 |
| 4 | 4 | 8 | 0.04167 |
| 4 | 5 | 9 | 0.04167 |
| 4 | 6 | 10 | 0.04167 |
To find the pdf of W we need to combine the same values of W.Following is the pdf of W:
| W | P(W=w) |
| 2 | 0.04167 |
| 3 | 0.08334 |
| 4 | 0.12501 |
| 5 | 0.16668 |
| 6 | 0.16668 |
| 7 | 0.16668 |
| 8 | 0.12501 |
| 9 | 0.08334 |
| 10 | 0.04167 |
b)
The expected value of W is
| W | P(W=w) | W*P(W=w) |
| 2 | 0.04167 | 0.08334 |
| 3 | 0.08334 | 0.25002 |
| 4 | 0.12501 | 0.50004 |
| 5 | 0.16668 | 0.8334 |
| 6 | 0.16668 | 1.00008 |
| 7 | 0.16668 | 1.16676 |
| 8 | 0.12501 | 1.00008 |
| 9 | 0.08334 | 0.75006 |
| 10 | 0.04167 | 0.4167 |
| Total | 1.00008 | 6.00048 |
So the expected value is
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