Question 1178071
Let's break down this problem step by step.

**Understanding the Experiment**

1.  **First Roll:** Roll a six-sided die once. Let the outcome be X. X can be 1, 2, 3, 4, 5, or 6.
2.  **Subsequent Rolls:** Roll the die X more times.
3.  **Wins:** A win occurs if a "3" is rolled in any of the subsequent rolls.
4.  **Goal:** Find the expected number of wins.

**Calculating Probabilities and Expected Wins**

1.  **Probability of First Roll Outcomes:**
    * P(X = 1) = P(X = 2) = P(X = 3) = P(X = 4) = P(X = 5) = P(X = 6) = 1/6

2.  **Probability of a Win in a Single Roll:**
    * P(Win) = 1/6
    * P(No Win) = 5/6

3.  **Expected Wins Given the First Roll (X):**
    * Let W be the number of wins.
    * If X = 1, W ~ Bernoulli(1/6), E[W|X=1] = 1/6
    * If X = 2, W ~ Binomial(2, 1/6), E[W|X=2] = 2 * (1/6) = 2/6
    * If X = 3, W ~ Binomial(3, 1/6), E[W|X=3] = 3 * (1/6) = 3/6
    * If X = 4, W ~ Binomial(4, 1/6), E[W|X=4] = 4 * (1/6) = 4/6
    * If X = 5, W ~ Binomial(5, 1/6), E[W|X=5] = 5 * (1/6) = 5/6
    * If X = 6, W ~ Binomial(6, 1/6), E[W|X=6] = 6 * (1/6) = 6/6 = 1

4.  **Expected Number of Wins (E[W]):**
    * E[W] = Σ [E[W|X=x] * P(X=x)]
    * E[W] = (1/6 * 1/6) + (2/6 * 1/6) + (3/6 * 1/6) + (4/6 * 1/6) + (5/6 * 1/6) + (6/6 * 1/6)
    * E[W] = (1/36) + (2/36) + (3/36) + (4/36) + (5/36) + (6/36)
    * E[W] = (1 + 2 + 3 + 4 + 5 + 6) / 36
    * E[W] = 21 / 36
    * E[W] = 7 / 12

**Therefore, the expected number of wins from this experiment is 7/12.**