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:
- Die \(X\) can roll any number from \(1\) to \(6\) (the probability of each number is \(\frac{1}{6}\)).
- 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.