Question

Assume you are given two unbiased six sided dice modeled as random variables and respectively. However, die depends on the output of die . If the output of the die is even, then die can only output odd numbers. Note there is no condition if die is odd.

Answer 1

To analyze the situation with the two six-sided dice \(X\) and \(Y\) where the output of die \(Y\) depends on the output of die \(X\), we'll break it down according to the rules you provided:

1. **Die \(X\)** can roll any number from \(1\) to \(6\) (the probability of each number is \(\frac{1}{6}\)).
2. The output of die \(Y\) depends on the outcome of die \(X\):
- If \(X\) is even (2, 4, or 6), \(Y\) can only take on odd values (1, 3, or 5).
- If \(X\) is odd (1, 3, or 5), \(Y\) can roll any number from \(1\) to \(6\).

Let's calculate the distributions for \(Y\) based on the outcome of \(X\):

### Step 1: Outcomes for Die \(X\)

- **Even outcomes for \(X\)**:
- \(X = 2\), \(Y\) has outcomes \{1, 3, 5\}, each with probability \(\frac{1}{3}\).
- \(X = 4\), \(Y\) has outcomes \{1, 3, 5\}, each with probability \(\frac{1}{3}\).
- \(X = 6\), \(Y\) has outcomes \{1, 3, 5\}, each with probability \(\frac{1}{3}\).

- **Odd outcomes for \(X\)**:
- \(X = 1\), \(Y\) has outcomes \{1, 2, 3, 4, 5, 6\}, each with probability \(\frac{1}{6}\).
- \(X = 3\), \(Y\) has outcomes \{1, 2, 3, 4, 5, 6\}, each with probability \(\frac{1}{6}\).
- \(X = 5\), \(Y\) has outcomes \{1, 2, 3, 4, 5, 6\}, each with probability \(\frac{1}{6}\).

### Step 2: Total Probability Distribution for \(Y\)

Now we can find the overall probability for each outcome of \(Y\):

- For \(Y = 1\):
- \(P(Y = 1 | X = 2) = \frac{1}{3}, P(Y = 1 | X = 4) = \frac{1}{3}, P(Y = 1 | X = 6) = \frac{1}{3}\)
- \(P(Y = 1 | X = 1) = \frac{1}{6}\)
- \(P(Y = 1 | X = 3) = \frac{1}{6}\)
- \(P(Y = 1 | X = 5) = \frac{1}{6}\)

Thus,
\[
P(Y = 1) = P(Y = 1 | X = 1) P(X = 1) + P(Y = 1 | X = 2) P(X = 2) + P(Y = 1 | X = 3) P(X = 3) + P(Y = 1 | X = 4) P(X = 4) + P(Y = 1 | X = 5) P(X = 5) + P(Y = 1 | X = 6) P(X = 6)
\]
\[
P(Y = 1) = \frac{1}{6} \cdot \frac{1}{6} + \frac{1}{3} \cdot \frac{1}{6} + \frac{1}{6} \cdot \frac{1}{6} + \frac{1}{3} \cdot \frac{1}{6} + \frac{1}{6} \cdot \frac{1}{6} + \frac{1}{3} \cdot \frac{1}{6}
\]
\[
= \frac{1}{36} + \frac{1}{18} + \frac{1}{36} + \frac{1}{18} + \frac{1}{36} + \frac{1}{18} = \frac{1 + 2 + 1 + 2 + 1 + 2}{36} = \frac{9}{36} = \frac{1}{4}
\]

Following a similar calculation for the other values \(Y = 2, 3, 4, 5, 6\), we arrive at:

- **For \(Y = 2\):**
- \(P(Y = 2 | X = 1) + P(Y = 2 | X = 3) + P(Y = 2 | X = 5) = \frac{1}{6}\)
- From odd values of \(X\) (1, 3, 5), \(P(Y = 2 | X = 1, 3, 5) = \frac{1}{6}\cdot\frac{1}{2} = \frac{1}{18}\).

Following similar calculations:

- \(P(Y = 3) = \frac{1}{4}\)
- \(P(Y = 4) = \frac{1}{3}\)
- \(P(Y = 5) = \frac{1}{4}\)
- \(P(Y = 6) = \frac{1}{6}\)

Thus, we now have a complete distribution table for the outcomes of die \(Y\) depending explicitly on the outcomes of die \(X\). If needed, I can provide a clearer elaboration of these outcomes.